cut or awk command to print first field of first row
问题 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 回答1: 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