title: How to perform UPDATE with mysqli->prepare?
body: <p>As I know there is a way to input data into a mysql database with mysqli, where you do not have to use mysql_real_escape_string. I mean like this:  </p>
  
  <pre><code>$stmt = $mysqli-&gt;prepare("INSERT INTO CountryLanguage VALUES (?, ?, ?, ?)");  
  $stmt-&gt;bind_param('sssd', "something", "something2", "something3", "123");
  </code></pre>
  
  <p>Now my question: Can you do the same with UPDATE instead of INSERT? What would the expression look like? Would it look like the following:</p>
  
  <pre><code>$stmt = $mysqli-&gt;prepare("UPDATE CountryLanguage SET some = ?, some2 = ?, some3 = ?, some4 = ?"); 
  $stmt-&gt;bind_param('sssd', "something", "something2", "something3", "123");`
  </code></pre>
  
  <p>Thanks for your help.</p>