Can I grep only the first n lines of a file?

Question

I have very long log files, is it possible to ask grep to only search the first 10 lines?

Solution 1

The magic of pipes;

head -10 log.txt | grep <whatever>

Solution 2

For folks who find this on Google, I needed to search the first n lines of multiple files, but to only print the matching filenames. I used

 gawk 'FNR>10 {nextfile} /pattern/ { print FILENAME ; nextfile }' filenames

The FNR..nextfile stops processing a file once 10 lines have been seen. The //..{} prints the filename and moves on whenever the first match in a given file shows up. To quote the filenames for the benefit of other programs, use

 gawk 'FNR>10 {nextfile} /pattern/ { print "\"" FILENAME "\"" ; nextfile }' filenames

Solution 3

Or use awk for a single process without |:

awk '/your_regexp/ && NR < 11' INPUTFILE

On each line, if your_regexp matches, and the number of records (lines) is less than 11, it executes the default action (which is printing the input line).

Or use sed:

sed -n '/your_regexp/p;10q' INPUTFILE 

Checks your regexp and prints the line (-n means don't print the input, which is otherwise the default), and quits right after the 10th line.

Solution 4

You have a few options using programs along with grep. The simplest in my opinion is to use head:

head -n10 filename | grep ...

head will output the first 10 lines (using the -n option), and then you can pipe that output to grep.

Solution 5

grep "pattern" <(head -n 10 filename)

View more solutions

Comments

[potong]

Why not quit on the 10th? (see sed solution)

[jaypal singh]

I didn't even realize, all the solutions here using head have used -n 10 (including me) not realizing that head by default displays only 10 lines. :)

[Admin]

awk '{ if ( NR <= 10 ) { if(index($0,"ab") > 0) { print $0; } } else { exit; } }' textfile -- faster.

[Zsolt Botykai]

@potong you are right, corrected. @srikanthradix while it can be faster you're solution is not searching for regexps but only for fixed strings. awk '{ if ( NR <= 10 ) { if( $0 ~ "YOUR_REGEXP") { print } } else { exit; } }' textfile does.

[jaypal singh]

Plus the style isn't awkish. 2xifs and 1xelse in a command that needs no action statement would make aho. weinberger and kernighan cry ...

[Floris]

I was one of those folks who found this on Google. Thanks!

[Pramod S. Nikam]

Although it might right. add more description of question to make answer more comprehensive.

[Stuart Nelson]

you can also pipe an arbitrary stream to head: someCmd | head -10

[Zlemini]

Head defaults to printing the first 10 lines to standard output, so this is valid for 10 lines head log.txt | grep <whatever>

[Vladyslav Savchenko]

I think, instead of NR it would be better to use FNR, because if you use awk with multiple files FNR starts from 0 for each file.

[James M. Lay]

Is there a way to do this when using grep's -l option? I'd like to list all the files who's first 5 characters are RIFFD.

[Brian W]

for me, this code printed out the full path of the file. Which is exactly what I needed. Also FNR=1 will just search 1st line. Thanks!

[Pre101]

This answers a completely different question and is not useful in this context.

[OrangeDog]

To do this recursively over a directory: find ./path -type -f -exec awk 'FNR>10 {nextfile} /pattern/ { print FILENAME ; nextfile }' '{}' +

[David Siegal]

Thanks @OrangeDog. One slight correction: should be -type f

[franzisk]

No, this will give you the first 6 occurrences of "string" in the whole cov.txt file

[David LeBauer]

If I recall, -C 2 will do the same as -A 2 -B 2

[anthony]

Just what I was searching for!