How to use regex as field separator in awk?

Question

I'm trying to use regex as a field seperator in awk. From my reading this seems possible but I can't get the syntax right.

rpm -qa | awk '{ 'FS == [0-9]' ; print $1 }'
awk: cmd. line:1: { FS
awk: cmd. line:1:     ^ unexpected newline or end of string

Thoughts? The goal if not obviouse is to get a list of software without version number.

Answer 1

You have mucked up your quotes and syntax. To set the input field separator, the easiest way to do it is with the -F option on the command line:

awk -F '[0-9]' '{ print $1 }'

or

awk -F '[[:digit:]]' '{ print $1 }'

This would use any digit as the input field separator, and then output the first field from each line.

The [0-9] and [[:digit:]] expressions are not quite the same, depending on your locale. See "Difference between [0-9], [[:digit:]] and \d".

One could also set FS in the awk program itself. This is usually done in a BEGIN block as it's a one-time initialisation:

awk 'BEGIN { FS = "[0-9]" } { print $1 }'

Note that single quotes can't be used in a single-quoted string in the shell, and that awk strings always use double quotes.

Answer 2

+1 for Kusalananda's answer. Alternately, the FS variable can be set in the BEGIN block:

awk 'BEGIN {FS="[0-9]"} {print $1}'

Changing FS in a action block won't take effect until the next line is read

$ printf "%s\n" "abc123 def456" "ghi789 jkl0" | awk '{FS="[0-9]"; print $1}'
abc123
ghi

The other errors in the question:

• can't use single quotes inside a single-quoted string
• == is a comparison operator, = is for variable assignment