title: Best way to complete this Perl regex one-liner
body: <p>I'm trying to use a Perl one-liner to munge some output from grepping svn diff, so I can automatically test the files.  We have a run_test.sh script that can take multiple PHP files prepended with 'Test" as its arguments.</p>
  
  <p>So far I have the following which successfully prepends 'Test' to the file names</p>
  
  <pre><code>[gjempty@gjempty-rhel4 classes]$ svn diff | grep '(revision' | perl -wpl -e 's/(.*)\/(.*)$/$1\/Test$2/'
  --- commerce/TestLCart.php      (revision 104387)
  --- commerce/manufacturing/TestLRoutingData.php (revision 104387)
  </code></pre>
  
  <p>Now I'd just like to grab the file/path to pass it to our run_test.sh.  I can finish it off with awk as below, but am trying to improve my Perl/one-liner skills.  So how do I revise the perl one-liner to additionally extract only the file path?</p>
  
  <pre><code> svn diff | grep '(revision' | perl -wpl -e 's/(.*)\/(.*)$/$1\/Test$2/' | awk '{print $2}' | xargs run_test.sh
  </code></pre>