bash - remove the last part of a string

Question

I want to know the ruby version being run from a bash script.

That can be done with ruby -v:

$ ruby -v
ruby 1.9.3p545 (2014-02-24) [i686-linux] Brightbox

My problem is that I just want the version number. How do I transform that output into this?

1.9.3p545

Is the p545 important? If not, you could do ruby -e 'puts RUBY_VERSION' — Tom Fenech
– Tom Fenech, Commented Sep 17, 2014 at 11:24
@TomFenech cool!. Without newline character ruby -e '$><<RUBY_VERSION' — Roman Kiselenko
– Roman Kiselenko, Commented Sep 17, 2014 at 11:33

Avinash Raj · Accepted Answer · 2014-09-17 11:08:55Z

3

Through cut,

$ ruby -v | cut -d" " -f2
1.9.3p484

Through awk,

$ ruby -v | awk '{print $2}'
1.9.3p484

answered Sep 17, 2014 at 11:08

Avinash Raj

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Comments

Component 10 · Accepted Answer · 2014-09-17 11:10:08Z

2

Try this:

ruby -v | cut -d' ' -f2

answered Sep 17, 2014 at 11:10

Component 10

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cdarke · Accepted Answer · 2014-09-17 12:17:27Z

2

Some solutions use awk or cut, those waste a child process. Try this:

read p ver rest < <(ruby -v)
echo $ver

$p will contain ruby, and $rest the trailing fields. This solution only uses one child process (ruby), not two.

answered Sep 17, 2014 at 12:17

cdarke

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