Print a file, skipping the first X lines, in Bash [duplicate]

Answer 1

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

$ sed 1,10d file.txt

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

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

Answer 4

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 5

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 6

If you want to see the first 10 lines 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