AJAX request is sent, but $_POST doesn't update

Question

The AJAX request goes through correctly, I checked with chrome's developer tools, there is a request on quiz.php page, but when I check for $_POST['risultato'] it looks doesn't exist. I noticed though that in Chrome's dev tools there's 2 quiz.php elements (one xhr the other document)

I tried changing the code in several ways, but it seems like it doesn't work

<?php
    if(isset($_POST['risultato'])){
        print($_POST['risultato']);
    }
?>

---

<script>    
    function inviaRisultati(ris){
            $.ajax({
                url: "quiz.php",
                type: "POST",
                cache: false,
                data: {risultato: ris},
                success: function(){
                    alert("INVIATI");
                }
            })
    }

The program is expected to return the result on quiz.php page (the same page where ajax request is fired), and it's supposed to print it somewhere

EDIT: I fixed it

<?php
    file_get_contents('php://input');
    if(isset($_POST['risultato'])){
        print($_POST['risultato']);
    }
?>

---

function inviaRisultati(param) {
   return $.ajax({
         url:"quiz.php",
         method:"POST",
         data:{action: "SLC", risultato :param},
         dataType:"text"
    });

}

---

inviaRisultati(1).done(function(response){``
    document.open(); 
    document.write(response);
});

Answer 1

In Ajax, the data attribute is in JSON format. your data attribute will be like this

data: {risultato: ris}

Answer 2

Hi you can do it this way:

your php script:

     if (isset($_POST["action"])) {
    $action = $_POST["action"];
    switch ($action) {
        case 'SLC':
        if (isset($_POST["risultato"])) {
            $response = $_POST["risultato"];
            echo $response;
        }
    }
    break;
}

Where action is a command you want to do SLC, UPD, DEL etc and risultato is a parameter

then in your ajax:

 var param = $('#yourinputid').val();
function getInfo(param) {
   return $.ajax({
         url:"quiz.php",
         method:"POST",
         data:{action: "SLC", risultato :param},
         dataType:"text"
});

}

call it like this:

getInfo(param).done(function(response){
alert(response);
//do something with your response here
})

Hope it helps

Answer 3

HTML CODE

<script>
        jQuery(document).ready(function(){
            ris = 'abcd';
            jQuery.ajax({
                url: "quiz.php",
                type: "POST",
                cache: false,
                data: {risultato: ris},
                success: function(){

                }
            })
        });
    </script>

PHP CODE

<?php

file_get_contents('php://input');

print($_POST['risultato']);

Console Output