sed not working with a single long line file [duplicate] - sed

I'm trying to use sed to clean up lines of URLs to extract just the domain.
So from:
http://www.suepearson.co.uk/product/174/71/3816/
I want:
http://www.suepearson.co.uk/
(either with or without the trailing slash, it doesn't matter)
I have tried:
sed 's|\(http:\/\/.*?\/\).*|\1|'
and (escaping the non-greedy quantifier)
sed 's|\(http:\/\/.*\?\/\).*|\1|'
but I can not seem to get the non-greedy quantifier (?) to work, so it always ends up matching the whole string.

Neither basic nor extended Posix/GNU regex recognizes the non-greedy quantifier; you need a later regex. Fortunately, Perl regex for this context is pretty easy to get:
perl -pe 's|(http://.*?/).*|\1|'

In this specific case, you can get the job done without using a non-greedy regex.
Try this non-greedy regex [^/]* instead of .*?:
sed 's|\(http://[^/]*/\).*|\1|g'

With sed, I usually implement non-greedy search by searching for anything except the separator until the separator :
echo "http://www.suon.co.uk/product/1/7/3/" | sed -n 's;\(http://[^/]*\)/.*;\1;p'
Output:
http://www.suon.co.uk
this is:
don't output -n
search, match pattern, replace and print s/<pattern>/<replace>/p
use ; search command separator instead of / to make it easier to type so s;<pattern>;<replace>;p
remember match between brackets \( ... \), later accessible with \1,\2...
match http://
followed by anything in brackets [], [ab/] would mean either a or b or /
first ^ in [] means not, so followed by anything but the thing in the []
so [^/] means anything except / character
* is to repeat previous group so [^/]* means characters except /.
so far sed -n 's;\(http://[^/]*\) means search and remember http://followed by any characters except / and remember what you've found
we want to search untill the end of domain so stop on the next / so add another / at the end: sed -n 's;\(http://[^/]*\)/' but we want to match the rest of the line after the domain so add .*
now the match remembered in group 1 (\1) is the domain so replace matched line with stuff saved in group \1 and print: sed -n 's;\(http://[^/]*\)/.*;\1;p'
If you want to include backslash after the domain as well, then add one more backslash in the group to remember:
echo "http://www.suon.co.uk/product/1/7/3/" | sed -n 's;\(http://[^/]*/\).*;\1;p'
output:
http://www.suon.co.uk/

Simulating lazy (un-greedy) quantifier in sed
And all other regex flavors!
Finding first occurrence of an expression:
POSIX ERE (using -r option)
Regex:
(EXPRESSION).*|.
Sed:
sed -r ‍'s/(EXPRESSION).*|./\1/g' # Global `g` modifier should be on
Example (finding first sequence of digits) Live demo:
$ sed -r 's/([0-9]+).*|./\1/g' <<< 'foo 12 bar 34'
12
How does it work?
This regex benefits from an alternation |. At each position engine tries to pick the longest match (this is a POSIX standard which is followed by couple of other engines as well) which means it goes with . until a match is found for ([0-9]+).*. But order is important too.
Since global flag is set, engine tries to continue matching character by character up to the end of input string or our target. As soon as the first and only capturing group of left side of alternation is matched (EXPRESSION) rest of line is consumed immediately as well .*. We now hold our value in the first capturing group.
POSIX BRE
Regex:
\(\(\(EXPRESSION\).*\)*.\)*
Sed:
sed 's/\(\(\(EXPRESSION\).*\)*.\)*/\3/'
Example (finding first sequence of digits):
$ sed 's/\(\(\([0-9]\{1,\}\).*\)*.\)*/\3/' <<< 'foo 12 bar 34'
12
This one is like ERE version but with no alternation involved. That's all. At each single position engine tries to match a digit.
If it is found, other following digits are consumed and captured and the rest of line is matched immediately otherwise since * means
more or zero it skips over second capturing group \(\([0-9]\{1,\}\).*\)* and arrives at a dot . to match a single character and this process continues.
Finding first occurrence of a delimited expression:
This approach will match the very first occurrence of a string that is delimited. We can call it a block of string.
sed 's/\(END-DELIMITER-EXPRESSION\).*/\1/; \
s/\(\(START-DELIMITER-EXPRESSION.*\)*.\)*/\1/g'
Input string:
foobar start block #1 end barfoo start block #2 end
-EDE: end
-SDE: start
$ sed 's/\(end\).*/\1/; s/\(\(start.*\)*.\)*/\1/g'
Output:
start block #1 end
First regex \(end\).* matches and captures first end delimiter end and substitues all match with recent captured characters which
is the end delimiter. At this stage our output is: foobar start block #1 end.
Then the result is passed to second regex \(\(start.*\)*.\)* that is same as POSIX BRE version above. It matches a single character
if start delimiter start is not matched otherwise it matches and captures the start delimiter and matches the rest of characters.
Directly answering your question
Using approach #2 (delimited expression) you should select two appropriate expressions:
EDE: [^:/]\/
SDE: http:
Usage:
$ sed 's/\([^:/]\/\).*/\1/g; s/\(\(http:.*\)*.\)*/\1/' <<< 'http://www.suepearson.co.uk/product/174/71/3816/'
Output:
http://www.suepearson.co.uk/
Note: this will not work with identical delimiters.

sed does not support "non greedy" operator.
You have to use "[]" operator to exclude "/" from match.
sed 's,\(http://[^/]*\)/.*,\1,'
P.S. there is no need to backslash "/".

sed - non greedy matching by Christoph Sieghart
The trick to get non greedy matching in sed is to match all characters excluding the one that terminates the match. I know, a no-brainer, but I wasted precious minutes on it and shell scripts should be, after all, quick and easy. So in case somebody else might need it:
Greedy matching
% echo "<b>foo</b>bar" | sed 's/<.*>//g'
bar
Non greedy matching
% echo "<b>foo</b>bar" | sed 's/<[^>]*>//g'
foobar

Non-greedy solution for more than a single character
This thread is really old but I assume people still needs it.
Lets say you want to kill everything till the very first occurrence of HELLO. You cannot say [^HELLO]...
So a nice solution involves two steps, assuming that you can spare a unique word that you are not expecting in the input, say top_sekrit.
In this case we can:
s/HELLO/top_sekrit/ #will only replace the very first occurrence
s/.*top_sekrit// #kill everything till end of the first HELLO
Of course, with a simpler input you could use a smaller word, or maybe even a single character.
HTH!

