conditional coloring in grep

Question

I have a simple bash script. One line of my script matches a "pattern" with grep and colors the match

grep -i --color=always -- "$1"

I need to modify it, so that it only colors the match, if the line is not a comment (i.e. if it does not start with a #). In other words: I still want grep to match the commented line, but without coloring it.

Let's say I have following text file (with added line numbers)

1 aaa PATTERN
2 bbb ccc PATTERN ddd
3 # eee PATTERN
4 fff ggg

I need my grep expression to match lines 1,2,3 but only color the lines which are not commented (1 and 2)

Answer 1

Using GNU grep with -P option:

$ grep PATTERN file | grep -P '^[^#]*\KPATTERN|^'
aaa PATTERN
bbb ccc PATTERN ddd
# eee PATTERN

\K causes all thing in the left of PATTERN don't include in matched string. This solution assumes that your grep is aliased to grep --color=auto

Answer 2

It's not possible to have grep do conditional coloring in this manner. It can only highlight the "PATTERN" that it's matching.

Answer 3

just a small hack: grep -E 'PATTERN' <filename> | grep --color -E "^[^#]*|$" -

you can use this: first grep matches the lines with PATTERN and second grep colors lines that does not start with '#'.

I don't think grep --color can do conditional coloring.