title: How to get all column from all tables including associated table along with computed field
body: <p>In cakephp 3 document <a href="http://book.cakephp.org/3.0/en/orm/query-builder.html#selecting-all-fields-from-a-table" rel="nofollow">selecting-all-fields-from-a-table</a></p>
  
  <pre><code>// Only all fields from the articles table including
  // a calculated slug field.
  $query = $articlesTable-&gt;find();
  $query
     -&gt;select(['slug' =&gt; $query-&gt;func()-&gt;concat(['title', '-', 'id'])])
     -&gt;select($articlesTable); // Select all fields from articles
  </code></pre>
  
  <p>i have query where i have 5-6 table in contain for joining to bring data along with computed field.</p>
  
  <p>now from your document example passing table instance can bring all records along with computed field.</p>
  
  <p>now my question is that for all the joining table how i can bring their data along with computed column.</p>
  
  <p>I tried to pass joining table class instance within select but i am getting error.</p>
  
  <blockquote>
    <p>SQLSTATE[42S22]: Column not found: 1054 Unknown column 'Table2.id' in 'field list'</p>
  </blockquote>
  
  <p>to create table classes instance first then have to pass them in select to be able to be able to bring joined table data?</p>
  
  <p>my query</p>
  
  <pre><code>        $table2 = TableRegistry::get('Table2');
          $query = $this-&gt;table1
              -&gt;find('all')
              -&gt;select(['computed_column' =&gt; 'A subquery here'])
              -&gt;select($this-&gt;table1)
              -&gt;select($table2)
              -&gt;contain(['Table2','Table3','Table4','Table5'])
              -&gt;where($conditions); 
  </code></pre>