对所有下属编写递归函数查询

Question

我在laravel 9

中有这个功能

private function userDirectSubordinates($user_id, $company_id) 
{
    return User::select(['users.id', 'users.name as label', 'avatar', 'departments.name as department'])
        ->join('department_user', 'users.id', '=', 'department_user.user_id')
        ->join('departments', 'departments.id', '=', 'department_user.department_id')
        ->where('departments.company_id', $company_id)
        ->whereIn('users.id', function($query) use($user_id) {
            $query->select('user_id')
                ->from('company_user')
                ->where('superior_id', $user_id);
        })
        ->get();
}

此功能return id 为$user_id 的用户的直接下属来自所选公司$company_id。示例：

[
    ["id" => 880, 'label' => 'User 880', 'avatar' => null, 'department' => 'IT'],
    ["id" => 41, 'label' => 'User 41', 'avatar' => null, 'department' => 'Finance'],
    ...
]

我如何制作一个递归函数来从每个用户从 directSubordinates 获取...所有下属。最终数组应如下所示：

[
    ["id" => 880, 'label' => 'User 880', 'avatar' => null, 'department' => 'IT',
        'children' => [
            ["id" => 32, 'label' => 'User 32', 'avatar' => null, 'department' => 'IT', 'children' => []],
            ["id" => 56, 'label' => 'User 56', 'avatar' => null, 'department' => 'IT',
                'children' => [
                    ["id" => 21, 'label' => 'User 21', 'avatar' => null, 'department' => 'Maintenance', 'children' => []],
                    ["id" => 687, 'label' => 'User 687', 'avatar' => null, 'department' => 'Development',
                        'children' => [
                            ["id" => 334, 'label' => 'User 334', 'avatar' => null, 'department' => 'Development', 'children' => []],
                            ["id" => 335, 'label' => 'User 335', 'avatar' => null, 'department' => 'Development', 'children' => []],
                        ]
                    ]
                ],
            ]
        ],
    ],
   ...  
];

Answer 1

大概你正在使用 MySQL 8 和 Laravel 9 这意味着你可以这样做（递归语法）：

SQL:

select users.id, users.username as username, departments.name as dept, company_user.superior_id as superior_id
FROM users
INNER JOIN department_user on users.id = department_user.user_id
INNER JOIN departments on departments.id = department_user.department_id
INNER JOIN company_user on company_user.user_id = users.id
WHERE
    departments.company_id = 1
    AND
    users.id IN (
      with recursive cte (user_id) as (
        select     user_id
        from       company_user
        where      superior_id = 1
        union all
        select     parent.user_id
        from       company_user parent
        inner join cte
        on parent.superior_id = cte.user_id
      )
      select * from cte
    )

就个人而言，我不会为此使用 Laravel 查询构建器（只需使用原始数据）。

你得到的结果是这样的：

请注意父（目标用户）不存在，您应该先获取它。

现在您可以遍历数据并将子项附加到 PHP 中的父项并创建一棵树：

$user = findUserById();
$subordinates = findAllSubordinates();

function buildTree(array &$subordinates, $parentId) {

    $branch = array();

    foreach ($subordinates as &$subordinate) {

        if ($subordinate['superior_id'] == $parentId) {
            $children = buildTree($subordinates, $subordinate['id']);
            if ($children) {
                $subordinate['children'] = $children;
            }
            $branch[] = $subordinate;
            unset($subordinate);
        }
    }
    return $branch;
}

$user['children'] = buildTree($subordinates, $user['id']);

对所有下属编写递归函数查询

write recursive function query with all subordinates

php

laravel