Print nth line in a file

Shell Programming and Scripting
1

Bash/Oracle Linux 6.4

A basic requirement.

How can I get nth line of a file printed ? Can I use grep in this case ?

Example:
In the below file, 12th line is "Kernel parameter check passed for rmem_max" . I just want the 12 line to be printed.

# cat sometext.txt
Kernel version check passed
Kernel parameter check passed for semmsl
Kernel parameter check passed for semmns
Kernel parameter check passed for semopm
Kernel parameter check passed for semmni
Kernel parameter check passed for shmmax
Kernel parameter check passed for shmmni
Kernel parameter check passed for shmall
Kernel parameter check passed for file-max
Kernel parameter check passed for ip_local_port_range
Kernel parameter check passed for rmem_default
Kernel parameter check passed for rmem_max
Kernel parameter check passed for wmem_default
Kernel parameter check passed for wmem_max
Kernel parameter check passed for aio-max-nr
Package existence check passed for binutils
#
2 
awk 'NR==12' sometext.txt

OR

sed -n 12p sometext.txt
1 Like
3

Hello,

Simply one more approach :slight_smile:

awk 'NR != 12 {next}1' file_name

Output will be as follows.

Kernel parameter check passed for rmem_max

Thanks,
R. Singh