cut or awk command to print first field of first row

linux bash unix awk
user3331975 picture user3331975 · Mar 5, 2014 · Viewed 197.1k times · Source

I am trying print the first field of the first row of an output. Here is the case. I just need to print only SUSE from this output. 

# cat /etc/*release

SUSE Linux Enterprise Server 11 (x86_64)
VERSION = 11
PATCHLEVEL = 2

Tried with cat /etc/*release | awk {'print $1}' but that print the first string of every row

SUSE
VERSION
PATCHLEVEL

Answer

devnull picture devnull · Mar 5, 2014

Specify NR if you want to capture output from selected rows:

awk 'NR==1{print $1}' /etc/*release

An alternative (ugly) way of achieving the same would be:

awk '{print $1; exit}'

An efficient way of getting the first string from a specific line, say line 42, in the output would be:

awk 'NR==42{print $1; exit}'

Related questions

Looping through the content of a file in Bash

How do I iterate through each line of a text file with Bash? With this script: echo "Start!" for p in (peptides.txt) do echo "${p}" done I get this output on the screen: Start! ./runPep.sh: line 3: syntax error …

Read More
How to permanently set $PATH on Linux/Unix?

I'm trying to add a directory to my path so it will always be in my Linux path. I've tried: export PATH=$PATH:/path/to/dir This works, however each time I exit the terminal and start a new terminal …

Read More
How to run a shell script on a Unix console or Mac terminal?

I know it, forget it and relearn it again. Time to write it down.

Read More