Print a file skipping first X lines in Bash

Question

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 any option to do this. I am looking for a command to do this or a simple bash program.

Answer 1

Easiest way I found to remove the first ten lines of a file:

$ sed 1,10d file.txt

Answer 2

Just to propose a sed alternative. :) To skip first one million lines, try |sed '1,1000000d'.

Example:

$ perl -wle 'print for (1..1_000_005)'|sed '1,1000000d'
1000001
1000002
1000003
1000004
1000005

Answer 3

A less verbose version with AWK:

awk 'NR > 1e6' myfile.txt

But I would recommend using integer numbers.

Answer 4

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}"

Answer 5

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

tail -n +1000001 huge-file.log

It's the + character that does what you want. To quote from the man page:

If the first character of K (the number of bytes or lines) is a `+', print beginning with the Kth item from the start of each file.

Thus, as noted in the comment, putting +1000001 starts printing with the first item after the first 1,000,000 lines.

Answer 6

You'll need tail. Some examples:

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

If you really need to SKIP a particular number of "first" lines, use

$ tail -n +<N+1> <filename>
< filename, excluding first N lines. >

That is, if you want to skip N lines, you start printing line N+1. Example:

$ tail -n +11 /tmp/myfile
< /tmp/myfile, starting at line 11, or skipping the first 10 lines. >

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

$ tail -n <N> <filename>
< last N lines of file. >