How to match once per file in grep?

Question

Is there any grep option that let's me control total number of matches but stops at first match on each file?

Example:

If I do this grep -ri --include '*.coffee' 're' . I get this:

./app.coffee:express = require 'express' ./app.coffee:passport = require 'passport' ./app.coffee:BrowserIDStrategy = require('passport-browserid').Strategy ./app.coffee:app = express() ./config.coffee:    session_secret: 'nyan cat' 

And if I do grep -ri -m2 --include '*.coffee' 're' ., I get this:

./app.coffee:config = require './config' ./app.coffee:passport = require 'passport' 

But, what I really want is this output:

./app.coffee:express = require 'express' ./config.coffee:    session_secret: 'nyan cat' 

Doing -m1 does not work as I get this for grep -ri -m1 --include '*.coffee' 're' .

./app.coffee:express = require 'express' 

Tried not using grep e.g. this find . -name '*.coffee' -exec awk '/re/ {print;exit}' {} \; produced:

config = require './config'     session_secret: 'nyan cat' 

UPDATE: As noted below the GNU grep -m option treats counts per file whereas -m for BSD grep treats it as global match count

Answer 1

So, using grep, you just need the option -l, --files-with-matches.

All those answers about find, awk or shell scripts are away from the question.