对象数组返回相同的值

Question

我使用此函数返回用户及其角色和组。

我想返回具有正确索引的对象数组，这就是我创建 $index 计数器的原因。

这里的问题是，如果一个用户有超过 1 个组，我会得到重复的用户。

例如，如果我有一个用户有 3 个组，我将获得该用户 3 次。

如何避免重复用户？

我知道我需要以某种方式检查该用户是否已经存在，就像我检查了角色一样，但我不确定在哪里。

如果我将 users['$index'] 替换为 users['$id'] 则此代码将起作用，但这样我将无法获得具有正确索引的数组，而这正是我所需要的。

    $stmt = $mysqli->prepare("
    SELECT u.id
         , u.firstName
         , u.lastName
         , u.email
         , u.phoneNumber
         , u.address
         , u.birthDate
         , ur.roleName
         , cg.id
         , cg.name 
      FROM users as u 
      LEFT
      JOIN user_role as ur 
         ON u.id = ur.userId 
      LEFT 
      JOIN user_group as ug 
         on ug.userId = u.id 
      LEFT 
      JOIN control_group as cg 
         on cg.id = ug.groupId 
     WHERE u.id != ?");
    $stmt->bind_param("i", $_SESSION["id"]);
    $stmt->execute();
    $stmt->bind_result($id, $firstName, $lastName, $email, $phoneNumber, 
        $address, $birthDate, $roleName, $groupId, $groupName);
    $users = array();

    $index = 0;

    while ($stmt->fetch()) {
        if (empty($users[$index])) {
            $users[$index] = array(
                'id' => $id,
                'firstName' => $firstName,
                'lastName' => $lastName,
                'email' => $email,
                'phoneNumber' => $phoneNumber,
                'address' => $address,
                'birthDate' => $birthDate,
                'roles' => array(),
                'groups' => array()
            );
        }
        if ($roleName) {
            $found = false;
            foreach ($users[$index]['roles'] as $role) {
                if($role['roleName'] == $roleName){
                    $found = true;
                    break;
                }
            }
            if($found == false)
                $users[$index]['roles'][] = array(
                    'roleName' => $roleName
                );
         }

        if ($groupId) {
            $found = false;
            foreach ($users[$index]['groups'] as $group) {
                if($group['groupName'] == $groupName){
                    $found = true;
                    break;
                }
            }
            if($found == false)
                $users[$index]['groups'][] = array(
                    'groupName' => $groupName
                );
         }
         $index++;
    }

    $stmt->close();
    $mysqli->close();
    echo json_encode($users);

所以基本上我希望这样的对象

{  
      "id":2,
      "firstName":"Jon",
      "lastName":"Doe",
      "email":"jon.doe@email.com",
      "phoneNumber":"0621-123-444",
      "address":"Address 12a",
      "birthDate":"1976-01-01",
      "roles":['list of role objects'],
      "groups":['list of group objects']
   }

我也不确定我是否以正确的方式生成对象，我想知道是否有人可以告诉我什么是正确的方法以及如何正确生成像这样的对象。

Answer 1

在我看来，一旦您使用联接获得用户和数据库组，您就可以为同一个用户拥有多个索引，并且您无法改变这一事实。

所以，您有两个选择。 - 可以使用 user_id 并统一数组中的组和角色 - 或者您可以使用 user_id 代替索引，并为用户构建一个索引列表，就像您使用角色列表一样

但如果你真的需要知道你在数据库中获得的每个用户的索引，你只需要这样做。

我需要了解上下文或用户故事才能有更准确的想法

好的，我试试这样的：

$index = 0;
$temp_index = 0;
$user_index =[]
while ($stmt->fetch()) {

    if (empty($user_index[$id])){
        $user_index[$id] = $index;
    }else{
        $temp_index = $index;
        $index = $user_index[$id];
    }   

    if (empty($users[$index])) {        
        $users[$index] = array(
            'id' => $id,
            'firstName' => $firstName,
            'lastName' => $lastName,
            'email' => $email,
            'phoneNumber' => $phoneNumber,
            'address' => $address,
            'birthDate' => $birthDate,
            'roles' => array(),
            'groups' => array()
        );
    }
    if ($roleName) {
        $found = false;
        foreach ($users[$index]['roles'] as $role) {
            if($role['roleName'] == $roleName){
                $found = true;
                break;
            }
        }
        if($found == false)
            $users[$index]['roles'][] = array(
                'roleName' => $roleName
            );
     }

    if ($groupId) {
        $found = false;
        foreach ($users[$index]['groups'] as $group) {
            if($group['groupName'] == $groupName){
                $found = true;
                break;
            }
        }
        if($found == false)
            $users[$index]['groups'][] = array(
                'groupName' => $groupName
            );
     }
     if ($tempo_index != 0){
         $index = $temp_index;
         $temp_index = 0;
     }else{
        $index ++;
        $temp_index = 0;
}

Answer 2

如果可以选择执行多个查询，您可以先获取所有用户，然后为每个用户获取角色和组。

$user_stmt = $mysqli->prepare("SELECT id, firstName, lastName, email, phoneNumber, address, birthDate FROM users WHERE id != ?");
$user_stmt->bind_param("i", $_SESSION["id"]);
$user_stmt->execute();
$user_stmt->bind_result($id, $firstName, $lastName, $email, $phoneNumber, $address, $birthDate);

$role_stmt = $mysqli->prepare("SELECT roleName FROM user_role WHERE userId = ?");
$group_stmt = ...

$users = array();

while($user_stmt->fetch())
{
    $role_stmt->bind_param("i", $id);
    $role_stmt->execute();
    $role_stmt->bind_result($roleName);

    $roles = array();

    while($role_stmt->fetch())
    {
        $roles[] = array("roleName" => $roleName);
    }

    $groups = array();

    // Same as for roles

    $users[] = array(
            'id' => $id,
            'firstName' => $firstName,
            'lastName' => $lastName,
            'email' => $email,
            'phoneNumber' => $phoneNumber,
            'address' => $address,
            'birthDate' => $birthDate,
            'roles' => $roles,
            'groups' => $groups
        );
    }
}

$user_stmt->close();
$role_stmt->close();
$group_stmt->close();
$mysqli->close();
echo json_encode($users);