jQuery ajax not working when trying to call php function

Question

I'm trying to call php function from my JavaScript script. I've read several questions on how to call a php function using jQuery ajax. I simply copied most of the content and adapted it to fit my script, but somehow it's not working.

Here's my code:

    var var1 = document.getElementById("tag").value;
    var id = tag.id;
    var var3 = tag.checked;
    var test = $.ajax({
        url: '../PHP/phpFunc.php',
        type: 'GET',
        data: {param1: var1, param2: id, param3: var3},
        complete: function() {
            if(window.console) {
                console.log("success");
            }
        },
        error: function() {
            if(window.console) {
                console.log("Error");
            }
        }
    })

and the php code:

<?php

echo "<script>
     console.log(\"test\")
     </script>";
    $var1= $_GET['param1'];
    $var2 = $_GET['param2'];
    $var3 = $_GET['param3];
    $var4 = '../XML/user/'. $var1.'.xml';
   
    

?>

the "success" is being printed when the JS function is executed, but php code never returns the desired output.

Your PHP doesn't return anything, because you've not written the code to do it - or if you have then you've accidentally removed it from your sample so we can't debug it. Also, you're using the `complete` callback of `$.ajax()` . If you want to receive data it makes more sense to use `success` and then accept the response as an argument to that handler function. See the documentation for more details: https://api.jquery.com/jQuery.ajax/ — Rory McCrossan, Jul 08 '22 at 16:30
[What is the difference between client-side and server-side programming?](https://stackoverflow.com/questions/13840429/what-is-the-difference-between-client-side-and-server-side-programming) So you cannot execute PHP code in the browser — RiggsFolly, Jul 08 '22 at 16:30
For example: `$.get( "../PHP/phpFunc.php", function( data ) { console.log(data); });` Also your php code seems to be missing a closing quote symbol before closing the php code block. `$var4 = '../XML/user/'. $username .'.xml'; ";` — George G, Jul 08 '22 at 16:45
You never do anything with the response back in the Ajax call. calling the php does not magically do anything in the browser. — epascarello, Jul 08 '22 at 17:04

jQuery ajax not working when trying to call php function

0 Answers0