Can I grep only the first n lines of a file?
The magic of pipes;
head -10 log.txt | grep <whatever>
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.
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
.
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