How to use a regex for the field separator in AWK?
First of all echo doesn't auto escape and outputs a literal \n. So you'll need to add -e to enable escapes. Second of all awk doesn't support \d so you have to use [0-9] or [[:digit:]].
echo -e "1 2 foo\n2 3 bar\n42 2 baz" | awk -F '[0-9]+ [0-9]+ ' '{ print $2 }'or
echo -e "1 2 foo\n2 3 bar\n42 2 baz" | awk -F '[[:digit:]]+ [[:digit:]]+ ' '{ print $2 }'Both outputs:
foobarbaz
Just replace \d with [0-9]:
With this you can print all the fields and you can see the fields immediatelly:
$ echo -e "1 2 foo\n2 3 bar\n42 2 baz" |awk -v FS="[0-9]+ [0-9]+" '{for (k=1;k<=NF;k++) print k,$k}'1 2 foo1 2 bar1 2 bazSo just use [0-9] in your command:
$ echo -e "1 2 foo\n2 3 bar\n42 2 baz" |awk -v FS="[0-9]+ [0-9]+" '{print $2}' foo bar baz