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

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. >

Answer 2

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 3

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

$ sed 1,10d file.txt

Answer 4

A less verbose version with AWK:

awk 'NR > 1e6' myfile.txt

But I would recommend using integer numbers.

Answer 5

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 6

if you want to skip first two line
tail -n +3 <filename>

if you want to skip first x line
tail -n +$((x+1)) <filename>

Answer 7

If you want to see first 10 line you can use sed as below:

sed -n '1,10 p' myFile.txt

or if you want to see lines from 20 to 30 you can use:

sed -n '20,30 p' myFile.txt

Answer 8

Use the sed delete command with a range address. For example:

$ sed 1,100d file.txt # Print file.txt omitting lines 1-100.

Alternatively, if you want to only print a known range use the print command with the -n flag:

$ sed -n 201,300p file.txt # Print lines 201-300 from file.txt

This solution should work reliably on all UNIX systems, regardless of the presence of GNU utilities.

Answer 9

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 10

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 11

sed -n '1d;p'

this command will delete the first line and print the rest

Answer 12

cat < File > | awk '{if(NR > 6) print $0}'

Answer 13

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}"