title: inner join wont work with mysqli prepared statement in php
body: <p>I can't seem to get this statement or statements alike to work with prepared queries, the code works just fine below:</p>
  
  <pre><code>$DBH = getDBH();
      $stmt = $DBH-&gt;prepare("SELECT a.id, a.title, a.photo FROM tag t INNER JOIN tag_reference atx ON t.tag_id = atx.tag_id
          INNER JOIN articles a
              ON atx.article_id = a.id
      WHERE t.tag_name = 'example'");
      $stmt-&gt;execute();
      $stmt-&gt;bind_result($id,$title,$photo);
      $stmt-&gt;fetch();
  </code></pre>
  
  <p>but when I change <code>t.tag_name = '?'</code> it gives me an error that the amount of parameters do not match. This is the statement that does not work.</p>
  
  <pre><code>$DBH = getDBH();
      $stmt = $DBH-&gt;prepare("SELECT a.id, a.title, a.photo FROM tag t INNER JOIN tag_reference atx ON t.tag_id = atx.tag_id
          INNER JOIN articles a
              ON atx.article_id = a.id
      WHERE t.tag_name = '?'");
          $stmt-&gt;bind_param('s',$example);
      $stmt-&gt;execute();
      $stmt-&gt;bind_result($id,$title,$photo);
      $stmt-&gt;fetch();
  </code></pre>
  
  <p>Can anyone please help?</p>