Ajax read PHP generated json

Question

So I want to send a json to an ajax call but I don't know how to read the json when is sent. And also my json looks strange because of the backslashes...

This is my ajax:

    function find(){
        var type = $('#object_type').val();
            $.ajax({
                type : 'POST',
                url : 'get_user.php',
                data : {
                    'type' : type
                },
                dataType : 'json',
                error : function(response){
                    alert('SOMETHING WENT WRONG');
                },
                success : function(response){

This is what I get as a response:
"[{\"name\":\"Test\",\"link\":\"test.php\"},{\"name\":\"Test2\",\"link\":\"test2
.php\"}]"
                }
            });     
    }

This is my PHP function:

    $type = $_POST['type'];
    $user_array;
    $user = mysqli_query($conn, "SELECT name,link FROM user WHERE `type` LIKE '%".$type."%'") or die();

            while ($row = mysqli_fetch_array($user, MYSQLI_ASSOC)) {

                $row_array['name'] = $row['name'];
                $row_array['link'] = $row['link'];
                array_push($user_array, $row_array);
            }
        mysqli_close($conn);
        $result = json_encode($user_array, 128);
        echo json_encode($result);

The response is probably valid. But is rendered this way. If you look at your dev tool (F12) in your browser you can see the actual respone text in the network tab. — Mouser, Aug 10 '15 at 13:24
Mah I don't know if it is really, I read that this backslashes can cause problems..and anyway I don't know how to access the data inside json.. — DJack, Aug 10 '15 at 13:26
Try doing this in your success function to see if it works: `alert(response[0].name)` — Mouser, Aug 10 '15 at 13:26
I use FireBug and the response in HTML is the as the response I wrote in my question.. — DJack, Aug 10 '15 at 13:28
As an aside from the main issue at hand, you should probably add the content type header `header('Content-Type: application/json');`. While not required, it is better to explicitly state this. See http://stackoverflow.com/questions/4064444/ for more details... — War10ck, Aug 10 '15 at 13:36

Mouser · Answer 1 · 2015-08-10T13:46:58.873

As provided in techierishi's answer. Do not encode your response twice. Just do

 echo $result;

The JSON should be valid and parsed, why?

json_encode is used on a valid PHP array returned from a MySQLi query.
dataType is used, explicitly forcing jQuery to JSON.parse the response and no errors where raised during that process.

The JSON that is returned has the following structure

ROOT
   ARRAY
       ARRAY OBJECT
             OBJECT PROPERTY

To address something like that:

root[array index].property

Will give you the array's value, which is an object. That object has multiple properties.

[] means array
{} means object

So [{name:"value 1"},{name:"value 2"}] is actually

ARRAY
   [0].name = value 1
   [1].name = value 2

So retrieving information like name from your JSON will be:

response[0].name; // = test

If I do var name = response[0].name; and then alert(name), I get Undefined — DJack, Aug 10 '15 at 13:45
See Techierishi's answer and the approriate edit I've made to my answer. — Mouser, Aug 10 '15 at 13:48

score 4 · Accepted Answer · answered Aug 10 '15 at 13:44

First change that you should make : You don't need to encode two times

$result = json_encode($user_array, 128);
echo json_encode($result);

should be

 $result = json_encode($user_array, 128);
 echo $result;

Now the response should look like

[{"name":"Test","link":"test.php"},{"name":"Test2","link":"test2
.php"}]

which is a valid Javascript Array and can be accessed using following code :

var len = response.length;
for(i=0;i<len;i++){

  console.log(response[i].name);
  console.log(response[i].link);

}

Ajax read PHP generated json

2 Answers2