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

I am going to answer the title of the question directly instead of considering the detail of the question itself:

"grep a pattern and output non-matching part of line"

The title to this question is important to me because the pattern I am searching for contains characters that sed will assign special meaning to. I want to use grep because I can use -F or --fixed-strings to cause grep to interpret the pattern literally. Unfortunately, sed has no literal option, but both grep and bash have the ability to interpret patterns without considering any special characters.

Note: In my opinion, trying to backslash or escape special characters in a pattern appears complex in code and is unreliable because it is difficult to test. Using tools which are designed to search for literal text leaves me with a comfortable 'that will work' feeling without considering POSIX.

I used both grep and bash to produce the result because bash is slow and my use of fast grep creates a small output from a large input. This code searches for the literal twice, once during grep to quickly extract matching lines and once during =~ to remove the match itself from each line.

    while IFS= read -r || [[ -n "$RESULT" ]]; do
        if [[ "$REPLY" =~ (.*)("$LITERAL_PATTERN")(.*) ]]; then
            printf '%s\n' "${BASH_REMATCH[1]}${BASH_REMATCH[3]}"
        else
            printf "NOT-REFOUND" # should never happen
            exit 1
        fi
    done < <(grep -F "$LITERAL_PATTERN" < "$INPUT_FILE")

Explanation:

IFS= Reassigning the input field separator is a special prefix for a read statement. Assigning IFS to the empty string causes read to accept each line with all spaces and tabs literally until end of line (assuming IFS is default space-tab-newline).

-r Tells read to accept backslashes in the input stream literally instead of considering them as the start of an escape sequence.

$REPLY Is created by read to store characters from the input stream. The newline at the end of each line will NOT be in $REPLY.

|| [[ -n "$REPLY" ]] The logical or causes the while loop to accept input which is not newline terminated. This does not need to exist because grep always provides a trailing newline for every match. But, I habitually use this in my read loops because without it, characters between the last newline and the end of file will be ignored because that causes read to fail even though content is successfully read.

=~ (.*)("$LITERAL_PATTERN")(.*) ]] Is a standard bash regex test, but anything in quotes in taken as a literal. If I wanted =~ to consider the regex characters in contained in $PATTERN, then I would need to eliminate the double quotes.

"${BASH_REMATCH[@]}" Is created by [[ =~ ]] where [0] is the entire match and [N] is the contents of the match in the Nth set of parentheses.

Note: I do not like to reassign stdin to a while loop because it is easy to error and difficult to see what is happening later. I usually create a function for this type of operation which acts typically and expects file_name parameters or reassignment of stdin during the call.

Answer 2

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.

---

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.

Answer 3

Your $tags will never have a value because you send it to /dev/null. Besides from that little problem, there is no input to grep.

echo hello |grep "^he" -q ; 
ret=$? ; 
if [ $ret -eq 0 ]; 
then 
echo there is he in hello; 
fi

a successful return code is 0.

...here is 1 take at your 'problem':

pat="most of "; 
data="The apples are ripe. I will use most of them for jam.";  
echo $data |grep "$pat" -q; 
ret=$?; 
[ $ret -eq 0 ] && echo $data |sed "s/$pat//"
The apples are ripe. I will use them for jam.

... exact same thing?:

echo The apples are ripe. I will use most of them for jam. | sed ' s/most\ of\ //'

It seems to me you have confused the basic concepts. What are you trying to do anyway?

Answer 4

How about using a combination of grep, sed and $PIPESTATUS to get the correct exit-status?

$ echo Humans are not proud of their ancestors, and rarely invite
  them round to dinner | grep dinner | sed -n "/dinner/s/dinner//p"
Humans are not proud of their ancestors, and rarely invite them round to 

$ echo $PIPESTATUS[1]
0[1]

The members of the $PIPESTATUS array hold the exit status of each respective command executed in a pipe. $PIPESTATUS[0] holds the exit status of the first command in the pipe, $PIPESTATUS[1] the exit status of the second command, and so on.