title: How do I get a part of the output of a command in Linux Bash?
body: <p>How do I get a part of the output of a command in Bash?</p>
  <p>For example, the command <code>php -v</code> outputs:</p>
  <pre><code>PHP 5.3.28 (cli) (built: Jun 23 2014 16:25:09)
  Copyright (c) 1997-2013 The PHP Group
  Zend Engine v2.3.0, Copyright (c) 1998-2013 Zend Technologies with the ionCube PHP Loader v4.6.1, Copyright (c) 2002-2014, by ionCube Ltd.
  </code></pre>
  <p>And I only want to output the <code>PHP 5.3.28 (cli)</code> part. How do I do that?</p>
  <p>I've tried <code>php -v | grep 'PHP 5.3.28'</code>, but that outputs: <code>PHP 5.3.28 (cli) (built: Jun 23 2014 16:25:09)</code> and that's not what I want.</p>