This can be done using cut:
echo "http://www.suepearson.co.uk/product/174/71/3816/" | cut -d'/' -f1-3

another way, not using regex, is to use fields/delimiter method eg
string="http://www.suepearson.co.uk/product/174/71/3816/"
echo $string | awk -F"/" '{print $1,$2,$3}' OFS="/"

sed certainly has its place but this not not one of them !
As Dee has pointed out: Just use cut. It is far simpler and much more safe in this case. Here's an example where we extract various components from the URL using Bash syntax:
url="http://www.suepearson.co.uk/product/174/71/3816/"
protocol=$(echo "$url" | cut -d':' -f1)
host=$(echo "$url" | cut -d'/' -f3)
urlhost=$(echo "$url" | cut -d'/' -f1-3)
urlpath=$(echo "$url" | cut -d'/' -f4-)
gives you:
protocol = "http"
host = "www.suepearson.co.uk"
urlhost = "http://www.suepearson.co.uk"
urlpath = "product/174/71/3816/"
As you can see this is a lot more flexible approach.
(all credit to Dee)

sed 's|(http:\/\/[^\/]+\/).*|\1|'

There is still hope to solve this using pure (GNU) sed. Despite this is not a generic solution in some cases you can use "loops" to eliminate all the unnecessary parts of the string like this:
sed -r -e ":loop" -e 's|(http://.+)/.*|\1|' -e "t loop"
-r: Use extended regex (for + and unescaped parenthesis)
":loop": Define a new label named "loop"
-e: add commands to sed
"t loop": Jump back to label "loop" if there was a successful substitution
The only problem here is it will also cut the last separator character ('/'), but if you really need it you can still simply put it back after the "loop" finished, just append this additional command at the end of the previous command line:
-e "s,$,/,"

sed -E interprets regular expressions as extended (modern) regular expressions
Update: -E on MacOS X, -r in GNU sed.

Because you specifically stated you're trying to use sed (instead of perl, cut, etc.), try grouping. This circumvents the non-greedy identifier potentially not being recognized. The first group is the protocol (i.e. 'http://', 'https://', 'tcp://', etc). The second group is the domain:
echo "http://www.suon.co.uk/product/1/7/3/" | sed "s|^\(.*//\)\([^/]*\).*$|\1\2|"
If you're not familiar with grouping, start here.

I realize this is an old entry, but someone may find it useful.
As the full domain name may not exceed a total length of 253 characters replace .* with .\{1, 255\}

This is how to robustly do non-greedy matching of multi-character strings using sed. Lets say you want to change every foo...bar to <foo...bar> so for example this input:
$ cat file
ABC foo DEF bar GHI foo KLM bar NOP foo QRS bar TUV
should become this output:
ABC <foo DEF bar> GHI <foo KLM bar> NOP <foo QRS bar> TUV
To do that you convert foo and bar to individual characters and then use the negation of those characters between them:
$ sed 's/#/#A/g; s/{/#B/g; s/}/#C/g; s/foo/{/g; s/bar/}/g; s/{[^{}]*}/<&>/g; s/}/bar/g; s/{/foo/g; s/#C/}/g; s/#B/{/g; s/#A/#/g' file
ABC <foo DEF bar> GHI <foo KLM bar> NOP <foo QRS bar> TUV
In the above:
s/#/#A/g; s/{/#B/g; s/}/#C/g is converting { and } to placeholder strings that cannot exist in the input so those chars then are available to convert foo and bar to.
s/foo/{/g; s/bar/}/g is converting foo and bar to { and } respectively
s/{[^{}]*}/<&>/g is performing the op we want - converting foo...bar to <foo...bar>
s/}/bar/g; s/{/foo/g is converting { and } back to foo and bar.
s/#C/}/g; s/#B/{/g; s/#A/#/g is converting the placeholder strings back to their original characters.
Note that the above does not rely on any particular string not being present in the input as it manufactures such strings in the first step, nor does it care which occurrence of any particular regexp you want to match since you can use {[^{}]*} as many times as necessary in the expression to isolate the actual match you want and/or with seds numeric match operator, e.g. to only replace the 2nd occurrence:
$ sed 's/#/#A/g; s/{/#B/g; s/}/#C/g; s/foo/{/g; s/bar/}/g; s/{[^{}]*}/<&>/2; s/}/bar/g; s/{/foo/g; s/#C/}/g; s/#B/{/g; s/#A/#/g' file
ABC foo DEF bar GHI <foo KLM bar> NOP foo QRS bar TUV

Have not yet seen this answer, so here's how you can do this with vi or vim:
vi -c '%s/\(http:\/\/.\{-}\/\).*/\1/ge | wq' file &>/dev/null
This runs the vi :%s substitution globally (the trailing g), refrains from raising an error if the pattern is not found (e), then saves the resulting changes to disk and quits. The &>/dev/null prevents the GUI from briefly flashing on screen, which can be annoying.
I like using vi sometimes for super complicated regexes, because (1) perl is dead dying, (2) vim has a very advanced regex engine, and (3) I'm already intimately familiar with vi regexes in my day-to-day usage editing documents.

Since PCRE is also tagged here, we could use GNU grep by using non-lazy match in regex .*? which will match first nearest match opposite of .*(which is really greedy and goes till last occurrence of match).
grep -oP '^http[s]?:\/\/.*?/' Input_file
Explanation: using grep's oP options here where -P is responsible for enabling PCRE regex here. In main program of grep mentioning regex which is matching starting http/https followed by :// till next occurrence of / since we have used .*? it will look for first / after (http/https://). It will print matched part only in line.

echo "/home/one/two/three/myfile.txt" | sed 's|\(.*\)/.*|\1|'
don bother, i got it on another forum :)

sed 's|\(http:\/\/www\.[a-z.0-9]*\/\).*|\1| works too

Here is something you can do with a two step approach and awk:
A=http://www.suepearson.co.uk/product/174/71/3816/
echo $A|awk '
{
var=gensub(///,"||",3,$0) ;
sub(/\|\|.*/,"",var);
print var
}'
Output:
http://www.suepearson.co.uk
Hope that helps!

