如何在grep中匹配每个文件一次？

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 eg this find . -name '*.coffee' -exec awk '/re/ {print;exit}' {} \\; 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.

Answer 2

I think you can just do something like

grep -ri -m1 --include '*.coffee' 're' . | head -n 2

to eg pick the first match from each file, and pick at most two matches total.

Note that this requires your grep to treat -m as a per-file match limit; GNU grep does do this, but BSD grep apparently treats it as a global match limit.

Answer 3

I would do this in awk instead.

find . -name \*.coffee -exec awk '/re/ {print FILENAME ":" $0;exit}' {} \;

If you didn't need to recurse, you could just do it with awk:

awk '/re/ {print FILENAME ":" $0;nextfile}' *.coffee

Or, if you're using a current enough bash, you can use globstar:

shopt -s globstar
awk '/re/ {print FILENAME ":" $0;nextfile}' **/*.coffee

Answer 4

using find and xargs. find every .coffee files and excute -m1 grep to each of them

find . -print0 -name '*.coffee'|xargs -0 grep -m1 -ri 're'

test without -m1

linux# find . -name '*.txt'|xargs grep -ri 'oyss'
./test1.txt:oyss
./test1.txt:oyss1
./test1.txt:oyss2
./test2.txt:oyss1
./test2.txt:oyss2
./test2.txt:oyss3

add -m1

linux# find . -name '*.txt'|xargs grep -m1 -ri 'oyss'
./test1.txt:oyss
./test2.txt:oyss1

Answer 5

find . -name \\*.coffee -exec grep -m1 -i 're' {} \\;

find的-exec选项为每个匹配的文件运行一次命令（除非你使用+而不是\\; ;，这使它像xargs一样）。

Answer 6

You can do this easily in perl, and no messy cross platform issues!

use strict;
use warnings;
use autodie;

my $match = shift;

# Compile the match so it will run faster
my $match_re = qr{$match};

FILES: for my $file (@ARGV) {
    open my $fh, "<", $file;

    FILE: while(my $line = <$fh>) {
        chomp $line;

        if( $line =~ $match_re ) {
            print "$file: $line\n";
            last FILE;
        }
    }
}

The only difference is you have to use Perl style regular expressions instead of GNU style. They're not much different .

You can do the recursive part in Perl using File::Find , or use find feed it files.

find /some/path -name '*.coffee' -print0 | xargs -0 perl /path/to/your/program