How do I get a part of the output of a command in Linux Bash?

Question

How do I get a part of the output of a command in Bash?

For example, the command php -v outputs:

PHP 5.3.28 (cli) (built: Jun 23 2014 16:25:09         


        

Answer 1

a classic "million ways to skin a cat" question...

These methods seem to filter by spaces... If the versions/notes contain spaces, this fails.

The ( brackets, however, seem consistent across all my platforms so I've used the following:

For example, on Debian:

root@host:~# php -v  | head -1
PHP 5.3.28-1~dotdeb.0 with Suhosin-Patch (cli) (built: Dec 13 2013 01:38:56)
root@host:~# php -v  | head -1 | cut -d " " -f 1-2
PHP 5.3.28-1~dotdeb.0

So here I trim everything before the second (:

root@host:~# php -v  | head -1 | cut -d "(" -f 1-2
PHP 5.3.28-1~dotdeb.0 with Suhosin-Patch (cli)

Note: there will be a trailing white-space (blank space at the end)

Alternatively, you could always use your package manager to determine this (recommended):

root@debian-or-ubuntu-host:~# dpkg -s php5 | grep 'Version'
Version: 5.3.28-1~dotdeb.0

...or on a CentOS, Red Hat Linux, or Scientific Linux distribution:

[root@rpm-based-host ~]# rpm -qa | grep php-5
php-5.4.28-1.el6.remi.x86_64