Laravel 在单个子句中使用多个 where 和 sum

Question

在我的数据库中我有 instagram_actions_histories table 其中我有 action_type 列，在列中我有不同的数据，例如 1 或 2 或 3

我正在尝试获取此 table 关系数据并对存储在列中的此值求和，例如

$userAddedPagesList = auth()->user()->instagramPages()->with([
        'history' => function ($query)  {
            $query->select(['action_type as count'])->whereActionType(1)->sum('action_type');
        }
    ]
)->get();

顺便说一句，这段代码不正确，因为我想在

中得到所有 history 和多个 sum

whereActionType(1)->sum('action_type')
whereActionType(2)->sum('action_type')
whereActionType(3)->sum('action_type')

例如（伪码）：

$userAddedPagesList = auth()->user()->instagramPages()->with([
        'history' => function ($query)  {
            $query->select(['action_type as like'])->whereActionType(1)->sum('action_type');
            $query->select(['action_type as follow'])->whereActionType(2)->sum('action_type');
            $query->select(['action_type as unfollow'])->whereActionType(3)->sum('action_type');
        }
    ]
)->get();

更新:

$userAddedPagesList = auth()->user()->instagramPages()->with([
        'history' => function ($query) {
            $query->select('*')
                ->selectSub(function ($query) {
                    return $query->selectRaw('SUM(action_type)')
                        ->where('action_type', '=', '1');
                }, 'like')
                ->selectSub(function ($query) {
                    return $query->selectRaw('SUM(action_type)')
                        ->where('action_type', '=', '2');
                }, 'follow')
                ->selectSub(function ($query) {
                    return $query->selectRaw('SUM(action_type)')
                        ->where('action_type', '=', '3');
                }, 'followBack');
        }
    ]
)->get();

错误:

Syntax error or access violation: 1140 Mixing of GROUP columns (MIN(),MAX(),COUNT(),...) with no GROUP columns is illegal if there is no GROUP BY clause (SQL: select *, (select SUM(action_type) where `action_type` = 1) as `like`, (select SUM(action_type) where `action_type` = 2) as `follow`, (select SUM(action_type) where `action_type` = 3) as `followBack` from `instagram_actions_histories` where `instagram_actions_histories`.`account_id` in (1, 2, 3))

我如何实施此解决方案？

更新：

Instagram 账号class:

class InstagramAccount extends Model
{
    ...

    public function schedule()
    {
        return $this->hasOne(ScheduleInstagramAccounts::class, 'account_id');
    }

    public function history()
    {
        return $this->hasMany(InstagramActionsHistory::class, 'account_id');
    }
}

InstagramActionsHistory class:

class InstagramActionsHistory extends Model
{
    protected $guarded=['id'];

    public function page(){
        return $this->belongsTo(InstagramAccount::class);
    }
}

用户class:

class User extends Authenticatable
{
    use Notifiable;

    ...

    public function instagramPages()
    {
        return $this->hasMany(InstagramAccount::class);
    }
}

Answer 1

selectSub() 创建您在其中使用聚合 (SUM) 的子查询，但 select() 是一个缩放器 returns 一个缩放器结果；除非你使用 grouping.If 你想 return 每个用户的结果，否则不允许在同一级别混合聚合与缩放器查询，尝试添加 groupBy，这里我假设 id 是userstable

上的主键

$userAddedPagesList = auth()->user()->instagramPages()->with([
        'history' => function ($query) {
            $query->select('*')
                ->selectSub(function ($query) {
                    return $query->selectRaw('SUM(action_type)')
                        ->where('action_type', '=', '1');
                }, 'like')
                ->selectSub(function ($query) {
                    return $query->selectRaw('SUM(action_type)')
                        ->where('action_type', '=', '2');
                }, 'follow')
                ->selectSub(function ($query) {
                    return $query->selectRaw('SUM(action_type)')
                        ->where('action_type', '=', '3');
                }, 'followBack');

            $query->groupBy('users.id');   //<-- add this
        }
    ]
)->get();

Answer 2

另一种获取不同类型操作的条件总和的方法，您可以在 InstagramAccount 模型中定义 hasOne() 关系，例如

public function history_sum()
{
    return $this->hasOne(InstagramActionsHistory::class, 'account_id')
        ->select('account_id',
            DB::raw('sum(case when action_type = 1 then 0 END) as `like`'),
            DB::raw('sum(case when action_type = 2 then 0 END) as `follow`'),
            DB::raw('sum(case when action_type = 3 then 0 END) as `followBack`')
        )->groupBy('account_id');
}

然后你可以预先加载相关数据为

$userAddedPagesList = auth()->user()->instagramPages()->with('history_sum')->get();

使用这种方法只会执行一个额外的查询，根据您的条件得到 3 个不同的总和结果

select `account_id`,
sum(case when action_type = 1 then action_type else 0 END) as `like`, 
sum(case when action_type = 2 then action_type else 0 END) as `follow`, 
sum(case when action_type = 3 then action_type else 0 END) as `followBack` 
from `instagram_actions_histories` 
where `instagram_actions_histories`.`account_id` in (?, ?, ?) 
group by `account_id`

虽然与使用 withCount 的其他方法（这也是一个有效答案）相比，将为每个操作类型添加 3 个相关的相关子查询，这可能会导致性能开销，但生成的查询看起来像以下

select `instagram_account`.*, 
(select sum(action_type) from `instagram_actions_histories` where `instagram_account`.`id` = `instagram_actions_histories`.`account_id` and `action_type` = ?) as `like`, 
(select sum(action_type) from `instagram_actions_histories` where `instagram_account`.`id` = `instagram_actions_histories`.`account_id` and `action_type` = ?) as `follow`,
(select sum(action_type) from `instagram_actions_histories` where `instagram_account`.`id` = `instagram_actions_histories`.`account_id` and `action_type` = ?) as `followBack`
from `instagram_account` 
where `instagram_account`.`user_id` = ? 
and `instagram_account`.`user_id` is not null

要检查生成的查询，请参阅

Laravel 在单个子句中使用多个 where 和 sum

Laravel use multiple where and sum in single clause

php

laravel

laravel-5.6