在 mysql 中獲取 root 用戶左右孩子的個數

Question

有什么方法可以通過其父 ID 獲取左右子節點的總數，直到子級別的“N”個。

這是我的用戶表，我在其中存儲帶有腿（左/右）position 的父子信息

在哪里：

1. referral_id : 是子用戶的父用戶的id
2. left_child_id ：是左腿加入用戶的子用戶的id
3. right_child_id ：是右腿加入用戶的子用戶的 id。
4. position_to_referral : 是他加入父用戶的 position 名稱（左/右腿）

歡迎任何幫助或任何建議。

我可以使用下面的 php 代碼獲取計數，但我想直接從 mysql 獲取計數

function countChildren($parentId, $Nlevel, $tempLevel = 0)
{
    if ($tempLevel < $Nlevel) {
        $tempLevel = $tempLevel + 1;
        $children = User::where('referral_id', $parentId)->get()->pluck('id');
        $count = count($children);
        foreach ($children as $userId) {
            $count += $this->countChildren($userId, $Nlevel, $tempLevel);
        }
        return $count;
    }
}

編輯 2

更新了代碼以在有或沒有 N 級別的情況下向左、向右

public function countChildren($parentId, $Nlevel = 0, $legPosition = 0, $tempLevel = 0)
{
    if ($Nlevel) {
        if ($tempLevel < $Nlevel) {
            $tempLevel = $tempLevel + 1;
            if ($tempLevel == 1 && $legPosition) {
                $children = User::where('referral_id', $parentId)
                    ->where('position_to_referral', $legPosition)
                    ->get()->pluck('id');
            } else {
                $children = User::where('referral_id', $parentId)->get()->pluck('id');
            }
            $count = count($children);
            foreach ($children as $userId) {
                $count += $this->countChildren($userId, $Nlevel, $legPosition, $tempLevel);
            }
            return $count;
        }
    } else {
        if ($legPosition) {
            $children = User::where('referral_id', $parentId)
                ->where('position_to_referral', $legPosition)
                ->get()->pluck('id');
        } else {
            $children = User::where('referral_id', $parentId)->get()->pluck('id');
        }
        $count = count($children);
        foreach ($children as $userId) {
            $count += $this->countChildren($userId);
        }
        return $count;
    }
}

調用上面的 function 得到計數：

$total = $profileService->countChildren(1); // get total children count of user_id = 1
$totalWith3Level = $profileService->countChildren(1, 3); // get total children count till 3 level
$totalLeft = $profileService->countChildren(1, '', 'left'); // get total children count of left leg
$totalRight = $profileService->countChildren(1, '', 'right'); // get total children count of right leg
$totalLeftWith3Level = $profileService->countChildren(1, 3, 'left'); // get total children count of left leg till 3 level
$totalRightWith3Level = $profileService->countChildren(1, 3, 'right'); // get total children count of right leg till 3 level

Answer 1

嵌套集 model 不是更有意義嗎？

            1 A 14
              |
      +-------+--------+
      |                |
    2 B 7            8 C 13
      |                |
  +---+---+       +----+----+
  |       |       |         |
3 D 4   5 E 6   9 F 10   11 G 12


user lft rgt
A      1  14
B      2   7
C      8  13
D      3   4
E      5   6
F      9  10
G     11  12