Another sed version:
sed 's|/[:alnum:].*||' file.txt
It matches / followed by an alphanumeric character (so not another forward slash) as well as the rest of characters till the end of the line. Afterwards it replaces it with nothing (ie. deletes it.)

#Daniel H (concerning your comment on andcoz' answer, although long time ago): deleting trailing zeros works with
s,([[:digit:]]\.[[:digit:]]*[1-9])[0]*$,\1,g
it's about clearly defining the matching conditions ...

You should also think about the case where there is no matching delims. Do you want to output the line or not. My examples here do not output anything if there is no match.
You need prefix up to 3rd /, so select two times string of any length not containing / and following / and then string of any length not containing / and then match / following any string and then print selection. This idea works with any single char delims.
echo http://www.suepearson.co.uk/product/174/71/3816/ | \
sed -nr 's,(([^/]*/){2}[^/]*)/.*,\1,p'
Using sed commands you can do fast prefix dropping or delim selection, like:
echo 'aaa #cee: { "foo":" #cee: " }' | \
sed -r 't x;s/ #cee: /\n/;D;:x'
This is lot faster than eating char at a time.
Jump to label if successful match previously. Add \n at / before 1st delim. Remove up to first \n. If \n was added, jump to end and print.
If there is start and end delims, it is just easy to remove end delims until you reach the nth-2 element you want and then do D trick, remove after end delim, jump to delete if no match, remove before start delim and and print. This only works if start/end delims occur in pairs.
echo 'foobar start block #1 end barfoo start block #2 end bazfoo start block #3 end goo start block #4 end faa' | \
sed -r 't x;s/end//;s/end/\n/;D;:x;s/(end).*/\1/;T y;s/.*(start)/\1/;p;:y;d'

If you have access to gnu grep, then can utilize perl regex:
grep -Po '^https?://([^/]+)(?=)' <<< 'http://www.suepearson.co.uk/product/174/71/3816/'
http://www.suepearson.co.uk
Alternatively, to get everything after the domain use
grep -Po '^https?://([^/]+)\K.*' <<< 'http://www.suepearson.co.uk/product/174/71/3816/'
/product/174/71/3816/

The following solution works for matching / working with multiply present (chained; tandem; compound) HTML or other tags. For example, I wanted to edit HTML code to remove <span> tags, that appeared in tandem.
Issue: regular sed regex expressions greedily matched over all the tags from the first to the last.
Solution: non-greedy pattern matching (per discussions elsewhere in this thread; e.g. https://stackoverflow.com/a/46719361/1904943).
Example:
echo '<span>Will</span>This <span>remove</span>will <span>this.</span>remain.' | \
sed 's/<span>[^>]*>//g' ; echo
This will remain.
Explanation:
s/<span> : find <span>
[^>] : followed by anything that is not >
*> : until you find >
//g : replace any such strings present with nothing.
Addendum
I was trying to clean up URLs, but I was running into difficulty matching / excluding a word - href - using the approach above. I briefly looked at negative lookarounds (Regular expression to match a line that doesn't contain a word) but that approach seemed overly complex and did not provide a satisfactory solution.
I decided to replace href with ` (backtick), do the regex substitutions, then replace ` with href.
Example (formatted here for readability):
printf '\n
<a aaa h href="apple">apple</a>
<a bbb "c=ccc" href="banana">banana</a>
<a class="gtm-content-click"
data-vars-link-text="nope"
data-vars-click-url="https://blablabla"
data-vars-event-category="story"
data-vars-sub-category="story"
data-vars-item="in_content_link"
data-vars-link-text
href="https:example.com">Example.com</a>\n\n' |
sed 's/href/`/g ;
s/<a[^`]*`/\n<a href/g'
apple
banana
Example.com
Explanation: basically as above. Here,
s/href/` : replace href with ` (backtick)
s/<a : find start of URL
[^`] : followed by anything that is not ` (backtick)
*` : until you find a `
/<a href/g : replace each of those found with <a href

Unfortunately, as mentioned, this it is not supported in sed.
To overcome this, I suggest to use the next best thing(actually better even), to use vim sed-like capabilities.
define in .bash-profile
vimdo() { vim $2 --not-a-term -c "$1" -es +"w >> /dev/stdout" -cq! ; }
That will create headless vim to execute a command.
Now you can do for example:
echo $PATH | vimdo "%s_\c:[a-zA-Z0-9\\/]\{-}python[a-zA-Z0-9\\/]\{-}:__g" -
to filter out python in $PATH.
Use - to have input from pipe in vimdo.
While most of the syntax is the same. Vim features more advanced features, and using \{-} is standard for non-greedy match. see help regexp.

Related

replace first occurence after a match on all lines for all matches using sed [duplicate]

