dongyukui8330 2017-05-22 01:43
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注意:在第12行的C:\ xampp \ htdocs \ auth egister.php中只能通过引用传递变量成功[复制]

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<?php 

require 'database.php';


if (isset($_POST['email']) && isset($_POST['password'])):



    // Enter the new user in the database

    $sql = "INSERT INTO users (email, password) VALUES (:email, :password)"; 

    $stmt = $conn->prepare($sql);



    $stmt->bindParam(':email', $_POST['email']);

    $stmt->bindParam(':password', password_hash($_POST['password'], PASSWORD_BCRYPT));

    if( $stmt->execute() ):

        die('Success');

    else:

        die('Fail');

    endif;


endif;


 ?>

It successfully add the user and password in the database but it gives the error below

Only variables should be passed by reference in C:\xampp\htdocs\authegister.php on line 12

</div>
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1条回答 默认 最新

  • doucheng3811 2017-05-22 01:50
    关注

    The format for bindParam is as follows:

    public bool PDOStatement::bindParam ( mixed $parameter , mixed &$variable [, int $data_type = PDO::PARAM_STR [, int $length [, mixed $driver_options ]]] )
    

    Notice this part mixed &$variable is passed by reference. To fix this, just change:

    $stmt->bindParam(':password', password_hash($_POST['password'], PASSWORD_BCRYPT));
    

    To:

    $password = password_hash($_POST['password'], PASSWORD_BCRYPT);
    $stmt->bindParam(':password', $password);
    

    Since you are then passing a variable itself by reference, not just the string returned by password_hash().

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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