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I use grep very often and am familiar with it's ability to return matching lines (by default) and non-matching lines (using the -v parameter). However, I want to be able to grep a file once to separate matching and non-matching lines.
If this is not possible, please let me know. I realize I could do this easily in perl or awk, but am curious if it is possible with grep.
Thanks!
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To display only the lines that do not match a search pattern, use the -v ( or --invert-match ) option. The -w option tells grep to return only those lines where the specified string is a whole word (enclosed by non-word characters). By default, grep is case-sensitive.
Displaying only the matched pattern : By default, grep displays the entire line which has the matched string. We can make the grep to display only the matched string by using the -o option. 6. Show line number while displaying the output using grep -n : To show the line number of file with the line matched.
For BSD or GNU grep you can use -B num to set how many lines before the match and -A num for the number of lines after the match. If you want the same number of lines before and after you can use -C num . This will show 3 lines before and 3 lines after. Save this answer.
If it does NOT have to be grep - this is a single pass split based on a pattern -- pattern found > file1 pattern not found > file2
awk '/pattern/ {print $0 > "file1"; next}{print $0 > "file2"}' inputfile
I had the exact same problem and I wrote a small Perl script for that [1]. It only accepts one argument: the regex to grep input on.
[1] https://gist.github.com/tonejito/c9c0bffd75d8c81483f9107c609439e1
It reads STDIN by line and checks against the given regex, matched lines go to STDOUT and not matched go to STDERR.
I made it this way because this tool sits in the middle of a pipeline and I use shell redirection to save the files on their final location.
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