grep how to suppress display of non-matched file?

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.

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.

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).