How to parse json data received from a jQuery ajax call in php

Question

I'm calling a php script (getNum.php) via ajax after creating an object and using jquery.json to turn it to json. Now i want to handle the object on the php side. print_r($_POST['data']) doesn't work nor anything else i've tried.

This is my code:

// create object
    var bing= new Object();
    bing.id = 99;
    bing.nameList = getBingList();

    //create pdf
    $.ajax({
    type: "POST",
    contentType: "application/json; charset=utf-8",
    url: "getNum.php",
    dataType: "html",
    data: $.toJSON(bing),
    success: function(data){
        alert(data);
        window.location = "generateBing.php?num="+data
    }

    });

what file do you have the print_r statement in? your page that contains the javascript or the page that is called by the javascript (generateBing.php)? — jedwards
– jedwards, Commented Apr 29, 2012 at 0:40
generateBing.php isn't related. getNum.php contains the print_r code — Tom
– Tom, Commented Apr 29, 2012 at 0:41
How about print_r($_POST)? I guess it will contain two keys: id and nameList. — Dmytro Shevchenko
– Dmytro Shevchenko, Commented Apr 29, 2012 at 0:45

adeneo · Accepted Answer · 2012-04-29 00:47:54Z

2

If you're using print_r($_POST['data']) to show the content, you'll need to send it as "data" as well.

$.ajax({
    type: "POST",
    url: "getNum.php",
    data: {data: $.toJSON(bing)},
    success: function(data){
        alert(data);
        window.location = "generateBing.php?num="+data
    }
});

Otherwise you have to do print_r($_POST)

answered Apr 29, 2012 at 0:47

adeneo

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3 Comments

Dmytro Shevchenko Over a year ago

I doubt jQuery will handle posting hierarchical data.

loganfsmyth Over a year ago

You'll want to remove contentType, since this way stops it from being application/json.

adeneo Over a year ago

@loganfsmyth - removed both, as they are normally not needed. If it's a valid json object, jQuery will figure it out.

loganfsmyth · Accepted Answer · 2012-04-29 00:57:55Z

0

Since you are posting a JSON object directly, there is no argument name for $_POST. You'll need to read the raw contents of the POST request. Try this:

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

edited Apr 29, 2012 at 0:57

answered Apr 29, 2012 at 0:43

loganfsmyth

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1 Comment

loganfsmyth Over a year ago

There, should work now. As you mentioned, the contentType effects all of this. Since you had JSON content-type, PHP cannot automatically process it. It is up to you to read the request data and process it into PHP vars.

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