$_POST error after saving information that was returned via Ajax

Question

Have issue with PHP $_POST. Then I save information and check $_POST['g_kcal']; returned

Notice: Undefined index: g_kcal in

This field g_kcal count's all input fields values after input change. It returns new input with name = 'g_kcal' via ajax.

JavaScript:

$('.box-body').change(function(){
    var totalPointsKcal = 0;

    $(this).find('input[name="kcal[]"]').each(function(i,n){
        totalPointsKcal += parseFloat($(n).val(),10); 
    });

    $.ajax({
        type: 'post',
        url: 'ajaxLoader.php',
        data: {
            galutinis_kcal:totalPointsKcal,
        },
        success: function (response) {
            document.getElementById("galutinis").innerHTML=response;
        }
    });
});

ajaxLoader.php:

if(isset($_POST['galutinis_kcal'])) {
    $kcal = $_POST['galutinis_kcal'];

    echo "<input class='skaiciuokle kcal1' type='text' name='g_kcal' value=". round($kcal, 1) ." disabled>";
}

Can't find their the problem is. Maybe POST can't receive information that was send via ajax ?

the place is n file there input field is. Have more fields where I enter values and it return my entered values, but g_kcal notice as undefined. — Kintamasis, Apr 12 '16 at 16:29
What happens when you execute `$('.kcal1').trigger('change');`? Do you submit the received input? — Eduardo Galván, Apr 12 '16 at 16:31
it was from my old code, this don't do anything, I remove this. — Kintamasis, Apr 12 '16 at 16:35
I think you should be doing `g_kcal:totalPointsKcal` in your JS OR you should do `$_POST['galutinis_kcal']` in your php — AdityaParab, Apr 12 '16 at 16:40
the bracket's shouldn't be inside `input[name="kcal[]"]` but in the input itself, as in `name='g_kcal[]'` plus the names don't match up here, so your question is unclear. — Funk Forty Niner, Apr 12 '16 at 17:05
possible duplicate of [PHP: Notice: Undefined variable and Notice: Undefined index](http://stackoverflow.com/questions/4261133/php-notice-undefined-variable-and-notice-undefined-index) — Funk Forty Niner, Apr 12 '16 at 17:10
@Fred-ii- input[name="kcal[]"] then input with name = "kcal[]" change, it count and return other input. — Kintamasis, Apr 12 '16 at 17:11

score 0 · Answer 1 · answered Apr 12 '16 at 16:29

0

Try setting the Content Type in your Ajax POST:

$.ajax({
    type: 'post',
    url: 'ajaxLoader.php',
    data: {
        galutinis_kcal:totalPointsKcal,
    },
    contentType: "application/json",
    success: function (response) {
        document.getElementById("galutinis").innerHTML=response;
        $('.kcal1').trigger('change');
    }
});

If you don't set the content type of the Ajax call the body (data) won't be sent.

answered Apr 12 '16 at 16:29

docksteaderluke

2,195
5
27
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not working, now the input field disappears and ajax loader not return new field. – Kintamasis Apr 12 '16 at 16:39
Can you verify that $_POST["galutinis_kcal"] is still not set? You may also need to add `dataType: "text/html"` to your ajax request. `dataType` is the content type that your ajax call expects from the server. – docksteaderluke Apr 12 '16 at 16:45
Without dataType and contentType I receive information and new input with counted value. But with dataType and contentType set it didn't enter success function, I just check with alert(); – Kintamasis Apr 12 '16 at 16:54

Jose Rojas · Answer 2 · 2016-04-12T16:55:24.227

0

At the server side use this code to get data sent via AJAX Request:

$info = json_decode(file_get_contents("php://input"));

if(isset($info->galutinis_kcal)){
    $kcal = $info->galutinis_kcal;
    echo "<input class='skaiciuokle kcal1' type='text' name='g_kcal' value=". round($kcal, 1) ." disabled>";
}

at the Client side use JSON.stringify before send your request

$.ajax({
    type: 'post',
    url: 'ajaxLoader.php',
    data: JSON.stringify({
        galutinis_kcal:totalPointsKcal
    }),
    success: function (response) {
        document.getElementById("galutinis").innerHTML=response;
    }
});

edited Apr 12 '16 at 16:55

answered Apr 12 '16 at 16:38

Jose Rojas

3,490
3
26
40

it returned new input value, but when I save, I still receive Notice: Undefined index: – Kintamasis Apr 12 '16 at 17:00
The php file returned the value?, where are you saving? – Jose Rojas Apr 12 '16 at 17:03
know I'm in test.php there are ajax request, javascript code. In test.php then input change, the code above count value and send to ajaxLoader.php and it innerHTML the #galutinis. And i see value in input. Then i click button to save. Form method="test.php" return in same page. And then I receive Notice: Undefined index: g_kcal in – Kintamasis Apr 12 '16 at 17:07

score 0 · Accepted Answer · answered Apr 12 '16 at 17:26

0

The problem was input field was set to DISABLE. Now it's working fine

answered Apr 12 '16 at 17:26

Kintamasis

265
6
27

$_POST error after saving information that was returned via Ajax

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