How do I pass a value to a PHP script using AJAX?

if you require the output (if it's html, say) on the same page, the better deal would be to use $("div").load("your_url.php?your_var="+your_var_value); where "div" is the div tag in which you're loading the processed html.

I have a feeling that the PHP script is encountering a fatal error in conn.php and not continuing after the error. I would make sure PHP error reporting and display errors feature is set in your PHP configuration file. That being said, I tried your code without including the database connection and the query. So we know the POSTed value is being received by the server. Try it out with the following code: JavaScript <script type="text/javascript"> $(function(){ $.ajax({ type: "POST", url: 'AjaxCallTest.php', data: ({Imgname:"13"}), success: function(data) { alert(data); } }); }); </script> PHP <?php $temp = $_POST['Imgname']; echo $temp; ?> Make sure you are using mysqli* family of functions securely. mysqli* family of functions support parameterization of input, this feature protects your web application against SQL Injection Attack: https://www.owasp.org/index.php/SQL_Injection https://www.owasp.org/index.php/SQL_Injection_Prevention_Cheat_Sheet

If you want to use pure JS to pass the value through AJAX, given below is the code if (window.XMLHttpRequest) { xmlhttp = new XMLHttpRequest(); xmlhttp.onreadystatechange = function () { if (xmlhttp.readyState === 4 && xmlhttp.status === 200) { alert('form was submitted'); } }; xmlhttp.open("POST", "phpcodes/AjaxCallTest.php", true); xmlhttp.send("Imgname=13");

Try this method: https://api.jquery.com/jQuery.post/

You should check these one with a working demo how to http://whats-online.info/science-and-tutorials/30/select-preview-rename-and-upload-image-using-jquery-Ajax/ .These post assisted me create a classic file chooser that allows submit a file to the server without page load. I think these a nice article for you

Edit: misread your code. Below is still valid. I imagine you are practicing / learning but please read up on sanitization. Your current script is very insecure.

try this $.ajax({ method:"POST", url:"phpcodes/AjaxCallTest.php", data:{Imgname:"13"}, success:function(data) { alert("data sent"); }, error:function(data) { alert("Data sending failed"); } });