Question

我在数据库中有一些用户，我可以编辑他们的名字和密码但是当我尝试编辑用户名时，查询失败了。

这是我的代码

    $user->username = $db->mysql_prep($_POST["username"]);
    $user->hashed_password = ($_POST["password"]);
    $user->firstname = $db->mysql_prep($_POST["firstname"]);
    $user->lastname = $db->mysql_prep($_POST["lastname"]);

    $user_query = $user->find_user_by_username($user->username);
    $user->id = $user_query["id"];

    $result = $user->change_user_by_id($user);
    //->id,$user->username,$user->hashed_password,$user->firstname,$user->lastname

    unset($user);

我的change_user_by_id方法：

public function change_user_by_id($user){
        global $db; 
        global $session;

        $query  = "UPDATE users SET ";
        $query .= "username = '{$user->username}', ";
        $query .= "first_name = '{$user->firstname}', ";
        $query .= "last_name = '{$user->lastname}' ";
        $query .= "WHERE id = {$user->id} ";
        $query .= "LIMIT 1";

        $result = mysqli_query($db->connection, $query);

        $db->confirm_query($result);

        if ($result && mysqli_affected_rows($db->connection) == 1) {
              // Success
            $session->message("User updated.");
            redirect_to("list.php");
        } else {
              // Failure
            $session->message("User update failed.");
        }
    }

我的find_user_by_username方法：

public static function find_user_by_username($username="default"){
        global $db; 

        $query  = "SELECT * ";
        $query .= "FROM users ";
        $query .= "WHERE username = '{$username}' ";
        $query .= "LIMIT 1";

        $user_set = mysqli_query($db->connection, $query);
        $db->confirm_query($user_set);

        if($user = mysqli_fetch_assoc($user_set)) {
            return $user;
        } else {
            return null;
        }   
    }

编辑：

我得到的唯一错误来自confirm_query函数我收到消息“数据库查询失败”这是函数：

public function confirm_query($result_set) {

    if (!$result_set) { 
      die("Database query failed."); 
    } 

}

编辑2：

添加了错误消息，这就是我得到的：

您的SQL语法有错误;检查手册对应于您的MySQL服务器版本，以便使用正确的语法在第1行'LIMIT 1'附近

Answer 1

我认为这是因为使用用户名来获取ID，这意味着更改用户名意味着无法检索到ID。所以我改变了我的代码并在我的表单中为id添加了一个隐藏字段。

旧代码：

$user->username = $db->mysql_prep($_POST["username"]);
$user->hashed_password = ($_POST["password"]);
$user->firstname = $db->mysql_prep($_POST["firstname"]);
$user->lastname = $db->mysql_prep($_POST["lastname"]);

$user_query = $user->find_user_by_username($user->username);
$user->id = $user_query["id"];

$result = $user->change_user_by_id($user);

unset($user);

替换为新代码：

$user->username = $db->mysql_prep($_POST["username"]);
$user->hashed_password = ($_POST["password"]);
$user->first_name = $db->mysql_prep($_POST["first_name"]);
$user->last_name = $db->mysql_prep($_POST["last_name"]);
$user->id = $db->mysql_prep($_POST["id"]);

$result = $user->change_user_by_id($user);

unset($user);

尝试更改用户名时，MySQL查询失败

1 个答案: