Print a file skipping X lines in Bash

Question

Hi

I have a very long file which I want to print but skipping the first 1e6 lines for example. I look into the cat man page but I did not see nay option to do this. I am looking for a command to do this or a simple bash program. I know how to do it using a program in C but I want to do it using the common commands. Any way to do it? Thanks a lot in advance..

Answer 1

you need tail.

$ tail great-big-file.log
< Last 10 lines of great-big-file.log >

if you really need to SKIP a particular number of lines, use

$ tail -n+<Lines to skip> <filename>
< filename, excluding first so many lines. >

If you want to just see the last so many lines, omit the "+":

$ tail -n<Lines to show> <filename>
< last so many lines of file. >

Answer 2

You can do this using the head and tail commands:

head -n <num> | tail -n <lines to print>

where num is 1e6 + the number of lines you want to print.

Answer 3

If you have GNU tail available on your system, you can do the following:

$ tail -n +1000000 huge-file.log

It's the + character that does what you want.

Answer 4

Hi,

This shell script works fine for me:

#!/bin/bash
awk -v initial_line=$1 -v end_line=$2 '{
    if (NR >= initial_line && NR <= end_line) 
    print $0
}' $3

Used with this sample file (file.txt):

one
two
three
four
five
six

The command (it will extract from second to fourth line in the file):

edu@debian5:~$./script.sh 2 4 file.txt

Output of this command:

two
three
four

Of course, you can improve it, for example by testing that all argument values are the expected :-)

Answer 5

I needed to do the same and found this thread.

I tried "tail -n +, but it just printed everything.

The more +lines worked nicely on the prompt, but it turned out it behaved totally different when run in headless mode (cronjob).

I finally wrote this myself:

skip=5
FILE="/tmp/filetoprint"
tail -n$((`cat "${FILE}" | wc -l` - skip)) "${FILE}"