grep how to suppress display of non-matched file?

Question

I am trying to find files containing a specific word using grep. There are many files in the directory (> 500)

Command I run

$ grep 'delete' *

Output

validate_data_stage1:0
validate_data_stage2:0
validate_data_stage3:0
validate_data_stage4:0
validate_data_stage5:0
validate_input_stage1:0
validate_input_stage2:0
validate_input_stage3:0
validate_input_stage4:0
.... and hundred of such lines

These are the files that don't contain the given match. I want to suppress those lines from displaying to stdout. I know of -q switch, but that would suppress the complete output, which I don't want.

How do I do that?

Answer 1

That's the behavior exhibited by grep -c.

Probably you have a file whose name starts with - and contains a c character and you're using GNU grep without setting the POSIXLY_CORRECT environment variable.

Use:

grep -- delete *

or better:

grep delete ./*

-- marks the end of options so that that filename will not be considered as an option (with a POSIX grep, it wouldn't since the non-option delete argument would have marked the end of options), but it wouldn't address the problem of a file called -. The grep delete ./* is more robust but has the drawback of outputting the extra ./ for matching files (though that may be considered a bonus since that helps identify file names that contain newline characters).

Answer 2

Add

| grep -v ':0'

Not very elegant but I think will do the job.

Adding -l to your command will give you only matches but it will also suppress printing number of matches for each file.

Answer 3

Using grep -l you will only get the files that contain at least one match.

Do you need information about how many matches there are in a file at all? Because the you could skip using -c which is used to count the number of matches in a file.

edit: And like warlock said using -I to suppress matches in binary files could also be a good idea.