How do I limit the number of results returned from grep?

Solution 1:

The -m option is probably what you're looking for:

grep -m 10 PATTERN [FILE]

From man grep:

-m NUM, --max-count=NUM
        Stop reading a file after NUM matching lines.  If the  input  is
        standard  input  from a regular file, and NUM matching lines are
        output, grep ensures that the standard input  is  positioned  to
        just  after the last matching line before exiting, regardless of
        the presence of trailing context lines.  This enables a  calling
        process  to resume a search.

Note: grep stops reading the file once the specified number of matches have been found!

Solution 2:

Another option is just using head:

grep ...parameters... yourfile | head

This won't require searching the entire file - it will stop when the first ten matching lines are found. Another advantage with this approach is that will return no more than 10 lines even if you are using grep with the -o option.

For example if the file contains the following lines:

112233
223344
123123

Then this is the difference in the output:

$ grep -o '1.' yourfile | head -n2
11
12

$ grep -m2 -o '1.'
11
12
12

Using head returns only 2 results as desired, whereas -m2 returns 3.

Solution 3:

Awk approach:

awk '/pattern/{print; count++; if (count==10) exit}' file

Solution 4:

For 2 use cases:

1. I only want n overall results, not n results per file, the grep -m 2 is per file max occurrence.
2. I often use git grep which doesn't take -m

A good alternative in these scenarios is grep | sed 2q to grep first 2 occurrences across all files. sed documentation: https://www.gnu.org/software/sed/manual/sed.html