passing JavaScript object to PHP not working

Question

I'm trying to send my javascript object to PHP via JSON.stringify()

Javascript:

$('#save').on('click touch', function(){
    obj = {
        "1" : {
            "1" : "hey",
            "2" : "hay"
            },              
        "2" : {
            "1" : "hey",
            "2" : "hay"
            }
    }

    var json = JSON.stringify( obj );
    console.log(json)
    $.ajax({
        type: 'POST',
        url: 'ajax.php',
        success: function(data) {
            alert(data);
            $("p").text(data);
        }
    });
});

ajax.php:

<?php 
    $obj = json_decode($json);
    echo $obj;
?>

But this code returns an error saying that $json is not defined. I have no idea why this is not working.

@csum: only if OP actually bothers to tell jquery to USE that json var in the ajax call... — Marc B
– Marc B, Commented Jan 13, 2016 at 20:46
@MarcB yup, true. It was more of a general statement. POST data is in $_POST — csum
– csum, Commented Jan 13, 2016 at 21:05

Community · Accepted Answer · 2020-06-20 09:12:55Z

6

There are 2 problems.

You are not sending any data with the request
That's not the way you'll get the value from a request in PHP

First, add this*:

$.ajax({
    type: 'POST',
    url: 'ajax.php',
    data : { json: json }, // <---------------------
    ...

_{^{* this works just because jQuery implementation will automatically convert any non-string data argument into a form-urlencoded query string. See the docs.}}

Then, in your PHP, you should do:

$jsonStr = $_POST['json'];
$json = json_decode($jsonStr);

Edit:

Another possible way:

$.ajax({
    type: 'POST',
    url: 'ajax.php',
    data : json , // <---------------------
    ...

This way, your data will not be a valid form-urlencoded input, so PHP will not parse it into $_POST, but you still can get the contents of your input doing this:

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

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Jan 13, 2016 at 20:46

Henrique Barcelos

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8 Comments

csum Over a year ago

Can you explain the reason behind the 2nd option not being a valid form-urlencoded input? Is it due to the contents of the json variable? The keys being numeric?

Henrique Barcelos Over a year ago

form-urlencoded values are in the form param1=value1&param2=value2&param3=value3...

Henrique Barcelos Over a year ago

Basically, it's a query string.

csum Over a year ago

So then wrapping your var in {json: ___ } solves this?

Henrique Barcelos Over a year ago

Solves because jQuery will automatically convert any object into a form-urlencoded string. Might not work in other Ajax libs.

|

tymeJV · Accepted Answer · 2016-01-13 20:39:14Z

2

Well - you never pass your data in the AJAX request!

$.ajax({
    type: 'POST',
    url: 'ajax.php',
    data: json //<---- RIGHT HERE
    success: function(data) {
        alert(data);
        $("p").text(data);
    }
});

answered Jan 13, 2016 at 20:39

tymeJV

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6 Comments

Steve Over a year ago

Thats half the problem

Henrique Barcelos Over a year ago

In addition, OP should assing $json = $_POST['json'] unless OP has register_globals=On

csum Over a year ago

don't forget the comma after json

Marc B Over a year ago

@HenriqueBarcelos: uh, php has no idea what json is. a json string is just a plain text string to php. php expects key=value pairs in POST, value by itself is utterly meaningless. register_globals never did, and never will,decode json...

Henrique Barcelos Over a year ago

I meant this part: $obj = json_decode($json). If register_globals=On and the data is passed under the 'json' key, like json={"foo":"bar"}, it would work.

|

samleurs · Accepted Answer · 2016-01-13 20:40:11Z

0

You have to send the obj with the ajax request

 $.ajax({
    type: 'POST',
    url: 'ajax.php',
    data : json,
    dataType : 'json' // for json response
    ...

answered Jan 13, 2016 at 20:40

samleurs

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2 Comments

Henrique Barcelos Over a year ago

dataType refers to the expected response, no the request.

samleurs Over a year ago

@HenriqueBarcelos in the comment is says 'for json response'

Paulo Henrique Queiroz · Accepted Answer · 2016-01-13 21:06:51Z

0

Check here for reference jQuery ajax

Data parameter: Specifies data to be sent to the server.

Try this:

$('#save').on('click touch', function(){
        obj = {
       "1" : {
           "1" : "hey",
           "2" : "hay"
        },              
    "2" : {
        "1" : "hey",
        "2" : "hay"
        }
}

var json = JSON.stringify( obj );

$.ajax({
    data : json,
    type: 'POST',
    url: 'ajax.php',
    success: function(data) {
        alert(data);
        $("p").text(data);
    }
});
});

edited Jan 13, 2016 at 21:06

answered Jan 13, 2016 at 20:42

Paulo Henrique Queiroz

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1 Comment

Henrique Barcelos Over a year ago

dataType refers to the expected response, no the request.

Abdul Rehman Khan · Accepted Answer · 2021-05-20 14:59:29Z

0

Replace your ajax code with this.

$.ajax({
    type: 'POST',
    url: 'ajax.php',
    data: json
    success: function(data) {
        alert(data);
        $("p").text(data);
    }
});

For ajax php

<?php 
    $obj = json_decode($_POST['data']);
    echo $obj;
?>

answered May 20, 2021 at 14:59

Abdul Rehman Khan

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