title: Open url given by stdout from node --inspect command
body: <p>Alright here's a fun one, we get the following stdout:</p>
  
  <pre><code>Debugger listening on port 9229.
  Warning: This is an experimental feature and could change at any time.
  To start debugging, open the following URL in Chrome:
      chrome-devtools://devtools/remote/serve_file/@60cd6e859b9f557d2312f5bf532f6aec5f284980/inspector.html?experiments=true&amp;v8only=true&amp;ws=127.0.0.1:9229/a7a2a58d-dfd1-4cc6-875c-47da78adeb1b
  </code></pre>
  
  <p>when we run a command like so:</p>
  
  <pre><code>node --inspect --debug-brk bin/www.js
  </code></pre>
  
  <p>I was thinking of creating a bash script like this:</p>
  
  <pre><code>#!/usr/bin/env bash
  
  NODE_DEBUG_OUTPUT=$(node --inspect --debug-brk bin/www.js)
  open -a "/Applications/Google Chrome.app"/ getUrl(${NODE_DEBUG_OUTPUT})
  </code></pre>
  
  <p>but here's where my bash skills end - how can I implement the <code>getUrl</code> function so that I can get the url from the output?</p>
  
  <p>I am pretty certain that if we pass a url to the open function in bash, it will open a browser, in this case though it has to be Google Chrome browser.</p>