grep - print line before, don't print match

Question

How to easily print line above the match and skip the match itself? grep -A, -B and -o opt do not solve it. Maybe some

Answer 1

awk '/foo/{print a}{a=$0}' your_file

Answer 2

awk '!/foo/ { line = $0 }
     /foo/ { print line }' foo.txt

The first clause saves each non-foo line in a variable. The second clause prints the most recent saved line when the line matches foo.

This also works (and handles the case where you have two foo lines in a row):

awk '/foo/ {print line}
     {line = $0}' foo.txt

With grep you can do:

grep -B 1 foo foo.txt | grep -vE 'foo|^--$'

The second command filters out the foo lines and the dividers that are printed between the matching blocks.

Answer 3

Just set p to the pattern you want:

$ awk '$0~p{print a}{a=$0}' p="foo" file
bar
baz
foo