Laravel 在单个子句中使用多个 where 和 sum
在我的数据库中,我有 instagram_actions_histories
表,其中有 action_type
列,在该列中我有不同的数据,例如 1
或2
或 3
in my database i have instagram_actions_histories
table which into that i have action_type
column, in the column i have unlike data such as 1
or 2
or 3
我正在尝试获取关系船中的此表数据并对存储在列中的这些值求和,例如
i'm trying to get this table data in relation ship and summing this values which stored in column, for example
$userAddedPagesList = auth()->user()->instagramPages()->with([
'history' => function ($query) {
$query->select(['action_type as count'])->whereActionType(1)->sum('action_type');
}
]
)->get();
顺便说一句,这段代码是不正确的,因为我想得到所有 history
里面有多个 sum
btw, this code is not correct, in that i want to get all history
with multiple sum
inside that
whereActionType(1)->sum('action_type')
whereActionType(2)->sum('action_type')
whereActionType(3)->sum('action_type')
例如(伪代码):
$userAddedPagesList = auth()->user()->instagramPages()->with([
'history' => function ($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))
我该如何实施这个解决方案?
how can i implementing this solution?
更新:
InstagramAccount 类:
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 类:
class InstagramActionsHistory extends Model
{
protected $guarded=['id'];
public function page(){
return $this->belongsTo(InstagramAccount::class);
}
}
用户类别:
class User extends Authenticatable
{
use Notifiable;
...
public function instagramPages()
{
return $this->hasMany(InstagramAccount::class);
}
}
推荐答案
另一种为不同类型的动作获取条件总和的方法,你可以在你的 中定义一个
模型如 hasOne()
关系InstagramAccount
Another approach to get conditional sum for your different types of action, you can define a hasOne()
relation in your InstagramAccount
model like
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');
}
然后您可以将相关数据预先加载为
Then you can eager load the related data as
$userAddedPagesList = auth()->user()->instagramPages()->with('history_sum')->get();
采用这种方法将仅执行一个额外的查询,以根据您的条件获得 3 个不同的总和结果
Going through with this approach will execute only one extra query to get 3 different sum results based on your criteria
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(这也是一个有效的答案)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,生成的查询将看起来有些东西如下图
While as compare to other approach (which is a valid answer also) using withCount
will add 3 dependent correlated sub queries for each action type which may result as a performance overhead, generated query will look something like below
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 5.3 - 如何记录页面上的所有查询?
To check the generated queries refer to Laravel 5.3 - How to log all queries on a page?
相关文章