我换了（$ newPassword到$ new），仍然是数据库中的空！ :(，[重复]

Question

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• “Notice: Undefined variable”, “Notice: Undefined index”, and “Notice: Undefined offset” using PHP 28 answers

`password is updating with empty value! old password is replaced with empty value!!

$email = $_POST['email'];

$password = $_POST['password']; //old password

$new = $POST['new'];  //new password

//table name is users

    $stmt = $db->prepare("SELECT * FROM users WHERE email=:email_id");


//checking email and password


    $stmt->execute(array(":email_id"=>$email));

    $userRow=$stmt->fetch(PDO::FETCH_ASSOC);

    if(($userRow['email'] === $email)&&($userRow['password'] === $password))

    {


        $doneQuery = $db->prepare("

        UPDATE users 

        SET password = :new


        WHERE email = :emailid" );


        $doneQuery->execute([":newPassword"=>$new,

                        ":emailid"=>$email]);

}

Answer 1

You are updating your database using $newpassword but you declare it at the top as $new, thus meaning if is a empty variable and entering it blank You also missed the under score from the post So change

$new = $POST['new'];

To

$newPassword = $_POST['new'];


$email = $_POST['email']; $password = $_POST['password']; //old password         $newPassword= $_POST['new']; //new password //table name is users $stmt = $db->prepare("SELECT * FROM users WHERE email=:email_id"); //checking email and password $stmt->execute(array(":email_id"=>$email)); $userRow=$stmt->fetch(PDO::FETCH_ASSOC); if(($userRow['email'] === $email)&&($userRow['password'] === $password)) { $doneQuery = $db->prepare(" UPDATE users SET password = :newPassword WHERE email = :emailid" ); $doneQuery->execute([":newPassword"=>$new, ":emailid"=>$email]); }