I have data that starts out like this in a .csv file
"684MF7","684MF7","RN"
The first field "684MF7" should only contain numeric characters; no alpha characters should be present in the first field. I have other checks for the second field, which in this case is also "684MF7", which is a legitimate value for the second field.
I want to find any alpha in the first field, and print that line. I invoke this sed file
{
/^".*[^0-9]*.*",/p
}
with -n and -f (for the file name).
What regular expression isolates the first field only? I am getting a match on everything, which isn't what I want. Is my problem because I am trying to match zero or more instead of 1 or more alpha characters?
The first field (any content) would be selected by:
/^"[^"]*"/
You want at least one of the characters in the field to be alpha (though it might be better regarded as 'non-digit'), in which case one of these should select what you're after:
/^"[^"]*[A-Za-z][^"]*"/
/^"[^"]*[^0-9"][^"]*"/
/^"[^"]*[[:alpha:]][^"]*"/
/^"[^"]*[^"[:digit:]][^"]*"/
Note that the negated classes must not match a double quote either (one reason for always testing answers — the first version of the script below listed both lines of input).
And converting one of those into a sed command:
sed -n '/^"[^"]*[^"[:digit:]][^"]*"/p' <<EOF
"684MF7","684MF7","RN"
"684007","684MF7","RN"
EOF
Another way of looking at the problem is "print any line where the first field is not all digits (with at least one digit present)". That is:
sed -n '/^"[[:digit:]]\{1,\}"/!p' <<EOF
"684MF7","684MF7","RN"
"684007","684MF7","RN"
EOF
On the whole, this is perhaps the better solution to use (and I shan't complain if you use [0-9] in place of [[:digit:]]).
Generally .* surrounding any other expressions tends to match more than expected. Try to write an expression that is more detailed with less large wildcard matches
I found this to work
> sed -n '/^".*[A-Z].*",".*",".*"/p' <(echo '"684MF7","684MF7","RN"')
> "684MF7","684MF7","RN"
> sed -n '/^".*[A-Z].*",".*",".*"/p' <(echo '"684117","684MF7","RN"')
>
It picks up all of the groups surrounded with "
Perhaps the following will work for you?
echo '"A84MF7","684MF7","3N"' | sed -n '/^"[^0-9,][^",]*"/p'
"A84MF7","684MF7","3N"
echo '"684MF7","684MF7","7N"' | sed -n '/^"[^0-9,][^",]*"/p'
--Nothing--
Related
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.
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.
The goal is to use sed to return only the url from each line of FF extension Mining Blocker which uses this format for its regex lines:
{"baseurl":"*://002.0x1f4b0.com/*", "suburl":"*://*/002.0x1f4b0.com/*"},
{"baseurl":"*://003.0x1f4b0.com/*", "suburl":"*://*/003.0x1f4b0.com/*"},
the result should be:
002.0x1f4b0.com
003.0x1f4b0.com
One way would be to keep everything after suburl":"*://*/ then remove each occurrence of /*"},
I found https://unix.stackexchange.com/questions/24140/return-only-the-portion-of-a-line-after-a-matching-pattern but the special characters are a problem.
this won't work:
sed -n -e s#^.*suburl":"*://*/##g hosts
Would someone please show me how to mark the 2 asterisks in the string so they are seen by regex as literal characters, not wildcards?
edit:
sed -n 's#.*://\*/\([^/]\+\)/.*#\1#p' hosts
doesn't work, unfortunately.
regarding character substitution, thanks for directing me to the references.
I reduced the searched-for string to //*/ and used ASCII character codes like this:
sed -n -e s#^.*\d047\d047\d042\d047##g hosts
Unfortunately, that didn't output any changes to the lines.
My assumptions are:
^.*something specifies everything up to and including the last occurrence of "something" in a line
sed -n -e s#search##g deletes (replace with nothing) "search" within a line
So, this line:
sed -n -e s#^.*\d047\d047\d042\d047##g hosts
Should output everything after //*/ in each line...except it doesn't.
What is incorrect with that line?
Regarding deleting everything including and after the first / AFTER that first operation, yes, that's wanted too.
This might work for you (GNU sed):
sed -n 's#.*://\*/\([^/]\+\)/.*#\1#p' file
Match greedily (the longest string that matches) all characters up to ://*/, followed by a group of characters (which will be referred to as \1) that do not match a /, followed by the rest of the line and replace it by the group \1.
