title: Ignore all fields (keys) into Mongo find
body: <p><a href="http://docs.mongodb.org/manual/reference/method/db.collection.find/" rel="nofollow">From documentation</a> <code>find()</code> is defined like bellow:</p>
  
  <blockquote>
    <p><code>db.collection.find(query, projection)</code></p>
  </blockquote>
  
  <p>where the <code>projection</code> is an object that <em>specifies the fields to return using projection operators</em>. </p>
  
  <blockquote>
    <p>To return all fields in the matching document, <strong>omit this parameter.</strong></p>
  </blockquote>
  
  <p>My question is that if there is any way to <strong>ignore all fields</strong> using <code>projection</code> object.</p>
  
  <p>I want to receive a response like this:</p>
  
  <blockquote>
    <p><code>[{}, {}, {}, {}]</code></p>
  </blockquote>
  
  <p>(an array with empty objects)</p>
  
  <p>To ignore a key I use: <code>{"key": 0}</code>.
  I've already tried <code>{"$all": 0}</code>  that seems that doesn't work.</p>
  
  <p>How can I ignore all fields?</p>
  
  <hr>
  
  <p>I know that I can use <code>count()</code> function, but this doesn't help me in this case.</p>