title: How to exclude _id without including other fields using the aggregation framework
body: <p>I would like to get the result of an aggregation pipeline without the _id filed. I know this is possible if you provide explicitly other fields that will be the output of the projection. But, ¿how can I mimic the $projec in a find call?  </p>
  
  <p>This is what I want (<strong>No field explicitly included</strong>):</p>
  
  <pre><code>db.col.find({},{_id:0})
  </code></pre>
  
  <p>But in the aggregation framework seems to be impossible:</p>
  
  <pre><code>db.col.aggregate([{'$project': {_id:0}}])
  
  Error: Printing Stack Trace
      at printStackTrace (src/mongo/shell/utils.js:37:15)
      at DBCollection.aggregate (src/mongo/shell/collection.js:927:9)
      at (shell):1:11
  2013-10-07T16:36:09.273+0200 aggregate failed: {
      "errmsg" : "exception: $projection requires at least one output field",
      "code" : 16403,
      "ok" : 0
  } at src/mongo/shell/collection.js:928
  </code></pre>
  
  <p>Any idea to workaround this issue ?</p>