N.B. the sed substitution delimiters are arbitrary, in this case chosen to be # so as make pattern matching / easier. Also the character * on the left hand side of the substitution command may be interpreted as a meta character that means zero or more of the previous character/group and so is quoted \* so that it does not mistakenly exert this property. Finally, using the option -n toggles off the usual printing of every thing in the pattern space after all the sed commands have been executed. The p flag on the substitution command, prints the pattern space following a successful substitution, therefore only URL's will appear in the output or nothing.
I'm trying to extract the name of the file name that has been generated by a Java program. This Java program spits out multiple lines and I know exactly what the format of the file name is going to be. The information text that the Java program is spitting out is as follows:
ABCASJASLEKJASDFALDSF
Generated file YANNANI-0008876_17.xml.
TDSFALSFJLSDJF;
I'm capturing the output in a variable and then applying a sed operator in the following format:
sed -n 's/.*\(YANNANI.\([[:digit:]]\).\([xml]\)*\)/\1/p'
The result set is:
YANNANI-0008876_17.xml.
However, my problem is that want the extraction of the filename to stop at .xml. The last dot should never be extracted.
Is there a way to do this using sed?
Let's look at what your capture group actually captures:
$ grep 'YANNANI.\([[:digit:]]\).\([xml]\)*' infile
Generated file YANNANI-0008876_17.xml.
That's probably not what you intended:
\([[:digit:]]\) captures just a single digit (and the capture group around it doesn't do anything)
\([xml]\)* is "any of x, m or l, 0 or more times", so it matches the empty string (as above – or the line wouldn't match at all!), x, xx, lll, mxxxxxmmmmlxlxmxlmxlm, xml, ...
There is no way the final period is removed because you don't match anything after the capture groups
What would make sense instead:
Match "digits or underscores, 0 or more": [[:digit:]_]*
Match .xml, literally (escape the period): \.xml
Make sure the rest of the line (just the period, in this case) is matched by adding .* after the capture group
So the regex for the string you'd like to extract becomes
$ grep 'YANNANI.[[:digit:]_]*\.xml' infile
Generated file YANNANI-0008876_17.xml.
and to remove everything else on the line using sed, we surround regex with .*\( ... \).*:
$ sed -n 's/.*\(YANNANI.[[:digit:]_]*\.xml\).*/\1/p' infile
YANNANI-0008876_17.xml
This assumes you really meant . after YANNANI (any character).
You can call sed twice: first in printing and then in replacement mode:
sed -n 's/.*\(YANNANI.\([[:digit:]]\).\([xml]\)*\)/\1/p' | sed 's/\.$//g'
the last sed will remove all the last . at the end of all the lines fetched by your first sed
or you can go for a awk solution as you prefer:
awk '/.*YANNANI.[0-9]+.[0-9]+.xml/{print substr($NF,1,length($NF)-1)}'
this will print the last field (and truncate the last char of it using substr) of all the lines that do match your regex.
I'm trying to extract a list of CentOS domain names only from http://mirrorlist.centos.org/?release=6.4&arch=x86_64&repo=os
Truncating prefix "http://" and "ftp://" to the first "/" character only resulting a list of
yum.phx.singlehop.com
mirror.nyi.net
bay.uchicago.edu
centos.mirror.constant.com
mirror.teklinks.com
centos.mirror.netriplex.com
centos.someimage.com
mirror.sanctuaryhost.com
mirrors.cat.pdx.edu
mirrors.tummy.com
I searched stackoverflow for the sed method but I'm still having trouble.
I tried doing this with sed
curl "http://mirrorlist.centos.org/?release=6.4&arch=x86_64&repo=os" | sed '/:\/\//,/\//p'
but doesn't look like it is doing anything. Can you give me some advice?
Here you go:
curl "http://mirrorlist.centos.org/?release=6.4&arch=x86_64&repo=os" | sed -e 's?.*://??' -e 's?/.*??'
Your sed was completely wrong:
/x/,/y/ is a range. It selects multiple lines, from a line matching /x/ until a line matching /y/
The p command prints the selected range
Since all lines match both the start and end pattern you used, you effectively selected all lines. And, since sed echoes the input by default, the p command results in duplicated lines (all lines printed twice).
In my fix:
I used s??? instead of s/// because this way I didn't need to escape all the / in the patterns, so it's a bit more readable this way
I used two expressions with the -e flag:
s?.*://?? matches everything up until :// and replaces it with nothing
s?/.*?? matches everything from / until the end replaces it with nothing
The two expressions are executed in the given order
In modern versions of sed you can omit -e and separate the two expressions with ;. I stick to using -e because it's more portable.