JSON PHP decode not working

Question

I have seen many examples, but for whatever reason, none seem to work for me.

I have the following sent from a app, via ajax, to a php file. This is how it looks when its sent:


    obj:{"ClientData":
    [{
        "firstName":"Master",
        "lastName":"Tester",
        "email":"[email protected]",
        "dob":"1973-01-22",
        "age":"51",
    }],
    "HealthData":
    [
        "condition : Prone to Fainting / Dizziness",
        "condition : Allergic Response to Plasters",
    ],
    "someData":
    [{
        "firstName":"Male",
        "lastName":"checking",
    }]
    }

Here is how it looks in debugger

Code as is:

{"ClientData":[{"firstName":"Master","lastName":"Tester","email":"[email protected]","dob":"1973-01-22","age":"51","pierceType":"Vici","street":"number of house","city":"here","county":"there","postcode":"everywhere"}],"HealthData":[["condtion : Prone to Fainting / Dizziness","condtion : Allergic Response to Plasters","condtion : Prone to Fainting / Dizziness"]],"PiercerData":[{"firstName":"Male","lastName":"checking","pierceDate":"2013-02-25","jewelleryType":"Vici","jewelleryDesign":"Vidi","jewellerySize":"Vici","idChecked":null,"medicalChecked":null,"notes":"This is for more info"}]}

This comes in one long line into a php file, here is the code:

<?php
header('Content-Type: application/json');
header("Access-Control-Allow-Origin: *");
//var_dump($_POST['obj']);

$Ojb = json_decode($_POST['obj'],true);

$clientData = $Ojb['ClientData'];
$healthData = $Ojb->HealthData;
$someData = $Ojb->someData;

print_r($clientData['firstName']);    
?>

No matter what I have tried, I am unable to see any of the information, I don't even get an error, just blank! Please can someone point me in the right direction.

Thank you :)

UPDATE

Here is the code that creates the object:

ClientObject = {

        ClientData : [
            {
                firstName : localStorage.getItem('cfn'),
                lastName : localStorage.getItem('cln'),
                email : localStorage.getItem('cem'),
                dob : localStorage.getItem('cdo'),
                age : localStorage.getItem('cag'),
                pierceType : localStorage.getItem('cpt'),
                street : localStorage.getItem('cst'),
                city : localStorage.getItem('cci'),
                county : localStorage.getItem('cco'),
                postcode : localStorage.getItem('cpc')
            }
        ],

        HealthData : health,

        PiercerData : [
        {
                firstName : localStorage.getItem('pfn'),
                lastName : localStorage.getItem('pln'),
                pierceDate : localStorage.getItem('pda'),
                jewelleryType : localStorage.getItem('pjt'),
                jewelleryDesign : localStorage.getItem('pjd'),
                jewellerySize : localStorage.getItem('pjs'),
                idChecked: localStorage.getItem('pid'),
                medicalChecked: localStorage.getItem('pmh'),
                notes: localStorage.getItem('poi')
        }
        ]

    };

And here is how its sent:

function senddata() {
    $.ajax({
        url: 'http://domain.com/app.php',
        type: 'POST',
        crossDomain: true,
        contentType: "application/json; charset=utf-8",
        dataType: 'jsonp',              
        data: 'obj='+JSON.stringify(ClientObject),

        success : function(res) {
            console.log(res);

        },
        error: function(err) {

        }
    });
}

what happens if you var_dump($Ojb)? If you get null, then json is invalid. If you still get blank page, then there is PHP error. You might not have the json module installed on the server. — quano
– quano, Commented Apr 2, 2014 at 13:31
@quano I do get Null, so that must mean the Json is not valid ? — Andrew Walker
– Andrew Walker, Commented Apr 2, 2014 at 13:40
try: $data = file_get_contents('php://input'); var_dump($data);. And post what you get (if anything). (I'm guessing the data is not sent in a way that php can populate the $_POST array with.) — Yoshi
– Yoshi, Commented Apr 2, 2014 at 13:42

Community · Accepted Answer · 2017-05-23 11:49:53Z

3

There are a few things that will cause problems:

why dataType: 'jsonp'? If you don't intend to utilize jsonp, don't instruct jQuery to do this. See the docs: https://api.jquery.com/jQuery.ajax/

"jsonp": Loads in a JSON block using JSONP. Adds an extra "?callback=?" to the end of your URL to specify the callback. Disables caching by appending a query string parameter, "_=[TIMESTAMP]", to the URL unless the cache option is set to true.
'obj='+JSON.stringify(ClientObject), this will guarantee invalid json.

For reference, have a look at this question: jQuery ajax, how to send JSON instead of QueryString on how to send json with jquery.

That said, try the following:

function senddata() {
  $.ajax({
    url: 'app.php',
    type: 'POST',
    crossDomain: true,
    contentType: 'application/json; charset=utf-8"',
    data: JSON.stringify(ClientObject),
    success : function(res) {
      console.log(res);
    },
    error: function(err) {
    }
  });
}

And in app.php use

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

to get the data. Use it like:

var_dump($input->ClientData[0]->firstName); // => string(6) "Master"

edited May 23, 2017 at 11:49

CommunityBot

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answered Apr 2, 2014 at 14:13

Yoshi

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2 Comments

Abhik Chakraborty Over a year ago

Yes this is exactly the issue is !!

Andrew Walker Over a year ago

@Yoshi Thank you for the above, and it did print what you put above. The reason I just jsonp is it get around the Access Control error i was getting. Which I again got with your code, but i changed the header and its working

Abhik Chakraborty · Accepted Answer · 2014-04-02 14:16:03Z

3

$Ojb = json_decode($_POST['obj'],true);

makes it array so u need to get them using array index instead of object

UPDATE1

With your update here how it could be done

$str ='{"ClientData":[{"firstName":"Master","lastName":"Tester","email":"[email protected]","‌dob":"1973-01-22","age":"51","pierceType":"Vici","street":"number of house","city":"here","county":"there","postcode":"everywhere"}],"HealthData":[["‌condtion : Prone to Fainting / Dizziness","condtion : Allergic Response to Plasters","condtion : Prone to Fainting / Dizziness"]],"PiercerData":[{"firstName":"Male","lastName":"checking","pierceDat‌e":"2013-02-25","jewelleryType":"Vici","jewelleryDesign":"Vidi","jewellerySize":"‌Vici","idChecked":null,"medicalChecked":null,"notes":"This is for more info"}]}' ;

$obj = json_decode($str,true);

echo $obj["ClientData"][0]["firstName"];

You can get other elements as above

UPDATE2

You are sending the data as JSONP and this will make the request as

?callback=jQuery17108448240196903967_1396448041102&{"ClientData"

Now you are also adding data: 'obj=' which is not correct.

You can simply send as json not jsonp

and on the php file you can do as

$Ojb = json_decode(file_get_contents('php://input'),true);

edited Apr 2, 2014 at 14:16

answered Apr 2, 2014 at 13:28

Abhik Chakraborty

44.9k6 gold badges56 silver badges63 bronze badges

6 Comments

Andrew Walker Over a year ago

so I would use something like: $Ojb = (array) json_decode($_POST['obj'],true); ?

Abhik Chakraborty Over a year ago

no no need to use (array) also it does not look like a valid json please post the json string that u received.

Andrew Walker Over a year ago

Okay @Yoshi But the real data is already above. But will post it without me cleaning it up. I only did that to make it easy to read.

Andrew Walker Over a year ago

That comes back with a blank screen :(

Abhik Chakraborty Over a year ago

did u use $Ojb = json_decode($_POST['obj'],true); ? since you are sending data as POST ?

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