I'm trying to use sed to clean up lines of URLs to extract just the domain.
So from:
http://www.suepearson.co.uk/product/174/71/3816/
I want:
http://www.suepearson.co.uk/
(either with or without the trailing slash, it doesn't matter)
I have tried:
sed 's|\(http:\/\/.*?\/\).*|\1|'
and (escaping the non-greedy quantifier)
sed 's|\(http:\/\/.*\?\/\).*|\1|'
but I can not seem to get the non-greedy quantifier (?) to work, so it always ends up matching the whole string.
Neither basic nor extended Posix/GNU regex recognizes the non-greedy quantifier; you need a later regex. Fortunately, Perl regex for this context is pretty easy to get:
perl -pe 's|(http://.*?/).*|\1|'
In this specific case, you can get the job done without using a non-greedy regex.
Try this non-greedy regex [^/]* instead of .*?:
sed 's|\(http://[^/]*/\).*|\1|g'
With sed, I usually implement non-greedy search by searching for anything except the separator until the separator :
echo "http://www.suon.co.uk/product/1/7/3/" | sed -n 's;\(http://[^/]*\)/.*;\1;p'
Output:
http://www.suon.co.uk
this is:
don't output -n
search, match pattern, replace and print s/<pattern>/<replace>/p
use ; search command separator instead of / to make it easier to type so s;<pattern>;<replace>;p
remember match between brackets \( ... \), later accessible with \1,\2...
match http://
followed by anything in brackets [], [ab/] would mean either a or b or /
first ^ in [] means not, so followed by anything but the thing in the []
so [^/] means anything except / character
* is to repeat previous group so [^/]* means characters except /.
so far sed -n 's;\(http://[^/]*\) means search and remember http://followed by any characters except / and remember what you've found
we want to search untill the end of domain so stop on the next / so add another / at the end: sed -n 's;\(http://[^/]*\)/' but we want to match the rest of the line after the domain so add .*
now the match remembered in group 1 (\1) is the domain so replace matched line with stuff saved in group \1 and print: sed -n 's;\(http://[^/]*\)/.*;\1;p'
If you want to include backslash after the domain as well, then add one more backslash in the group to remember:
echo "http://www.suon.co.uk/product/1/7/3/" | sed -n 's;\(http://[^/]*/\).*;\1;p'
output:
http://www.suon.co.uk/
Simulating lazy (un-greedy) quantifier in sed
And all other regex flavors!
Finding first occurrence of an expression:
POSIX ERE (using -r option)
Regex:
(EXPRESSION).*|.
Sed:
sed -r ‍'s/(EXPRESSION).*|./\1/g' # Global `g` modifier should be on
Example (finding first sequence of digits) Live demo:
$ sed -r 's/([0-9]+).*|./\1/g' <<< 'foo 12 bar 34'
12
How does it work?
This regex benefits from an alternation |. At each position engine tries to pick the longest match (this is a POSIX standard which is followed by couple of other engines as well) which means it goes with . until a match is found for ([0-9]+).*. But order is important too.
Since global flag is set, engine tries to continue matching character by character up to the end of input string or our target. As soon as the first and only capturing group of left side of alternation is matched (EXPRESSION) rest of line is consumed immediately as well .*. We now hold our value in the first capturing group.
POSIX BRE
Regex:
\(\(\(EXPRESSION\).*\)*.\)*
Sed:
sed 's/\(\(\(EXPRESSION\).*\)*.\)*/\3/'
Example (finding first sequence of digits):
$ sed 's/\(\(\([0-9]\{1,\}\).*\)*.\)*/\3/' <<< 'foo 12 bar 34'
12
This one is like ERE version but with no alternation involved. That's all. At each single position engine tries to match a digit.
If it is found, other following digits are consumed and captured and the rest of line is matched immediately otherwise since * means
more or zero it skips over second capturing group \(\([0-9]\{1,\}\).*\)* and arrives at a dot . to match a single character and this process continues.
Finding first occurrence of a delimited expression:
This approach will match the very first occurrence of a string that is delimited. We can call it a block of string.
sed 's/\(END-DELIMITER-EXPRESSION\).*/\1/; \
s/\(\(START-DELIMITER-EXPRESSION.*\)*.\)*/\1/g'
Input string:
foobar start block #1 end barfoo start block #2 end
-EDE: end
-SDE: start
$ sed 's/\(end\).*/\1/; s/\(\(start.*\)*.\)*/\1/g'
Output:
start block #1 end
First regex \(end\).* matches and captures first end delimiter end and substitues all match with recent captured characters which
is the end delimiter. At this stage our output is: foobar start block #1 end.
Then the result is passed to second regex \(\(start.*\)*.\)* that is same as POSIX BRE version above. It matches a single character
if start delimiter start is not matched otherwise it matches and captures the start delimiter and matches the rest of characters.
Directly answering your question
Using approach #2 (delimited expression) you should select two appropriate expressions:
EDE: [^:/]\/
SDE: http:
Usage:
$ sed 's/\([^:/]\/\).*/\1/g; s/\(\(http:.*\)*.\)*/\1/' <<< 'http://www.suepearson.co.uk/product/174/71/3816/'
Output:
http://www.suepearson.co.uk/
Note: this will not work with identical delimiters.
sed does not support "non greedy" operator.
You have to use "[]" operator to exclude "/" from match.
sed 's,\(http://[^/]*\)/.*,\1,'
P.S. there is no need to backslash "/".
sed - non greedy matching by Christoph Sieghart
The trick to get non greedy matching in sed is to match all characters excluding the one that terminates the match. I know, a no-brainer, but I wasted precious minutes on it and shell scripts should be, after all, quick and easy. So in case somebody else might need it:
Greedy matching
% echo "<b>foo</b>bar" | sed 's/<.*>//g'
bar
Non greedy matching
% echo "<b>foo</b>bar" | sed 's/<[^>]*>//g'
foobar
Non-greedy solution for more than a single character
This thread is really old but I assume people still needs it.
Lets say you want to kill everything till the very first occurrence of HELLO. You cannot say [^HELLO]...
So a nice solution involves two steps, assuming that you can spare a unique word that you are not expecting in the input, say top_sekrit.
In this case we can:
s/HELLO/top_sekrit/ #will only replace the very first occurrence
s/.*top_sekrit// #kill everything till end of the first HELLO
Of course, with a simpler input you could use a smaller word, or maybe even a single character.
HTH!
This can be done using cut:
echo "http://www.suepearson.co.uk/product/174/71/3816/" | cut -d'/' -f1-3
another way, not using regex, is to use fields/delimiter method eg
string="http://www.suepearson.co.uk/product/174/71/3816/"
echo $string | awk -F"/" '{print $1,$2,$3}' OFS="/"
sed certainly has its place but this not not one of them !
As Dee has pointed out: Just use cut. It is far simpler and much more safe in this case. Here's an example where we extract various components from the URL using Bash syntax:
url="http://www.suepearson.co.uk/product/174/71/3816/"
protocol=$(echo "$url" | cut -d':' -f1)
host=$(echo "$url" | cut -d'/' -f3)
urlhost=$(echo "$url" | cut -d'/' -f1-3)
urlpath=$(echo "$url" | cut -d'/' -f4-)
gives you:
protocol = "http"
host = "www.suepearson.co.uk"
urlhost = "http://www.suepearson.co.uk"
urlpath = "product/174/71/3816/"
As you can see this is a lot more flexible approach.
(all credit to Dee)
sed 's|(http:\/\/[^\/]+\/).*|\1|'
There is still hope to solve this using pure (GNU) sed. Despite this is not a generic solution in some cases you can use "loops" to eliminate all the unnecessary parts of the string like this:
sed -r -e ":loop" -e 's|(http://.+)/.*|\1|' -e "t loop"
-r: Use extended regex (for + and unescaped parenthesis)
":loop": Define a new label named "loop"
-e: add commands to sed
"t loop": Jump back to label "loop" if there was a successful substitution
The only problem here is it will also cut the last separator character ('/'), but if you really need it you can still simply put it back after the "loop" finished, just append this additional command at the end of the previous command line:
-e "s,$,/,"
sed -E interprets regular expressions as extended (modern) regular expressions
Update: -E on MacOS X, -r in GNU sed.
Because you specifically stated you're trying to use sed (instead of perl, cut, etc.), try grouping. This circumvents the non-greedy identifier potentially not being recognized. The first group is the protocol (i.e. 'http://', 'https://', 'tcp://', etc). The second group is the domain:
