优化 3 个 Eloquent 请求（请求持续时间 1.5 秒...）

Question

我想优化这 3 个查询，使我能够根据日期条件（当前周、上周、自开始以来）显示在我们网站上发布最多照片的用户。

用调试栏测得查询时间为1.5秒，真的很长！ 你知道怎么优化吗？

public function show()
{
    $currentWeek = User::whereHas('pictures')
        ->whereHas('pictures', fn ($q) => $q->whereBetween('created_at', [Carbon::now()->startOfWeek(), Carbon::now()->endOfWeek()]))
        ->withCount(['pictures' => fn ($q) => $q->whereBetween('created_at', [Carbon::now()->startOfWeek(), Carbon::now()->endOfWeek()])])
        ->orderBy('pictures_count', 'DESC')
        ->limit(10)
        ->get();

    $lastWeek = User::whereHas('pictures')
        ->whereHas('pictures', fn ($q) => $q->whereBetween('created_at', [Carbon::now()->startOfWeek()->subWeek(), Carbon::now()->endOfWeek()->subWeek()]))
        ->withCount(['pictures' => fn ($q) => $q->whereBetween('created_at', [Carbon::now()->startOfWeek()->subWeek(), Carbon::now()->endOfWeek()->subWeek()])])
        ->orderBy('pictures_count', 'DESC')
        ->limit(10)
        ->get();

    $overall = User::whereHas('pictures')
        ->whereHas('pictures')
        ->withCount('pictures')
        ->orderBy('pictures_count', 'DESC')
        ->limit(10)
        ->get();

    return view('users.leaderboard', [
        'currentWeek' => $currentWeek,
        'lastWeek' => $lastWeek,
        'overall' => $overall,
    ]);
}

Answer 1

首先，你已经在图片关系上调用了两次whereHas，所以你可以摆脱不合格的调用。

$currentWeek = User::whereHas('pictures', fn ($q) => $q->whereBetween('created_at', [now()->startOfWeek(), now()->endOfWeek()]))
    ->withCount(['pictures' => fn ($q) => $q->whereBetween('created_at', [now()->startOfWeek(), now()->endOfWeek()])])
    ->orderBy('pictures_count', 'DESC')
    ->limit(10)
    ->get();

这减少了 SQL 查询：

select `users`.*, (
    select count(*) from `pictures` where `users`.`id` = `pictures`.`user_id` and `created_at` between ? and ? and `pictures`.`deleted_at` is null
) as `pictures_count`
from `users`
where exists (select * from `pictures` where `users`.`id` = `pictures`.`user_id` and `pictures`.`deleted_at` is null)
    and exists (select * from `pictures` where `users`.`id` = `pictures`.`user_id` and `created_at` between ? and ? and `pictures`.`deleted_at` is null)
    and `users`.`deleted_at` is null
    order by `pictures_count` desc
    limit 10

到这里：

select `users`.*, (
    select count(*) from `pictures` where `users`.`id` = `pictures`.`user_id` and `created_at` between ? and ? and `pictures`.`deleted_at` is null
) as `pictures_count`
from `users`
where exists (select * from `pictures` where `users`.`id` = `pictures`.`user_id` and `created_at` between ? and ? and `pictures`.`deleted_at` is null)
    -- no second where exists clause
    and `users`.`deleted_at` is null
    order by `pictures_count` desc
    limit 10

现在，where 子句中只有一个条件。它选择在指定日期范围内拥有图片的用户。看起来更好，对吧？

---

但是，您已经在使用带有闭包的 withCount，因此您只计算日期范围内的图片。如果条件不匹配会怎样？它返回零。由于无论如何您都是按计数进行反向排序，因此对 whereHas 的另一个调用也可以进行。

$currentWeek = User::withCount(['pictures' => fn ($q) => $q->whereBetween('created_at', [now()->startOfWeek(), now()->endOfWeek()])])
    ->orderBy('pictures_count', 'DESC')
    ->limit(10)
    ->get();

现在您的 SQL 如下所示：

select `users`.*, (
    select count(*) from `pictures` where `users`.`id` = `pictures`.`user_id` and `created_at` between ? and ? and `pictures`.`deleted_at` is null
) as `pictures_count`
    from `users`
    where `users`.`deleted_at` is null
    -- no where exists clauses at all any more
    order by `pictures_count` desc
    limit 10

而且它应该运行得更快。这确实会稍微改变您的数据；结果集合总是有 10 个项目，即使其中一些是零。如果您不想在排行榜中出现零，只需将它们从集合中过滤掉即可。