Unable to get data returned from PHP via AJAX

Question

I'm trying to get data returned from a function in a PHP file via AJAX (using Chrome)

The JS:

    $.ajax({
            url: 'http://www.site.co.uk/api/file.php',
            data: data, 
            dataType: 'json', 
            type: 'POST', 
            contentType: 'application/json',
            async: false,
            success: function (jsonData){
                console.log("SUCCESS");
                var responseText = jQuery.parseJSON(jsonData.responseText);
                console.log(responseText);
            },
            error: function (jsonData){
                console.log("ERROR");
                var responseText = jQuery.parseJSON(jsonData.responseText);
                console.log(responseText);
            }
        });

Relevant part of the PHP:

public function post_method()
{
    $data['error_message'] = "Error message text";
    return json_encode($data);
}

In the console I'm getting:

POST http://www.site.co.uk/api/file.php 500 (Internal Server Error)
ERROR
Object {status: "{", message: "{"}

In the PHP, if I replace

return json_encode($data);

with

return $data;

the console shows

SUCCESS
Uncaught SyntaxError: Unexpected token u jquery.js:3

Not sure why I can't get 'error_message', and not sure why using 'json_encode' appears to produce a 500 error.

@Johannes Reuter - It's slightly complicated but I'm calling post_method() like this:

public function route_method()
{
   switch($this->resource['request_method'])
    {
        case 'POST':
            return $this->post_method();
            break;
        default:
            return FALSE;
            break;                
    }

}

This is part of some api framework code that I didn't write. '$this->resource' is essentially being passed into the class constructor. The same method is used successfully in other PHP files.

@Rory McCrossan - how should I define it? I thought it was OK just to create it as an array.

Can you show where you define $data in your PHP code and what it contains. The second error is because you've told jQuery to deserialise a JSON response, yet the response was not JSON. — Rory McCrossan
– Rory McCrossan, Commented Feb 27, 2015 at 11:02
Use die() instead of return - die(json_encode($data)); or echo and die combo. — marian0
– marian0, Commented Feb 27, 2015 at 11:06
Check if you are echo your data in your php file. You should echo it and not return it. — lmarcelocc
– lmarcelocc, Commented Feb 27, 2015 at 11:09
What data are you passing using data is it form.serialize() or else — Narendrasingh Sisodia
– Narendrasingh Sisodia, Commented Feb 27, 2015 at 11:25

Mohamad Shiralizadeh · Accepted Answer · 2015-02-27 14:06:14Z

1

If you inform dataType as json in the request, just dont need use jQuery.parseJSON function because it's json.

edited Feb 27, 2015 at 14:06

Mohamad Shiralizadeh

8,8348 gold badges65 silver badges99 bronze badges

answered Feb 27, 2015 at 12:39

Juan de Parras

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Comments

Alpesh · Accepted Answer · 2015-02-27 12:34:22Z

0

you will try.

echo json_encode($data);exit;

answered Feb 27, 2015 at 12:34

Alpesh

11 bronze badge

Comments

Suresh Vaishnav · Accepted Answer · 2015-02-27 12:48:49Z

0

use echo json_encode($data);die; instead of return json_encode($data);

answered Feb 27, 2015 at 12:48

Suresh Vaishnav

1013 bronze badges

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