Say I've got a bunch of files that are all >100 lines long. I'd like to trim off the top 14 lines and the bottom 9 lines, leaving only the lines in the middle. This command will trim off the top fourteen:
cat myfile.txt | tail -n +15
Is there another command I can pipe through to trim off the bottom 9 without explicitly passing the length of the file?
Edited to add: My version of head (Mac OS 10.5) doesn't accept a negative number of lines as a parameter.
The head command should do what you want. It woks just like tail but from the other end of the file.
You could use sed:
sed -n -e :a -e '1,9!{P;N;D;};N;ba' myfile.txt
You can also use sed for the first 15:
sed '1,15d' myfile.txt
Use a negative number of lines with the head command:
cat myfile.txt | head -n -9
That prints everything except the last 9 lines.
What jbourque said is completely right. He just wasn't too wordy about it:
cat myfile.txt | tail -n +15 | head -n -9
If you can recognize the last 9 lines by a distinctive pattern for the first of those lines, then a simple sed command would do the trick:
sed -e '1,15d' -e '/distinctive-pattern/,$d' $file
If you need a pure numeric offset from the bottom, standard (as opposed to GNU) sed won't help, but ed would:
ed $file <<'!'
1,15d
$-8,$d
w
q
!
This overwrites the original files. You'd have to script where the file is written to if you wanted to avoid that.
This will work on OS X and might be a bit more easily understandable than the sed example:
< myfile.txt tail -n +15 | tail -r | tail -n +10 | tail -r
Of course, if you can get your hands on GNU's version of head, it can be done even more elegantly:
< myfile.txt tail -n +15 | head -n -9
Be aware the tail starts at the nth line while head skips n lines of the input.
This should also work, and does things in a single process:
seq 15 |
awk -v N=5 '
{ lines[NR % N] = $0 }
END { i = NR-N+1; if (i<0) i=0; for (; i <= NR; ++i) print lines[i % N] }'
(The seq is just an easy way to produce some test data.)