title: Sequelize: delete JSONB property (with PostgreSQL)
body: <p>I'm using <a href="https://github.com/sequelize/sequelize" rel="noreferrer">sequelize</a> (v3.12.2), with <a href="https://github.com/brianc/node-postgres" rel="noreferrer">pg</a> (4.4.3), <a href="http://www.postgresql.org/" rel="noreferrer">PostgreSQL</a> (v9.4), and <a href="https://nodejs.org/en/" rel="noreferrer">Node</a> (v4.1.2).</p>
  
  <p>I have a model that inlcudes a JSONB data type field.</p>
  
  <pre><code>var User = {
    data: {
      type: Sequelize.JSONB
  }
  </code></pre>
  
  <p>Now I can do</p>
  
  <pre><code>User.findOne({where: {id: 12345}})
    .update({data: {x: 'foo'}});
  </code></pre>
  
  <p>And</p>
  
  <pre><code>User.findOne({where: {id: 12345}})
    .update({'data.y': 'bar'});
  </code></pre>
  
  <p>Now, if I want to delete just the <code>data.x</code> property how can I do it in one command?</p>
  
  <pre><code>User.findOne({where: {id: 12345}})
    .update({'data.x': null});
  </code></pre>
  
  <p>Obviously doesn't work. In fact, the resulting data object should be:</p>
  
  <pre><code>{y: 'bar'}
  </code></pre>
  
  <p>And not:</p>
  
  <pre><code>{x: null, y: 'bar'}
  </code></pre>
  
  <p>How can I do this with Sequelize? Thanks for your help!</p>