cut or awk command to print first field of first row

Question

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

User · Answer

Try  sed  NUMq d    etc  release   awk   print  1     where NUM is line number  ex  sed  1q d    etc  release   awk   print  1

User · Answer

sed -n 1p  etc  release  cut -d     -f1   if tab delimited    sed -n 1p  etc  release  cut -f1

User · Answer

try this   head -1  etc  release   awk   print  1

User · Answer

You could use the head instead of cat    head -n1  etc  release   awk   print  1

User · Answer

awk  sed  pipe  that s heavy  set  cat  etc  release   echo  1

User · Answer

Specify the Line Number using NR built-in variable    awk  NR  1 print  1    etc  release

User · Answer

You can kill the process which is running the container   With this command you can list the processes related with the docker container   ps -aux   grep   docker ps -a   grep container-name   awk   print  1      Now you have the process ids to kill with kill or kill -9

User · Answer

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

[linux] cut or awk command to print first field of first row

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