awk - 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 awk magic?

for example:

$ cat foo.txt
bar
foo
baz
foo

$ cat foo.txt | grep foo-SOMETHING
bar
baz

Edit

• in case when line 2 and 3 has "foo", then line 1 and 2 should be printed (although I am not very strict here)

Additional feature: consider the example:

bar
foo
baz
foo
foo

This should ideally return

bar
baz
foo

Answer 1

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.