echo "http://www.suon.co.uk/product/1/7/3/" | sed "s|^\(.*//\)\([^/]*\).*$|\1\2|"
If you're not familiar with grouping, start here.
I realize this is an old entry, but someone may find it useful.
As the full domain name may not exceed a total length of 253 characters replace .* with .\{1, 255\}
This is how to robustly do non-greedy matching of multi-character strings using sed. Lets say you want to change every foo...bar to <foo...bar> so for example this input:
$ cat file
ABC foo DEF bar GHI foo KLM bar NOP foo QRS bar TUV
should become this output:
ABC <foo DEF bar> GHI <foo KLM bar> NOP <foo QRS bar> TUV
To do that you convert foo and bar to individual characters and then use the negation of those characters between them:
$ sed 's/#/#A/g; s/{/#B/g; s/}/#C/g; s/foo/{/g; s/bar/}/g; s/{[^{}]*}/<&>/g; s/}/bar/g; s/{/foo/g; s/#C/}/g; s/#B/{/g; s/#A/#/g' file
ABC <foo DEF bar> GHI <foo KLM bar> NOP <foo QRS bar> TUV
In the above:
s/#/#A/g; s/{/#B/g; s/}/#C/g is converting { and } to placeholder strings that cannot exist in the input so those chars then are available to convert foo and bar to.
s/foo/{/g; s/bar/}/g is converting foo and bar to { and } respectively
s/{[^{}]*}/<&>/g is performing the op we want - converting foo...bar to <foo...bar>
s/}/bar/g; s/{/foo/g is converting { and } back to foo and bar.
s/#C/}/g; s/#B/{/g; s/#A/#/g is converting the placeholder strings back to their original characters.
Note that the above does not rely on any particular string not being present in the input as it manufactures such strings in the first step, nor does it care which occurrence of any particular regexp you want to match since you can use {[^{}]*} as many times as necessary in the expression to isolate the actual match you want and/or with seds numeric match operator, e.g. to only replace the 2nd occurrence:
$ sed 's/#/#A/g; s/{/#B/g; s/}/#C/g; s/foo/{/g; s/bar/}/g; s/{[^{}]*}/<&>/2; s/}/bar/g; s/{/foo/g; s/#C/}/g; s/#B/{/g; s/#A/#/g' file
ABC foo DEF bar GHI <foo KLM bar> NOP foo QRS bar TUV
Have not yet seen this answer, so here's how you can do this with vi or vim:
vi -c '%s/\(http:\/\/.\{-}\/\).*/\1/ge | wq' file &>/dev/null
This runs the vi :%s substitution globally (the trailing g), refrains from raising an error if the pattern is not found (e), then saves the resulting changes to disk and quits. The &>/dev/null prevents the GUI from briefly flashing on screen, which can be annoying.
I like using vi sometimes for super complicated regexes, because (1) perl is dead dying, (2) vim has a very advanced regex engine, and (3) I'm already intimately familiar with vi regexes in my day-to-day usage editing documents.
Since PCRE is also tagged here, we could use GNU grep by using non-lazy match in regex .*? which will match first nearest match opposite of .*(which is really greedy and goes till last occurrence of match).
grep -oP '^http[s]?:\/\/.*?/' Input_file
Explanation: using grep's oP options here where -P is responsible for enabling PCRE regex here. In main program of grep mentioning regex which is matching starting http/https followed by :// till next occurrence of / since we have used .*? it will look for first / after (http/https://). It will print matched part only in line.
echo "/home/one/two/three/myfile.txt" | sed 's|\(.*\)/.*|\1|'
don bother, i got it on another forum :)
sed 's|\(http:\/\/www\.[a-z.0-9]*\/\).*|\1| works too
Here is something you can do with a two step approach and awk:
A=http://www.suepearson.co.uk/product/174/71/3816/
echo $A|awk '
{
var=gensub(///,"||",3,$0) ;
sub(/\|\|.*/,"",var);
print var
}'
Output:
http://www.suepearson.co.uk
Hope that helps!
Another sed version:
sed 's|/[:alnum:].*||' file.txt
It matches / followed by an alphanumeric character (so not another forward slash) as well as the rest of characters till the end of the line. Afterwards it replaces it with nothing (ie. deletes it.)
#Daniel H (concerning your comment on andcoz' answer, although long time ago): deleting trailing zeros works with
s,([[:digit:]]\.[[:digit:]]*[1-9])[0]*$,\1,g
it's about clearly defining the matching conditions ...
You should also think about the case where there is no matching delims. Do you want to output the line or not. My examples here do not output anything if there is no match.
You need prefix up to 3rd /, so select two times string of any length not containing / and following / and then string of any length not containing / and then match / following any string and then print selection. This idea works with any single char delims.
echo http://www.suepearson.co.uk/product/174/71/3816/ | \
sed -nr 's,(([^/]*/){2}[^/]*)/.*,\1,p'
Using sed commands you can do fast prefix dropping or delim selection, like:
echo 'aaa #cee: { "foo":" #cee: " }' | \
sed -r 't x;s/ #cee: /\n/;D;:x'
This is lot faster than eating char at a time.
Jump to label if successful match previously. Add \n at / before 1st delim. Remove up to first \n. If \n was added, jump to end and print.
If there is start and end delims, it is just easy to remove end delims until you reach the nth-2 element you want and then do D trick, remove after end delim, jump to delete if no match, remove before start delim and and print. This only works if start/end delims occur in pairs.
echo 'foobar start block #1 end barfoo start block #2 end bazfoo start block #3 end goo start block #4 end faa' | \
sed -r 't x;s/end//;s/end/\n/;D;:x;s/(end).*/\1/;T y;s/.*(start)/\1/;p;:y;d'
If you have access to gnu grep, then can utilize perl regex:
grep -Po '^https?://([^/]+)(?=)' <<< 'http://www.suepearson.co.uk/product/174/71/3816/'
http://www.suepearson.co.uk
Alternatively, to get everything after the domain use
grep -Po '^https?://([^/]+)\K.*' <<< 'http://www.suepearson.co.uk/product/174/71/3816/'
/product/174/71/3816/
The following solution works for matching / working with multiply present (chained; tandem; compound) HTML or other tags. For example, I wanted to edit HTML code to remove <span> tags, that appeared in tandem.
Issue: regular sed regex expressions greedily matched over all the tags from the first to the last.
Solution: non-greedy pattern matching (per discussions elsewhere in this thread; e.g. https://stackoverflow.com/a/46719361/1904943).
Example:
echo '<span>Will</span>This <span>remove</span>will <span>this.</span>remain.' | \
sed 's/<span>[^>]*>//g' ; echo
This will remain.
Explanation:
s/<span> : find <span>
[^>] : followed by anything that is not >
*> : until you find >
//g : replace any such strings present with nothing.
Addendum
I was trying to clean up URLs, but I was running into difficulty matching / excluding a word - href - using the approach above. I briefly looked at negative lookarounds (Regular expression to match a line that doesn't contain a word) but that approach seemed overly complex and did not provide a satisfactory solution.
I decided to replace href with ` (backtick), do the regex substitutions, then replace ` with href.
Example (formatted here for readability):
printf '\n
<a aaa h href="apple">apple</a>
<a bbb "c=ccc" href="banana">banana</a>
<a class="gtm-content-click"
data-vars-link-text="nope"
data-vars-click-url="https://blablabla"
data-vars-event-category="story"
data-vars-sub-category="story"
data-vars-item="in_content_link"
data-vars-link-text
href="https:example.com">Example.com</a>\n\n' |
sed 's/href/`/g ;
s/<a[^`]*`/\n<a href/g'
apple
banana
Example.com
Explanation: basically as above. Here,
s/href/` : replace href with ` (backtick)
s/<a : find start of URL
[^`] : followed by anything that is not ` (backtick)
*` : until you find a `
/<a href/g : replace each of those found with <a href
Unfortunately, as mentioned, this it is not supported in sed.
To overcome this, I suggest to use the next best thing(actually better even), to use vim sed-like capabilities.
define in .bash-profile
vimdo() { vim $2 --not-a-term -c "$1" -es +"w >> /dev/stdout" -cq! ; }
That will create headless vim to execute a command.
Now you can do for example:
echo $PATH | vimdo "%s_\c:[a-zA-Z0-9\\/]\{-}python[a-zA-Z0-9\\/]\{-}:__g" -
to filter out python in $PATH.
Use - to have input from pipe in vimdo.
While most of the syntax is the same. Vim features more advanced features, and using \{-} is standard for non-greedy match. see help regexp.

