I know I can use -A, -B and -C to show surrounding lines but all of them also show the matching line. What I'm trying to make here is so, in this example file:
foo
bar
I'd be doing something like grep <option> "bar" file and my output should be
foo
Side note: I know the way of doing it with another grep or using sed but I would like to do it just by using one time grep
It's quite a job for sed:
$ printf 'foo\nbar\n' | sed -n '$!N;/\nbar$/P;D'
foo
'$!N;/\nbar$/P;D'.
$!N -> If not last line, append next line to the pattern space. So If you are in line 1, your pattern space will look like line 1\nline2. Then /\nbar$/ will match if line 2 is bar. If yes, Print line 1, then Delete the rest of pattern space \nline 2.
yet another solution
grep -B 1 "bar" test | head -n 1
it has the advantage that you don't need any additional regexp or matching test, it is therefore more efficient but this won't work for multiple match
Another possible solution:
grep -B 1 "bar" test | sed '/bar/d'
You could do another grep which ignores "bar":
grep -B 1 "bar" test.txt | grep -v "bar"
bar? what should it return ?