grep a pattern and output non-matching part of line

Question

I know it is possible to invert grep output with the -v flag. Is there a way to only output the non-matching part of the matched line? I ask because I would like to use the return code of grep (which sed won't have). Here's sort of what I've got:

tags=$(grep "^$PAT" >/dev/null 2>&1)
[ "$?" -eq 0 ] && echo $tags 

Answer 1

You could use sed:

$ sed -n "/$PAT/s/$PAT//p" $file

The only problem is that it'll return an exit code of 0 as long as the pattern is good, even if the pattern can't be found.

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Explanation

The -n parameter tells sed not to print out any lines. Sed's default is to print out all lines of the file. Let's look at each part of the sed program in between the slashes. Assume the program is /1/2/3/4/5:

1. /$PAT/: This says to look for all lines that matches pattern $PAT to run your substitution command. Otherwise, sed would operate on all lines, even if there is no substitution.
2. /s/: This says you will be doing a substitution
3. /$PAT/: This is the pattern you will be substituting. It's $PAT. So, you're searching for lines that contain $PAT and then you're going to substitute the pattern for something.
4. //: This is what you're substituting for $PAT. It is null. Therefore, you're deleting $PAT from the line.
5. /p: This final p says to print out the line.

Thus:

• You tell sed not to print out the lines of the file as it processes them.
• You're searching for all lines that contain $PAT.
• On these lines, you're using the s command (substitution) to remove the pattern.
• You're printing out the line once the pattern is removed from the line.