how to use sed/awk to remove words with multiple pattern count

I have a file of string records where one of the fields - delimited by "," - can contain one or more "-" inside it.
The goal is to delete the field value if it contains more than two "-".
i am trying to recoup my past knowledge of sed/awk but can't make much headway
==========
info,whitepaper,Data-Centers,yes-the-6-top-problems-in-your-data-center-lane
info,whitepaper,Data-Centers,the-evolution-center
info,whitepaper,Data-Centers,the-evolution-of-lan-technology-lanner
==========
expected outcome:
info,whitepaper,Data-Centers
info,whitepaper,Data-Centers,the-evolution-center
info,whitepaper,Data-Centers
thanks
Try
sed -r 's/(^|,)([^,-]+-){3,}[^,]+(,|$)/\3/g'
or if you're into slashes
sed 's/\(^\|,\)\([^,-]\+-\)\{3,\}[^,]\+\(,\|$\)/\3/g'
Explanation:
I'm using the most basic sed command: substitution. The syntax is: s/pattern/replacement/flags.
Here pattern is (^|,)([^,-]+-){3,}[^,]+(,|$), replacement is \3, flags is g.
The g flag means global replacement (all matching parts are replaced, not only the first in line).
In pattern:
brackets () create a group. Somewhat like in math. They also allow to refer to a group with a number later.
^ and $ mean beginning and end of the string.
| means "or", so (^|,) means "comma or beginning of the string".
square brackets [] mean a character class, ^ inside means negation. So [^,-] means "anything but comma or hyphen". Not that usually the hyphen has a special meaning in character classes: [a-z] means all lowercase letters. But here it's just a hyphen because it's not in the middle.
+ after an expression means "match it 1 or more times" (like * means match it 0 or more times).
{N} means "match it exactly N times. {N,M} is "from N to M times". {3,} means "three times or more". + is equivalent to {1,}.
So this is it. The replacement is just \3. This refers to the third group in (), in this case (,|$). This will be the only thing left after the substitution.
P.S. the -r option just changes what characters need to be escaped: without it all of ()-{}| are treated as regular chars unless you escape them with \. Conversely, to match literal ( with -r option you'll need to escape it.
P.P.S. Here's a reference for sed. man sed is your friend as well.
Let me know if you have further questions.
You could try perl instead of sed or awk:
perl -F, -lane 'print join ",", grep { !/-.*-.*-/ } #F' < file.txt
This might work for you:
sed 's/,\{,1\}[^,-]*\(-[^,]*\)\{3,\}//g file
sed 's/\(^\|,\)\([^,]*-\)\{3\}[^,]*\(,\|$\)//g'
This should work in more cases:
sed 's/,$/\n/g;s/\(^\|,\|\n\)\([^,\n]*-\)\{3\}[^,\n]*\(,\|\n\|$\)/\3/g;s/,$//;s/\n/,/g'

