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Home/ Questions/Q 7499663
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Editorial Team
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Editorial Team
Asked: May 29, 20262026-05-29T19:54:03+00:00 2026-05-29T19:54:03+00:00

I have the following HTML. Currently it relies on the user hitting the ‘Go’

  • 0

I have the following HTML.

Currently it relies on the user hitting the ‘Go’ button to Submit the Form.

Is it possible to change this so it submits each time the user selects a dropdown option?

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"
        "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">
<head>
    <meta http-equiv="content-type" content="text/html; charset=utf-8" />
    <title>Employees by Department</title>
    <script src="ajax.js" type="text/javascript"></script>
    <script src="dept.js" type="text/javascript"></script>
    <style type="text/css" media="all">@import "style.css";</style>
</head>
<body>
<!-- dept_form_ajax.html -->
<p>Select a department and click 'GO' to see the employees in that department.</p>
<form action="dept_results.php" method="get" id="dept_form">
<p>
<select id="did" name="did">
<option value="1">Human Resources</option>
<option value="2">Accounting</option>
<option value="3">Marketing</option>
<option value="4">Redundancy Department</option>
</select>
<input name="go" type="submit" value="GO" />
</p>
</form>

<select id="results"></select>
</body>
</html>

For the record, here is my dept.js file contents:

// dept.js

/*  This page does all the magic for applying
 *  Ajax to an employees listing form.
 *  The department_id is sent to a PHP 
 *  script which will return data in HTML format.
 */

// Have a function run after the page loads:
window.onload = init;

// Function that adds the Ajax layer:
function init() {

  // Get an XMLHttpRequest object:
  var ajax = getXMLHttpRequestObject();

  // Attach the function call to the form submission, if supported:
  if (ajax) {

    // Check for DOM support:
    if (document.getElementById('results')) {

      // Add an onsubmit event handler to the form:
      document.getElementById('dept_form').onsubmit = function() {

        // Call the PHP script.
        // Use the GET method.
        // Pass the department_id in the URL.

        // Get the department_id:
        var did = document.getElementById('did').value;

        // Open the connection:
        ajax.open('get', 'dept_results_ajax.php?did=' + encodeURIComponent(did));

        // Function that handles the response:
        ajax.onreadystatechange = function() {
          // Pass it this request object:
          handleResponse(ajax);
        }

        // Send the request:
        ajax.send(null);

        return false; // So form isn't submitted.

      } // End of anonymous function.

    } // End of DOM check.

  } // End of ajax IF.

} // End of init() function.

// Function that handles the response from the PHP script:
function handleResponse(ajax) {

  // Check that the transaction is complete:
  if (ajax.readyState == 4) {

    // Check for a valid HTTP status code:
    if ((ajax.status == 200) || (ajax.status == 304) ) {

      // Put the received response in the DOM:
      var results = document.getElementById('results');
      results.innerHTML = ajax.responseText;

      // Make the results box visible:
      results.style.display = 'block';

    } else { // Bad status code, submit the form.
      document.getElementById('dept_form').submit();
    }

  } // End of readyState IF.

} // End of handleResponse() function.

Many thanks for any pointers here.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-29T19:54:06+00:00Added an answer on May 29, 2026 at 7:54 pm

    This might give you an idea.

    replace:

     document.getElementById('dept_form').onsubmit = function() { 
    

    with this:

       $('#did').change(function () {
    

    or rather with this:

    document.getElementById('did').onchange = function() {
    
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