title: How to if isset checking in PHP switch?
body: <p>I have this code :</p>
  
  <pre><code>&lt;form method="POST"&gt;
  &lt;input type="text" name="my"&gt;
  &lt;input type="submit" name="submit"&gt;
  &lt;/form&gt;
  
  &lt;?php
  if (isset($_POST['my'])) {
  $ali = $_POST['my'];
  }
  switch ($ali) {
      case '1':
          echo "one";
          break;
      case '2':
          echo "two";
          break;
      case '3':
          echo "three";
          break;
  }
  ?&gt;
  </code></pre>
  
  <p>The problem is that i have error when i want to open the window ! </p>
  
  <blockquote>
    <p>Notice: Undefined variable: ali in C:\Program Files (x86)\EasyPHP-5.4.0RC4\www\test\index.php on line 11</p>
  </blockquote>
  
  <p>How to check the value into the switch ? </p>