sed to replace only matching part in search string

I have a file that contains:
Lorem ipsum dolem file1.jar.
file1.jar (MD5: 12345678901234567890123456789012)
file2.jar (MD5: 09876543210987654321098765432109)
file3.jar (MD5: 24681357902468135790246813579024)
and I'd like to replace the first MD5. This sed command does the job:
sed "s/file1.*MD5\:\(.*\)/file1.jar \(MD5\: `md5 file1.jar | awk '{print $4}'`\)/"
Is there a way to tell sed to replace only the matching group while leaving the rest of the line alone? For example:
sed "s/file1.*MD5\:\(.*\)/`md5 file1.jar | awk '{print $4}'`/"
You can use a search to specify the line to match, and then a simpler regex in the substitute:
sed "/file1\.jar (MD5: [0-9A-Fa-f]*)/s/(MD5: [^)]*)/(MD5: $(md5 file1.jar | awk '{print $4}'))/"
That uses the $(...) notation to run the command. The tricky bit in that is at the end, where the sequence ))/" appears. The first close parenthesis is the end of the $(...) notation; the second is a character in the replacement text.
The first regex /file1\.jar (MD5: [0-9A-Fa-f]*)/ specifies fairly precisely the line to be matched. Then, knowing it is the correct line, the pattern in the substitute can be simpler: the search part /(MD5: [^)]*)/ looks for just the parenthesized MD5 data, safe in the knowledge that even though many other lines contain the same pattern, the substitution will only be applied to the one desired line.
I might be inclined to use:
md5=$(md5 file1.jar | awk '{print $4}')
sed "/file1\.jar (MD5: [0-9A-Fa-f]*)/ s/(MD5: [^)]*)/(MD5: $md5)/"
which clarifies what's what considerably (and doesn't involve a horizontal scroll bar on SO). You could be even more precise in the line matching pattern:
md5=$(md5 file1.jar | awk '{print $4}')
sed "/^file1\.jar (MD5: [0-9A-Fa-f]\{32\})\$/ s/(MD5: [^)]*)/(MD5: $md5)/"
That insists on exactly 32 hex digits and the close parenthesis at the end of the line.
One of the comments asks:
Can sed operate in such a way that the replacement string replaces only the matching groups in the search pattern? For example, given 's/A B \(D\)/C/', it outputs A B C.
If I understand the (clarification of the) question, then you can do what you want with appropriate capturing - but the replacement part will have to specify exactly what you want as output (no shortcuts like you seem to be after). So, for the example, you would write something like:
s/\(A B \)\(D\)/\1C/
(where the capturing \(D\) does not need the capturing parentheses since the captured material is not used in the replacement, and you could write either of:
s/\(A B \)D/\1C/
s/\(A B\) D/\1 C/
You could also do:
/A B / s/D/C/
This has a search (for the A B sequence) and then the substitute looks for D and replaces it with C. This is basically what the main answer is suggesting. You can probably also do:
/\(A B\) D/ s//\1 C/
The 'empty search' should repeat the match, but the replacement has to be written out in full, and that is effectively the same as one of the previous commands:
s/\(A B\) D/\1 C/
This should do it (untested):
sed "s/(file1.jar \(MD5: )(..............................)/\1`md5 file1.jar | awk '{print $1}'`/"
That's 32 dots, mind you.

