PHP class variables as public giving an error

Question

I'm really new to php classes and now i have an error on my code. I have read some PHP documentations about classes and stuff, but something isn't just working correctly now.

Here's the code

public function change_salts($user_id) {
        global $mysqli_db;
        public $new_salt_one = "LOL"; //SaltyLogin::makesalt(60);
        private $new_salt_two = SaltyLogin::makesalt(60);
        mysqli_query($mysqli_db, "UPDATE `".SaltyLogin::sanitize(SALTY_MYSQLI_TB_SALTS)."` SET `".SaltyLogin::sanitize(SALTY_MYSQLI_TB_SALTS_SALT_ONE)."`='".$new_salt_one."' WHERE `".SaltyLogin::sanitize(SALTY_MYSQLI_TB_USER_ID)."` = '".SaltyLogin::sanitize($user_id)."'");
        mysqli_query($mysqli_db, "UPDATE `".SaltyLogin::sanitize(SALTY_MYSQLI_TB_SALTS)."` SET `".SaltyLogin::sanitize(SALTY_MYSQLI_TB_SALTS_SALT_TWO)."`='".$new_salt_two."' WHERE `".SaltyLogin::sanitize(SALTY_MYSQLI_TB_USER_ID)."` = '".SaltyLogin::sanitize($user_id)."'");
    }

Now this is the error I get all the time.

Parse error: syntax error, unexpected 'public' (T_PUBLIC) in C:\xampp\htdocs\GitHub\Salty-login\functions.php on line 60

For full source code, have a look at github and naturally branch wip-2.

Thanks in advance.

You shouldn't use encapsulation inside of the functions/methods, only in class declarations — deb0rian
– deb0rian, Commented Nov 20, 2013 at 18:30

John Conde · Accepted Answer · 2013-11-20 18:30:29Z

3

You can't declare class variables in side of a function. You have to either move them out of the method or make them local to the function only:

Option 1

public $new_salt_one = "LOL"; 
private $new_salt_two = '';
public function change_salts($user_id) {
    global $mysqli_db;
    $this->new_salt_two = SaltyLogin::makesalt(60);
    mysqli_query($mysqli_db, "UPDATE `".SaltyLogin::sanitize(SALTY_MYSQLI_TB_SALTS)."` SET `".SaltyLogin::sanitize(SALTY_MYSQLI_TB_SALTS_SALT_ONE)."`='".$this->new_salt_one."' WHERE `".SaltyLogin::sanitize(SALTY_MYSQLI_TB_USER_ID)."` = '".SaltyLogin::sanitize($user_id)."'");
    mysqli_query($mysqli_db, "UPDATE `".SaltyLogin::sanitize(SALTY_MYSQLI_TB_SALTS)."` SET `".SaltyLogin::sanitize(SALTY_MYSQLI_TB_SALTS_SALT_TWO)."`='".$this->new_salt_two."' WHERE `".SaltyLogin::sanitize(SALTY_MYSQLI_TB_USER_ID)."` = '".SaltyLogin::sanitize($user_id)."'");
}

Option 2

$new_salt_one = "LOL"; 
$new_salt_two = SaltyLogin::makesalt(60);

answered Nov 20, 2013 at 18:30

John Conde

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1 Comment

Marc B Over a year ago

plus class variables can't be initialized with expression results. constants only.

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