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);
});

How is `inviaRisultati()` being called? From a form submit? If so, are you preventing the default submit behavior (like you should be)? — Patrick Q, Apr 03 '19 at 17:58
ris is set when I call the function, anyway I had already tried data: {risultato: ris}, but it's the same — Marius Bauld, Apr 03 '19 at 18:00
@MariusBauld as RiggsFolly pointed out you are doing a POST so use the syntax he comment — stan chacon, Apr 03 '19 at 18:01
`success: function(aVariable){ alert(aVariable);}` then you can see what is ACTUALLY returned — RiggsFolly, Apr 03 '19 at 18:02
I'm pretty sure that passing a querystring is acceptable and jquery handles appropriately. — Patrick Q, Apr 03 '19 at 18:02
Click on the Request on the developer tools, go to Headers, and look for Request Payload. What is listed there? — aynber, Apr 03 '19 at 18:02
It is still not clear how this is being called, and that is what will get to the root of "I noticed though that in Chrome's dev tools there's 2 quiz.php elements (one xhr the other document)" — Patrick Q, Apr 03 '19 at 18:03
RiggsFolly I tried doing that, and it return 1, as It's supposed to — Marius Bauld, Apr 03 '19 at 18:08
I'm betting you are trying to open that php path in browser after the ajax is done. If so that won't work. Send back a response and inspect it in success callback — charlietfl, Apr 03 '19 at 18:12
The Request Payload isn't a tab. It's a section on the Headers tab — aynber, Apr 03 '19 at 18:13
charlietfl, I am sending the ajax request on the same page (quiz.php) — Marius Bauld, Apr 03 '19 at 18:16
There is no Request Payload section.General, Response Headers, Request Headers, Form Data (int the first element) General, Response Headers, Request Headers (in the second one) — Marius Bauld, Apr 03 '19 at 18:17
Fine but how are you checking that it's not in $_POST is what matters — charlietfl, Apr 03 '19 at 18:17
There are two types you can send data. 1. enctype="multipart/form-data" and 2.contentType:application/json...... if you send as enctype, your code will work, which is not preferred on ajax call. So you need use file_get_contents('php://input') which will return you an object. — Alaksandar Jesus Gene, Apr 03 '19 at 18:17
https://stackoverflow.com/questions/44291886/php-input-returning-string-for-json-ajax-request — Alaksandar Jesus Gene, Apr 03 '19 at 18:18
I get in it the second element, in fact that element displays the page correctly, but the open page is the first element — Marius Bauld, Apr 03 '19 at 18:19
@AlaksandarJesusGene actually there another and it is $.ajax default...`application/x-www-form-urlencoded` — charlietfl, Apr 03 '19 at 18:19
@MariusBauld starting to sound like you have a different problem and haven't shown us enough in question — charlietfl, Apr 03 '19 at 18:21
It says: risultato: 1 which is what it's supposed to say, but as I said there's 2 quiz.php elements, and the one that's open on my browser doesn't have the Form Data field — Marius Bauld, Apr 03 '19 at 18:24
I would suggest you to add below lines to your jQuery code: data: jQuery.param({"risultato": ris}), contentType: "application/x-www-form-urlencoded; charset=utf-8", dataType: "json", — slon, Apr 03 '19 at 18:27
@MariusBauld nothing will change in the page when you make that request. You have never identified how you check that `$_POST` doesn't contain what it should — charlietfl, Apr 03 '19 at 18:28
@slon both of those are done by default by `$.ajax`. That is pointless — charlietfl, Apr 03 '19 at 18:29
I'm still sticking with the fact that this is all due to a form submit without the necessary `preventDefault()`. Not going to wade through 100 more comments before that sinks in though. — Patrick Q, Apr 03 '19 at 18:32
@PatrickQ agree it is probably something trivial and OP thinks something else should be happening magically — charlietfl, Apr 03 '19 at 18:35
I am not submitting a form when firing inviaRisultato, I fire it with an EventListener — Marius Bauld, Apr 03 '19 at 18:52

