How do I 'grep -c' and avoid printing files with zero '0' count

Question

The command 'grep -c blah *' lists all the files, like below.

 % grep -c jill *
 file1:1
 file2:0
 file3:0
 file4:0
 file5:0
 file6:1
 % 

What I want is:

 % grep -c jill * | grep -v ':0'
 file1:1
 file6:1
 % 

Instead of piping and grep'ing the output like above, is there a flag to suppress listing files with 0 counts?

SJ

Answer 1

How to grep nonzero counts:

grep -rIcH 'string' . | grep -v ':0$'

• -r Recurse subdirectories.
• -I Ignore binary files ^{(thanks @tongpu, warlock)}.
• -c Show count of matches. Annoyingly, includes 0-count files.
• -H Show file name, even if only one file ^{(thanks @CraigEstey)}.
• 'string' your string goes here.
• . Start from the current directory.
• | grep -v ':0$' Remove 0-count files. ^{(thanks @LaurentiuRoescu)}

(I realize the OP was excluding the pipe trick, but this is what works for me.)

Answer 2

Just use awk. e.g. with GNU awk for ENDFILE:

awk '/jill/{c++} ENDFILE{if (c) print FILENAME":"c; c=0}' *