Question

这是我在Laravel中创建复杂查询的最后一次尝试。我有这个场景需要的3个表：photos, events, countries。每个活动都有很多或没有照片，每个国家/地区可能有多个带照片的活动。我的结果显示

PhotosByCountry, EventsByCountry, rContinent, rCountry.

这是运行良好的MySQL本机查询：

SELECT SUM( allPhotos ) AS PhotosByCountry, 
 COUNT( temp.Land ) AS EventsByCountry, 
 rContinent, 
 temp.Land AS rCountry
FROM (
 SELECT e.country AS Land, COUNT( p.id ) AS allPhotos, c.continent AS rContinent
 FROM  `photos` p
 INNER JOIN  `events` e ON e.id = p.eventID
 INNER JOIN countries c ON e.country = c.country
 GROUP BY p.eventID
 )temp
GROUP BY rCountry

谁可以帮助我将其翻译成没有DB::raw()或whereRaw()的“Laravel查询”构建器。构建该东西的主要问题是子查询。

我为所有表格Models Photo, Country, Sportevent (for table events (= legacy name), couldn't use Event)。

感谢您的努力，如果需要，我很乐意提供更多信息。

加成

表：

events

id | name        | country ... has more columns of course
1  | Eventname 1 | France
2  | Eventname 2 | Switzerland
3  | Eventname 3 | France

photos

id | eventID | path  ...
1  |   2     | .....
2  |   1     | .....
3  |   2     | .....
4  |   3     | .....
5  |   3     | .....
6  |   2     | .....

countries

id |  country      | continent (or geographical Region)  ...
1  |  France       | Europe
2  |  Switzerland  | Europe
3  |  Germany      | Europe
4  |  United States| North America
5  |  Australia    | Oceania
6  |   .....

Result

PhotosByCountry | EventsByCountry | rContinent | rCountry 
      3         |       2         | Europe     | France
      3         |       1         | Europe     | Switzerland

Answer 1

试试这个：

国家模式

protected function photosCount()
{
    return $this->hasManyThrough('photos', 'events', 'country', 'event_id')
        ->groupBy('countries.country')
        ->count();
}

public function getPhotosCount() {
    return $this->photosCount ? $this->photosCount->count : 0;
}

protected function eventsCount()
{
    return $this->hasMany('events', 'country', 'event_id')
        ->groupBy('countries.country')
        ->count();
}

public function getEventsCount() {
    return $this->eventsCount ? $this->eventsCount->count : 0;
}

像这样访问它：

$countries = Country::with('photosCount', 'eventsCount')->get();
$countries->first()->getPhotosCount();
$countries->first()->getEventsCount();

代码未经测试。

来源：http://laravel.io/forum/05-03-2014-eloquent-get-count-relation

这可能会使3个sql查询而不是1个（一个用于国家+ 2个WHERE IN查询计数）。但我认为这是一个非常好的方法来获得计数。

Answer 2

完成问题：

由于在Eloquent Query Builder中构建初始查询似乎不可能或太复杂，我在Photo模型中使用了以下函数（但我可以将它放在任何其他模型中）

public function allCountrysWithPhotos(){
    $rawQuery=" SELECT SUM( allPhotos ) AS PhotosByCountry, 
                 COUNT( temp.Land ) AS EventsByCountry, 
                 rContinent, 
                 temp.Land AS rCountry
                FROM (
                 SELECT e.country AS Land, 
                  COUNT( p.id ) AS allPhotos, 
                  c.continent AS rContinent
                 FROM  `photos` p
                 INNER JOIN  `events` e ON e.id = p.eventID
                 INNER JOIN countries c ON e.country = c.country
                 GROUP BY p.eventID
                 )temp
                GROUP BY rCountry
                ORDER BY rContinent, rCountry";

    return DB::select($rawQuery);   
}

如果将来需要调整某些内容（索引可能在某个阶段），这对SQLers来说很容易理解。

无论如何，感谢您的努力，我不再会问Eloquent问题了，而是会用原生查询来做。

复杂查询3个表，其中包含2个INNER JOIN，1个子查询，2个在Laravel中的Group By

2 个答案: