Question

Can someone show me how to make a If statement where if a drop down does not equal the default value of 0 (in my case) to automatically submit the form if possible?

session_start();
$current = isset($_SESSION['ClientNamefour']) ? $_SESSION['ClientNamefour'] : 0;
while ($row = mysql_fetch_array($result)) { 
    $id = $row["Client_Code"]; 
    $thing = $row["Client_Full_Name"];
    $value = "$id, $thing";

    $sel=($id==$current)?'SELECTED':'';

    $options4.="<OPTION $sel VALUE=\"$value\">$thing</option>";
} 
?>
<FORM name="form" action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST">

<SELECT NAME="ClientNamefour" OnChange="this.form.submit()"
)">
  <OPTION VALUE=0>Client
   <?php echo $options4?> 
</SELECT>
</FORM>

Answer 1

Use the onchange event and some JavaScript. In general your generated HTML should look something like this:

<form id="form" name="form" method="POST" action="http://example/script.php">
  <select id="select" name="select" onChange="document.getElementById('form').submit();">
    <option value="0" selected="selected">0</option>
    <option value="1">1</option>
    <option value="2">2</option>
  </select>
</form>

The onchange event is only fired when you select an unselected option, so no additional checks are required.

Compared with your solution:

• the id is missing in your <form> tag
• you need closing </option> tags
• you’d probably need to change the JavaScript in the onchange attribute of the <select> tag