Given a hostname in format of aaa0.bbb.ccc, I want to extract the first substring before ., that is, aaa0 in this case. I use following awk script to do so,
echo aaa0.bbb.ccc | awk '{if (match($0, /\./)) {print substr($0, 0, RSTART - 1)}}'
While the script running on one machine A produces aaa0, running on machine B produces only aaa, without 0 in the end. Both machine runs Ubuntu/Linaro, but A runs newer version of awk(gawk with version 3.1.8 while B with older awk (mawk with version 1.2)
I am asking in general, how to write a compatible awk script that performs the same functionality ...
You just want to set the field separator as . using the -F option and print the first field:
$ echo aaa0.bbb.ccc | awk -F'.' '{print $1}'
aaa0
Same thing but using cut:
$ echo aaa0.bbb.ccc | cut -d'.' -f1
aaa0
Or with sed:
$ echo aaa0.bbb.ccc | sed 's/[.].*//'
aaa0
Even grep:
$ echo aaa0.bbb.ccc | grep -o '^[^.]*'
aaa0