How to return encoded json with json object in json codeigniter rest server

Question

I am using the codeigniter rest by phil sturgeon.

I want to return a JSON object that contains another JSON object within.

My code looks like this

function volunteer_get()
    {
        if(!$this->get('id'))
        {
            $this->response(NULL, 400);
        }

        $this->load->model('user/users_model');

        $user = $this->users_model->get( $this->get('id') );
        $address = $this->address_model->getAddress( $this->get('id') );

        $user->address = $address;

        $userJson = json_encode($user);
        var_dump($userJson);

        /*if($user && $user->auth_level == 1)
        {
            $this->response($userJson, 200); // 200 being the HTTP response code
        }
        else
        {
            $this->response(NULL, 404);
        }*/
    }

It is not showing any result... If i do this without adding the php object in the other php object, it shows me the json!

D:\wamp\www\codeigniter\application\controllers\api\Users.php:37:string '{"user_id":"1","username":"abc","email":"abc","auth_level":"1","banned":null,"passwd":"abcdefg","passwd_recovery_code":"abcdefg","passwd_recovery_date":"2017-06-12 18:50:31","passwd_modified_at":"2016-11-18 21:20:30","last_login":"2017-08-30 15:10:36","created_at":"2016-09-02 12:01:46","modified_at":"2017-08-30 15:22:45","first_name":"aze","family_name":"'... (length=1354)

Is $user an array or an object? Try echo gettype($user); and let me know what it shows. — MonkeyZeus
– MonkeyZeus, Commented Oct 5, 2017 at 16:35

LSerni · Accepted Answer · 2017-10-05 16:26:59Z

2

In general, you need to check whether you got a JSON object (usually a PHP dictionary or object) or a JSON representation (a string).

You can not add a string to another string. And if you add a string to a dictionary or object, it won't be properly encoded as a JSON sub-object because it is, well, a string.

So if you have a representation, you have to decode it first:

// Here, $dataOne is a string, and $dataTwo is too.
// we decode to array rather than to object (see manual for json_encode)

$dataOneJSON = json_decode($dataOne, true);

$dataTwoJSON = json_decode($dataTwo, true);

$dataOneJSON['two'] = $dataTwoJSON;

$result = json_encode($dataOneJSON);

$this->response($result, 200);

answered Oct 5, 2017 at 16:26

LSerni

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