Send jQuery object to PHP through Ajax

Question

I'm very aware this is a possible duplicate, but none of the other questions/answers here on Stackoverflow solve my issue, and I've seen dozens!

Here's what I'm trying to do: Send the object in jQuery to PHP through Ajax, if possible as an array (not mandatory, but preferable).

My jQuery code:

 var category = {
 id: $(this).data('id'),
 name: $(this).data('name'),
 brief_description: $(this).data('briefdesc'),
 description: $(this).data('desc')
 };
 $.ajax({url: '/ajax.php',
 data: {action: 'removeCategory', category_info: JSON.stringify(category)},
 type: 'post',
 success: function (result) {
 console.log(result);
 },
 error: function () {
 console.log("Error"); 
 }, dataType: "json"
});

The variable category is working fine, every index has its value

Now my ajax.php code

$category = json_decode($_POST['category_info']);
//category['name'] should exist and have the value sent from ajax
echo "We did it?";

The problem is that the error function is called.

• jeroen: You forgot to mention what the problem is.
• xickoh: @jeroen it is not working, it's running the error function in the ajax call
• Romi Halasz: If the error function is called, can you check in the browser console to see the request status ?
• Harish Kommuri: You don't need to stringify while sending request.
• xickoh: @Harish removing the stringify and the «dataType: json» does solve my issue. How can I mark it as correct? :) Thanks a lot

Answer 1

You no need to stringify.

$.ajax({
 url: '/ajax.php',
 data: {
 action: 'removeCategory', 
 category_info:category
 },
 type: 'post',

On PHP side you will get it in $_POST['category_info']. No any need to decode

• xickoh: I actually had tried this before, but the browser cache might have interfered with the javascript code and I wasn't successful. This works though, in fact, not only I do not need to stringify, but if I try to stringify it will cause an error. Thank you!

Answer 2

 json_decode($_POST['category_info'], true);

to decode it as array, otherwise it will be object

http://php.net/manual/en/function.json-decode.php

• xickoh: Thank you, I didn't know that. It's not working yet though