score 0 · Answer 1 · answered Apr 03 '19 at 18:00

0

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

data: {risultato: ris}

answered Apr 03 '19 at 18:00

user2993497

525
1
7
21

I tried setting data this way aswell, but it doesn't work anyway – Marius Bauld Apr 03 '19 at 18:01
are you sure `ris` is set? before the ajax call, alert or console.log `ris`. And see if it displays. Also, set `risultato` to some string like `"test" `and see if it gets posted. – user2993497 Apr 03 '19 at 18:03
Yes, I set it when a button is clicked, and it prints it correctly – Marius Bauld Apr 03 '19 at 18:06
try running `inviaRisultati("test")` after declaring it and see if it works. It should. and that will tell you that the problem most likely is elsewhere in your code. – user2993497 Apr 03 '19 at 18:09
1

Can use string for data also. Easier to use object but nothing wrong with what OP was doing – charlietfl Apr 03 '19 at 18:10
I tried doing that, it does the same, the POST request goes through, I can see it in chrome dev tools, but as I said there's 2 quiz.php elements, one that's the correct one that displays the data I've sent, and another (the one that's open) that doesn't display the data as It's supposed to – Marius Bauld Apr 03 '19 at 18:14
1

@MariusBauld comment about *"the one that is open"* is telling me that you are expecting something magical to happen that shouldn't happen and misunderstanding the whole process – charlietfl Apr 03 '19 at 18:45

score 0 · Answer 2 · answered Apr 03 '19 at 18:19

0

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

answered Apr 03 '19 at 18:19

stan chacon

768
1
6
19

If I replace the whole html code with response it might work, I try it now – Marius Bauld Apr 03 '19 at 18:30
@charlietfl Since hes not being clear enought I decide to answer a sample code of a basic ajax request cause it seems what he is trying to do is that – stan chacon Apr 03 '19 at 18:31
1

ok but nothing you are doing would make their issue any different. Problem is it's not clear how they know it's not working...or if it even has a real problem – charlietfl Apr 03 '19 at 18:32
@MariusBauld glad to hear it =) – stan chacon Apr 03 '19 at 20:19

score -2 · Answer 3 · answered Apr 03 '19 at 18:34

-2

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

answered Apr 03 '19 at 18:34

Alaksandar Jesus Gene

6,523
12
52
83

`file_get_contents('php://input')` is not needed – charlietfl Apr 03 '19 at 18:36
Yep. I tested it now. – Alaksandar Jesus Gene Apr 03 '19 at 18:39
Well that's very strange...been using `$.ajax` with php for many years and never had problems with form encoded data not showing up in `$_POST`. Are you saying `$_POST` is empty without it? – charlietfl Apr 03 '19 at 18:39
Sorry. My mistake. I deleted that comment. It works without file_get_contents. – Alaksandar Jesus Gene Apr 03 '19 at 18:40
So this answer doesn't really provide anything relevant to aid OP other than it should work which everyone else already knows – charlietfl Apr 03 '19 at 18:46
The result in xhr is correct, it doesn't display on the page though, how am I supposed to display it? – Marius Bauld Apr 03 '19 at 18:50
My understanding is you are getting the result. But not displaying on the html page(chrome browser). If true, you need to create an element and using that element as reference, you can populate the data using jQuery. If false, can u show us your xhr call as i have shown in my answer image – Alaksandar Jesus Gene Apr 03 '19 at 18:55
@MariusBauld I think you assume that the php that generated the page should fill some element in the browser after the ajax post completes. If that is correct that's just not how ajax or php works. it sounds like you should study some ajax tutorials – charlietfl Apr 03 '19 at 19:03
charlietfl, I started using ajax today, and I didn't really know how to use it, I thought that as you sent the request the data would have been in $_POST or $_GET, but clearly it's not like that – Marius Bauld Apr 03 '19 at 19:08
@MariusBauld If you get the response as html codes then it will be easy. All you need to do is refer an element(lets say element with id="quiz") then $("#quiz").html(response) will give you the html output. If you get as data, you need handson experience with jQuery to populate it properly. – Alaksandar Jesus Gene Apr 03 '19 at 19:13

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

3 Answers3