I want both matches: Campus and Mode but the following code only gets Campus:
sed -n 's/.*<th colspan="7">Campus:\(.*\)<br \/> Mode:\(.*\)<\/th><\/tr>.*/\1/p'
Since your replacement string only includes the \1 back reference, it's only going to print the first matched group, i.e. the one from Campus:\(.*\). Try changing your replacement string to \1 \2 to include both...
Related
I am struggling to work out how to get a , out from inbetween various patterns such as:
500,000
xyz ,CA
I have tried something like:
sed -E "s/\([a-zA-Z]*\),([a-zA-Z]*\)/\([a-zA-Z]*\) ([a-zA-Z]*\)/g" $file -i
It picks up the first pattern, but then over writes it with the second pattern, I feel like I am missing something very simple and I can't work it out, any help really appreciated.
You're missing the notion of capture groups, I think. To refer to a parenthesized portion of the search within the replacement string, use \1 for the first group, \2 for the second group, etc.
The modified line would be:
sed -E "s/([a-zA-Z]),([a-zA-Z])/\1 \2/g" $file -i
Rather than replacing the part that matches the first ([a-zA-Z]) with the literal text "([a-zA-Z])", this modified line just copies the matched portion into the output (and likewise for the second group).
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 have a file with a lot of text, but I want to print only words that contain "#" at the beginning. Ex:
My name is #Laura and I live in #London. Name=#Laura. City=#London
How can I print all words that start with #?.I did this the following and it worked, but I want to do it using sed. I tried several patters, but I cannot make it print anything.
grep -o -E "#\w+" file.txt
Thanks
Use this sed command:
sed 's/[^#]*\(#[^ .]*\)/\1\n/g' file.txt
Explanation: we invoke the substitution command of sed. This has following structure: sed 's/regex/replace/options'. We will search for a regex and replace it using the g option. g makes sure the match is made multiple times per line.
We look for a series of non at chars followed by an # and a number of non-spaces #[^ ]*. We put this last part in a group \(\) and sub it during the replacement \1.
Note that we add a newline at the end of each match, you can also get the output on a single line by omitting the \n.
I have a list of usernames and i would like add possible combinations to it.
Example. Lets say this is the list I have
johna
maryb
charlesc
Is there is a way to use sed to edit it the way it looks like
ajohn
bmary
ccharles
And also
john_a
mary_b
charles_c
etc...
Can anyone assist me into getting the commands to do so, any explanation will be awesome as well. I would like to understand how it works if possible. I usually get confused when I see things like 's/\.(.*.... without knowing what some of those mean... anyway thanks in advance.
EDIT ... I change the username
sed s/\(user\)\(.\)/\2\1/
Breakdown:
sed s/string/replacement/ will replace all instances of string with replacement.
Then, string in that sed expression is \(user\)\(.\). This can be broken down into two
parts: \(user\) and \(.\). Each of these is a capture group - bracketed by \( \). That means that once we've matched something with them, we can reuse it in the replacement string.
\(user\) matches, surprisingly enough, the user part of the string. \(.\) matches any single character - that's what the . means. Then, you have two captured groups - user and a (or b or c).
The replacement part just uses these to recreate the pattern a little differently. \2\1 says "print the second capture group, then the first capture group". Which in this case, will print out auser - since we matched user and a with each group.
ex:
$ echo "usera
> userb
> userc" | sed "s/\(user\)\(.\)/\2\1/"
auser
buser
cuser
You can change the \2\1 to use any string you want - ie. \2_\1 will give a_user, b_user, c_user.
Also, in order to match any preceding string (not just "user"), just replace the \(user\) with \(.*\). Ex:
$ echo "marya
> johnb
> alfredc" | sed "s/\(.*\)\(.\)/\2\1/"
amary
bjohn
calfred
here's a partial answer to what is probably the easy part. To use sed to change usera to user_a you could use:
sed 's/user/user_/' temp
where temp is the name of the file that contains your initial list of usernames. How this works: It is finding the first instance of "user" on each line and replacing it with "user_"
Similarly for your dot example:
sed 's/user/user./' temp
will replace the first instance of "user" on each line with "user."
Sed does not offer non-greedy regex, so I suggest perl:
perl -pe 's/(.*?)(.)$/$2$1/g' file
ajohn
bmary
ccharles
perl -pe 's/(.*?)(.)$/$1_$2/g' file
john_a
mary_b
charles_c
That way you don't need to know the username before hand.
Simple solution using awk
awk '{a=$NF;$NF="";$0=a$0}1' FS="" OFS="" file
ajohn
bmary
ccharles
and
awk '{a=$NF;$NF="";$0=$0"_" a}1' FS="" OFS="" file
john_a
mary_b
charles_c
By setting FS to nothing, every letter is a field in awk. You can then easy manipulate it.
And no need to using capturing groups etc, just plain field swapping.
This might work for you (GNU sed):
sed -r 's/^([^_]*)_?(.)$/\2\1/' file
This matches any charactes other than underscores (in the first back reference (\1)), a possible underscore and the last character (in the second back reference (\2)) and swaps them around.