Undefined index in PHP post with AJAX

Question

In my local machine, I am trying to save data from a json to my mysql database, I am using Wampserver.

In my html page (saveInfo.php), I have this jquery code:

<script type="text/javascript">
            var jsObj = {"user_id":5, "login":"hsm"};
            var jsonobj = JSON.stringify(jsObj);            
            $.ajax({  
                type: "POST",  
                url: "json_handler.php",  
                data: { 'jsonobj':jsonobj },      
                success: function(){  
                  alert('success');
                  window.location = "http://localhost/quranMapping/php/json_handler.php";
                } 
            });
</script>

In the other side, I have my server-side php code (json_handler.php) like that:

<?php

$input = file_get_contents('php://input');
$input = $_POST['jsonobj'];

$result = json_decode($input);

echo $result->user_id;

?>

But when I run that code, I get this error: enter image description here

this could mean that $_POST['jsonobj'] is not getting set. You need to make sure that the value of jsonobj is getting passed to the php page. — Maximus2012, Aug 02 '13 at 17:43
It looks like the variable isn't making it as far as PHP. Can you `var_dump($_POST)` and output what comes out. Also, use Firefox or Chrome to get a dump of your outgoing POST request and post it up here too. — Code Whisperer, Aug 02 '13 at 17:44
Yeah it is not set, this is the problem, but how can garantee that jsonobj is getting passe plz? — Houssam Badri, Aug 02 '13 at 17:45
@Venkat It's not a deprecated error. And turning it off won't magically pass the POST value. Please people stop suggesting error silencing as a solution — Damien Pirsy, Aug 02 '13 at 17:46

Orangepill · Answer 1 · 2013-08-02T17:58:58.560

3

You should remove this:

var jsonobj = JSON.stringify(jsObj);

and change

data: { 'jsonobj':jsonobj },

to

data: jsObj,

On the php side to decode the data just use

$user_id = isset($_POST["user_id"])?$_POST["user_id"]:"";
$login   = isset($_POST["login"])?$_POST["login"]:"";

Also there is no need to do

$input = file_get_contents('php://input');

Since the form is being posted with an object as data the value will be application/x-www-form-urlencoded so it don't be valid json.

edited Aug 02 '13 at 17:58

answered Aug 02 '13 at 17:48

Orangepill

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Named `jsonobj` on the client, but `jsObj` on the server? – mishik Aug 02 '13 at 17:49
array (size=0) : this was the output – Houssam Badri Aug 02 '13 at 17:52
@HoussemBdr that is the value of `var_dump($_POST["jsonobj"]);` ? – Orangepill Aug 02 '13 at 17:55
@HoussemBdr See updates ... there is no reason to wrap the two object in another object... this should give your the data you want right in the $_POST array – Orangepill Aug 02 '13 at 17:59
echo "user is : ".$user_id; gave empty result : ( – Houssam Badri Aug 02 '13 at 18:02
the erro was "undefiened variable" – Houssam Badri Aug 02 '13 at 18:05
are you assigning $user_id from the $_POST array value? – Orangepill Aug 02 '13 at 18:09

Undefined index in PHP post with AJAX

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