How to use sed-awk-gawk to display a matched string

I've got a file called 'res' that's 29374 characters of http data in a one-line string. Inside it, there are several http links, but I only want to be display those that end in '/idNNNNNNNNN' where N is a digit. In fact I'm only interested in the string 'idNNNNNNNNN'.
I've tried with:
cat res | sed -n '0,/.*\(id[0-9]*\).*/s//\1/p'
but I get the whole file.
Do you know a way to do it?
perl -n -E 'say $1 while m!/id(\d{9})!g' input-file
should work. That assumes exactly 9 digits; that's the {9} in the above. You can match 8 or 9 ({8,9}), 8 or more ({8,}), up to 9 ({0,9}), etc.
Example of this working:
$ echo -n 'junk jumk http://foo/id231313 junk lalala http://bar/id23123 asda' | perl -n -E 'say $1 while m!id(\d{0,9})!g'
231313
23123
That's with the 0 to 9 variant, of course.
If you're stuck with a pre-5.10 perl, use -e instead of -E and print "$1\n" instead of say $1.
How it works
First is the two command-line arguments to Perl. -n tells Perl to read input from standard input or files given on the command line, line by line, setting $_ to each line. $_ is perl's default target for a lot of things, including regular expression matches. -E merely tells Perl that the next argument is a Perl one-liner, using the new language features (vs. -e which does not use the 5.10 extensions).
So, looking at the one liner: say means to print out some value, followed by a newline. $1 is the first regular expression capture (captures are made by parentheses in regular expressions). while is a looping construct, which you're probably familiar with. m is the match operator, the ! after it is the regular expression delimiter (normally, you see / here, but since the pattern contains / it's easier to use something else, so you don't have to escape the / as \/). /id(\d{9}) is the regular expression to match. Keep in mind that the delimiter is !, so the / is not special, it just matches a literal /. The parentheses form a capture group, so $1 will be the number. The ! is the delimiter, followed by g which means to match as many times as possible (as opposed to once). This is what makes it pick up all the URLs in the line, not just the first. As long as there is a match, the m operator will return a true value, so the loop will continue (and run that say $1, printing out the match).
Two-sed solution
I think this is one way to do this with only sed. Much more complicated!
echo 'junk jumk http://foo/id231313 junk lalala http://bar/id23123 asda' | \
sed 's!http://!\nhttp://!g' | \
sed 's!^.*/id\([0-9]*\).*$!\1!'
cat res | perl -ne 'chomp; print "$1\n" if m/\/(id\d*)/'
The trouble is that sed and grep and awk work on lines, and you've only got one line. So, you probably need to split things up so you have more than one line -- then you can make the normal tools work.
tr ':' '\012' < res |
sed -n 's%.*/\(id[0-9][0-9]*\).*%\1%p'
This takes advantage of URLs containing colons and maps colons to newlines with tr, then uses sed to pick up anything up to a slash, followed by id and one or more digits, followed by anything, and prints out the id and digit string (only). Since these only occur in URLs, they will only appear one per line and relatively near the start of the line too.
Here's a solution using only one invocation of sed:
sed -n 's| |\n|g;/^http/{s|http://[^/]*/id\([0-9]*\)|\1|;P};D' inputfile
Explanation:
s| |\n|g; - Divide and conquer
/^http/{ - If pattern space begins with "http"
s|http://[^/]*/id\([0-9]*\)|\1|; - capture the id
P - Print the string preceding the first newline
}; - end if
D - Delete the string preceding the first newline regardless of whether it contains "http"
Edit:
This version uses the same technique but is more selective.
sed -n 's|http://|\n&|g;/^\n*http/{s|\n*http://[^/]*/id\([0-9]*\)|\1\n|;P};D' inputfile

Confining Substitution to Match Space Using sed?

Is there a way to substitute only within the match space using sed?
I.e. given the following line, is there a way to substitute only the "." chars that are contained within the matching single quotes and protect the "." chars that are not enclosed by single quotes?
Input:
'ECJ-4YF1H10.6Z' ! 'CAP' ! '10.0uF' ! 'TOL' ; MGCDC1008.S1 MGCDC1009.A2
Desired result:
'ECJ-4YF1H10-6Z' ! 'CAP' ! '10_0uF' ! 'TOL' ; MGCDC1008.S1 MGCDC1009.A2
Or is this just a job to which perl or awk might be better suited?
Thanks for your help,
Mark
Give the following a try which uses the divide-and-conquer technique:
sed "s/\('[^']*'\)/\n&\n/g;s/\(\n'[^.]*\)\.\([^']*Z'\)/\1-\2/g;s/\(\n'[^.]*\)\.\([^']*uF'\)/\1_\2/g;s/\n//g" inputfile
Explanation:
s/\('[^']*'\)/\n&\n/g - Add newlines before and after each pair of single quotes with their contents
s/\(\n'[^.]*\)\.\([^']*Z'\)/\1-\2/g - Using a newline and the single quotes to key on, replace the dot with a dash for strings that end in "Z"
s/\(\n'[^.]*\)\.\([^']*uF'\)/\1_\2/g - Using a newline and the single quotes to key on, replace the dot with a dash for strings that end in "uF"
s/\n//g - Remove the newlines added in the first step
You can restrict the command to acting only on certain lines:
sed "/foo/{s/\('[^']*'\)/\n&\n/g;s/\(\n'[^.]*\)\.\([^']*Z'\)/\1-\2/g;s/\(\n'[^.]*\)\.\([^']*uF'\)/\1_\2/g;s/\n//g}" inputfile
where you would substitute some regex in place of "foo".
Some versions of sed like to be spoon fed (instead of semicolons between commands, use -e):
sed -e "/foo/{s/\('[^']*'\)/\n&\n/g" -e "s/\(\n'[^.]*\)\.\([^']*Z'\)/\1-\2/g" -e "s/\(\n'[^.]*\)\.\([^']*uF'\)/\1_\2/g" -e "s/\n//g}" inputfile
$ cat phoo1234567_sedFix.sed
#! /bin/sed -f
/'[0-9][0-9]\.[0-9][a-zA-Z][a-zA-Z]'/s/'\([0-9][0-9]\)\.\([0-9][a-zA-Z][a-zA-Z]\)'/\1_\2/
This answers your specific question. If the pattern you need to fix isn't always like the example you provided, they you'll need multiple copies of this line, with reg-expressions modified to match your new change targets.
Note that the cmd is in 2 parts, "/'[0-9][0-9].[0-9][a-zA-Z][a-zA-Z]'/" says, must match lines with this pattern, while the trailing "s/'([0-9][0-9]).([0-9][a-zA-Z][a-zA-Z])'/\1_\2/", is the part that does the substitution. You can add a 'g' after the final '/' to make this substitution happen on all instances of this pattern in each line.
The \(\) pairs in match pattern get converted into the numbered buffers on the substitution side of the command (i.e. \1 \2). This is what gives sed power that awk doesn't have.
If your going to do much of this kind of work, I highly recommend O'Rielly's Sed And Awk book. The time spent going thru how sed works will be paid back many times.
I hope this helps.
P.S. as you appear to be a new user, if you get an answer that helps you please remember to mark it as accepted, or give it a + (or -) as a useful answer.
this is a job most suitable for awk or any language that supports breaking/splitting strings.
IMO, using sed for this task, which is regex based , while doable, is difficult to read and debug, hence not the most appropriate tool for the job. No offense to sed fanatics.
awk '{
for(i=1;i<=NF;i++) {
if ($i ~ /\047/ ){
gsub(".","_",$i)
}
}
}1' file
The above says for each field (field seperator by default is white space), check to see if there is a single quote, and if there is , substitute the "." to "_". This method is simple and doesn't need complicated regex.