php - MySQLI 更新不起作用且未返回任何错误

Question

我正在尝试纠正脚本以更新数据库中的用户，这是我的代码:

我从表单中获取用户数据:

$user = new User();

if (isset($_POST['submit'])) {
  // Process the form

  // validations
  // $required_fields = array("username", "password");
  // validate_presences($required_fields);

  // $fields_with_max_lengths = array("username" => 30);
  // validate_max_lengths($fields_with_max_lengths);

  $errors = ""; //temp

  if (empty($errors)) {

    $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"]);

    print_r($user);

    $user_by_id = $user->find_user_by_id($user->id);

    print_r($user_by_id);

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

    //->id,$user->username,$user->hashed_password,$user->firstname,$user->lastname

    unset($user);
}

这是我的 find_user_by_id 方法:

public static function find_user_by_id($id=0){

global $db; 

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

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

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

最后是我的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->first_name}', ";
    $query .= "last_name = '{$user->last_name}' ";
    $query .= "WHERE id = {$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
        $session->message("User update failed.");
        redirect_to("change.php?id={$user->id}");

    }
}   

当我尝试更新用户时，我被重定向到change.php，并显示错误消息“用户更新失败”。我一直在试图理清逻辑，但我似乎找不到哪里出了问题。后来添加了 print_r 以查看对象是否具有正确的值，结果确实如此。

编辑1:

我更改了一些代码来查看 mysqli_affected_rows 返回什么，它返回 0。

Answer 1

几个小时后我终于想通了，我只需要把线拿出来:

$user_by_id = $user->find_user_by_id($user->id);

并更改行:

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

至:

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

该行是 the way I was doing it before 的剩余部分这导致我将原始数据而不是新数据发送到更改方法。