Laravel 拉雷维尔雄辩-到底在哪里

在Laravel4.2中，我试图实现一个返回所有用户的查询，这些用户拥有所有特定的活动。到目前为止，我有一个查询，该查询返回具有多个活动之一的所有用户：

//$selectedActivities being an array
        $userByActivities = User::with('activities')
                ->whereHas('activities', function($query) use($selectedActivities){
                    $query->whereIn('id', $selectedActivities);
                })->get();

更清楚地说：给定活动a、b、c。我正在寻找所有有活动a、b和c的用户。我的查询返回具有活动a、b或c的所有用户

谢谢你的帮助

编辑：

由提供的解决方案将导致以下查询：

select * from `users` where 
        (select count(*) from `activities` inner join `activity_user` on `activities`.`id` = `activity_user`.`activity_id` where `activity_user`.`user_id` = `users`.`id` and `id` = '7') >= 1 
        and (select count(*) from `activities` inner join `activity_user` on `activities`.`id` = `activity_user`.`activity_id` where `activity_user`.`user_id` = `users`.`id` and `id` = '3') >= 1 
        and (select count(*) from `activities` inner join `activity_user` on `activities`.`id` = `activity_user`.`activity_id` where `activity_user`.`user_id` = `users`.`id` and `id` = '1') >= 1 
        and (select count(*) from `activities` inner join `activity_user` on `activities`.`id` = `activity_user`.`activity_id` where `activity_user`.`user_id` = `users`.`id` and `id` = '2') >= 1

select * from `users` where (select count(distinct id) from `activities` 
    inner join `activity_user` on `activities`.`id` = `activity_user`.`activity_id` 
    where `activity_user`.`user_id` = `users`.`id` and `id` in ('7', '3', '1', '2')) = 4

鉴于以下机构提供的解决方案：

对于类似的请求，将生成以下查询：

select * from `users` where 
        (select count(*) from `activities` inner join `activity_user` on `activities`.`id` = `activity_user`.`activity_id` where `activity_user`.`user_id` = `users`.`id` and `id` = '7') >= 1 
        and (select count(*) from `activities` inner join `activity_user` on `activities`.`id` = `activity_user`.`activity_id` where `activity_user`.`user_id` = `users`.`id` and `id` = '3') >= 1 
        and (select count(*) from `activities` inner join `activity_user` on `activities`.`id` = `activity_user`.`activity_id` where `activity_user`.`user_id` = `users`.`id` and `id` = '1') >= 1 
        and (select count(*) from `activities` inner join `activity_user` on `activities`.`id` = `activity_user`.`activity_id` where `activity_user`.`user_id` = `users`.`id` and `id` = '2') >= 1

select * from `users` where (select count(distinct id) from `activities` 
    inner join `activity_user` on `activities`.`id` = `activity_user`.`activity_id` 
    where `activity_user`.`user_id` = `users`.`id` and `id` in ('7', '3', '1', '2')) = 4

---

为此，您必须添加多个

，其中有

：

$query = User::with('activities');
foreach($selectedActivities as $activityId){
    $query->whereHas('activities', function($q) use ($activityId){
        $q->where('id', $activityId);
    });
}
$userByActivities = $query->get();

---

如果您遇到基数冲突：1241个操作数应该包含2列，问题是嵌套的selectCount会添加到正常的select count（*）中，而不是覆盖现有的select，因此更改为

$query->distinct（）->其中（'id'，$selectedActivities）

为我做了这个把戏，或者改为

$query->select（DB:：raw（count（distinct id））

现在你这么说似乎合乎逻辑。非常感谢您的帮助。对于数量未知的相关项目来说，这太过分了。您应该将计数调整为

count（distinct id）

，并使

whereHas

查找数组中项目的确切数量。@JarekTkaczyk如何做？

$query->distinct（）->其中（'id'，$selectedActivities）为我工作，感谢Jarek Tkaczyk提出的解决方案比公认的解决方案优化得多