json_encode : values showing null

Question

<?php
require_once (realpath(dirname(__FILE__) . '/../includes/database.php'));

class User {
    public $email;
    public $password;

    public function find_email($email, $password) {
        global $database;
        $pswd = substr(md5($password), 0, 25);
        $results_array = self::find_by_sql("SELECT * FROM tbl_users where email_id='".$email."' AND password='".$pswd."'");
        return !empty($results_array)? array_shift($results_array) : false;
   }

    public static function find_by_sql($sql){
        global $database;
        $results = $database -> query($sql);
        $object_array = array();
        while($row = $database -> fetch_array($results)){
            $object_array[] = self::instantiate($row);
       }
        return $object_array;
    }

    public static function instantiate($row) {
         $event   =   new self;
         foreach($row as $attribute => $value) {
             if($event -> has_attribute($attribute)) {
                 $event -> $attribute = $value;
             }
         }
         return $event;
    }
    private function has_attribute($attribute) {
        $object_vars = get_object_vars($this);
        return array_key_exists($attribute, $object_vars);
    }
}

if (isset($_GET['email']) && isset($_GET['password'])) {
    $result = new User();
    $result->find_email($_GET['email'], $_GET['password']);
    echo json_encode($result);
}
?>

This is the login.php which is supposed to print out the json for the required user, but whenever I try to get the json, this is getting returned.

{"email":null,"password":null}

Any help would be appreciated. Thanks.

Hi, this calls for basic debugging first. At which point does what go wrong? Can you do test outputs to find out at which point the values become empty? — Pekka
– Pekka, Commented Jul 4, 2013 at 7:48
Please print_r($_GET), also, email_id, is that a number or a string? — Dave Chen
– Dave Chen, Commented Jul 4, 2013 at 7:49
first check $result data is it containing values,nothing to do with json_encode() it simply return json_object — Rajeev Ranjan
– Rajeev Ranjan, Commented Jul 4, 2013 at 7:50
@IgnacioVazquez-Abrams I've included a line on top require_once... that connects with the database. — Avanish Shrestha
– Avanish Shrestha, Commented Jul 4, 2013 at 7:51

MrCode · Accepted Answer · 2013-07-04 08:05:19Z

1

You don't do anything with the result of find_email. Your class doesn't update it's own properties when find_email is called. Instead, it returns a new instance of the class with the email and password properties set, so you need to capture the return value and encode that.

Change to:

$result = new User();
$user = $result->find_email($_GET['email'], $_GET['password']);
echo json_encode($user);

Side note: have a look at SOLID and Dependency Injection. DI would be preferred over having a global $Database.

edited Jul 4, 2013 at 8:05

answered Jul 4, 2013 at 7:53

MrCode

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4 Comments

T.J. Crowder Over a year ago

Better yet, refactor so you don't have to create the temporary User object just to call find_email.

MrCode Over a year ago

I agree there is a lot of room for refactoring here, the class is doing a bit too much. It seems the class is trying to be both a user entity and a user manager/factory.

Avanish Shrestha Over a year ago

This seems to fix the issue. Thanks a lot. Was working on some project, trying out php for the first time. This is a part of feeder btw, for an iOS app. Will look into those SOLID and Dependency Injection. Thanks again.

MrCode Over a year ago

No problem. SOLID and DI can get quite complex so don't let it overwhelm you. If you just understand the basics/reasoning behind them, it will help a lot.

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