Grep only the first match and stop

-m 1 means return the first match in any given file. But it will still continue to search in other files. Also, if there are two or more matched in the same line, all of them will be displayed.

You can use head -1 to solve this problem:

grep -o -a -m 1 -h -r "Pulsanti Operietur" /path/to/dir | head -1

explanation of each grep option:

-o, --only-matching, print only the matched part of the line (instead of the entire line)
-a, --text, process a binary file as if it were text
-m 1, --max-count, stop reading a file after 1 matching line
-h, --no-filename, suppress the prefixing of file names on output
-r, --recursive, read all files under a directory recursively

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You can pipe grep result to head in conjunction with stdbuf.

Note, that in order to ensure stopping after Nth match, you need to using stdbuf to make sure grep don't buffer its output:

stdbuf -oL grep -rl 'pattern' * | head -n1
stdbuf -oL grep -o -a -m 1 -h -r "Pulsanti Operietur" /path/to/dir | head -n1
stdbuf -oL grep -nH -m 1 -R "django.conf.urls.defaults" * | head -n1

As soon as head consumes 1 line, it terminated and grep will receive SIGPIPE because it still output something to pipe while head was gone.

This assumed that no file names contain newline.

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My grep-a-like program ack has a -1 option that stops at the first match found anywhere. It supports the -m 1 that @mvp refers to as well. I put it in there because if I'm searching a big tree of source code to find something that I know exists in only one file, it's unnecessary to find it and have to hit Ctrl-C.

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You can use below command if you want to print entire line and file name if the occurrence of particular word in current directory you are searching.

grep -m 1 -r "Not caching" * | head -1

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For anyone who lands here, as I did, perplexed as to why --max-count didn't seem to be working when acting on stdin...

TL;DR - --max-count n does NOT stop after finding n matches, it stops after finding all matches on n lines.

(And stdin, even if it's only a string, counts as one line.)

This is true despite the fact that, in zsh 5.8, at least, man grep describes the option this way:

-m num, --max-count=num
        Stop reading the file after num matches.

Longer Explanation

In my case, I was trying to grab just the first part of a relative path:

>  echo "some/path/here" | grep -E -o -m 1 '[^\/]+'

and was quite confused when it gave me back

some
path
here

Thanks to the comment from @harperville above, I finally figured out: It's not about the output, it's about the input.

Indeed, when I tried

>  echo "some/path/here\nanother/path/there" | grep -E -o -m 1 '[^\/]+'

I got the same result as above (i.e., only the parts before the \n in this second example).

Notes

For those who are less familiar with grep:

-E (--extended-regexp) tells it to use "extended" regular expressions, i.e., the ones you're used to from most other programming languages. The differences between "extended" and "basic" aren't big - it's just about which characters you need to escape in your regex - but as someone who's primarily a TS and Python developer, I always use -E because that way I never have to think about it. (Pro-tip: Add alias grep="grep -E" to your .zshrc and you'll never have to worry about it again!)

-o (--only-matching) tells it to only print the matches, rather than each line on which it found a match.

-m n (--max-count n) restricts it to searching n lines. (If you've read this far you clearly already know that, though! 😛)

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Reading the grep manual (man grep) this is the minimum command to find first match with Extended regexp. Example getting the ethernet name that in my laptop is NOT eth0 !

$ ifconfing | grep -E -o -m 1 "^[a-z0-9]+"

Explanation: -E for extended regexp, -o to return only the match, -m 1 to look only one line

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A single liner, using find:

find -type f -exec grep -lm1 "PATTERN" {} \; -a -quit

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