Ada*_*ann 3 php laravel eloquent
** 更新帖子:我认为使用Eloquent实际上是不可能的,所以我采取了另一种方法,谢谢!**
考虑以下简化表:
content
---------------
| id | name |
---------------
| 1 | news |
| 2 | review |
---------------
games
-------------
| id | name |
-------------
| 8 | halo |
| 9 | gta |
-------------
releases
-------------------------------
| id | name | game_id |
-------------------------------
| 14 | halo for ps3 | 8 |
| 15 | halo for wii | 8 |
| 16 | gta for ps4 | 9 |
-------------------------------
content_releases
-------------------------------------
| content_id | release_id | game_id |
-------------------------------------
| 1 | 14 | 8 |
| 1 | 15 | 8 |
| 2 | 15 | 8 |
| 2 | 16 | 9 |
-------------------------------------
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content可以releases通过content_games表格将许多项目链接到许多项目。releases属于一个gamecontent项目链接到releases它们,它们也固有地链接到games。我的模型如下所示:
class Content extends \Eloquent
{
public function games()
{
return $this->belongsToMany('Models\Game', 'content_game');
}
}
class Game extends \Eloquent
{
public function releases()
{
return $this->belongsToMany('Models\Release', 'content_game');
}
}
class Release extends \Eloquent
{
public function content()
{
return $this->belongsToMany('Models\Content', 'content_game');
}
}
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我正在尝试编写一个相对较快的延迟加载的口才查询,该查询可以使我满意,并在其下嵌套相关游戏,然后在其下嵌套相关发行版。所以我希望输出数组看起来像这样:
Array
(
[0] => Array
(
[id] => 1
[name] => News
[games] => Array
(
[0] => Array
(
[id] => 8
[name] => Halo
[releases] => Array
(
[0] => Array
(
[id] => 14
[name] => Halo for PS3
)
[1] => Array
(
[id] => 15
[name] => Halo for Wii
)
)
)
)
)
[1] => Array
(
[id] => 2
[name] => Review
[games] => Array
(
[0] => Array
(
[id] => 8
[name] => Halo
[releases] => Array
(
[0] => Array
(
[id] => 15
[name] => Halo for Wii
)
)
)
[1] => Array
(
[id] => 9
[name] => GTA
[releases] => Array
(
[0] => Array
(
[id] => 16
[name] => GTA for PS4
)
)
)
)
)
)
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Content::with('games', 'games.releases')->toArray();用3列表将content_releases它们全部连接在一起不能充分发挥作用,因为Eloquent仅基于两个列进行查询。因此,我需要添加一个约束games.releases。
它可以工作(下面的代码),但是效率不高:
Content::with(['games' => function ($query) {
$query->groupBy(['game_id','game.id','pivot_content_id']);
},
'games.releases' => function ($query) {
$query->has('content');
}]);
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它产生以下查询:
select * from "content" limit 10 offset 0
select "game".*, "content_game"."content_id" as "pivot_content_id", "content_game"."game_id" as "pivot_game_id"
from "game"
inner join "content_game" on "game"."id" = "content_game"."game_id"
where "content_game"."content_id" in (?, ?, ?, ?, ?, ?, ?, ?, ?, ?)
group by "game_id", "game"."id", "pivot_content_id"
select "releases".*, "content_game"."game_id" as "pivot_game_id", "content_game"."release_id" as "pivot_release_id"
from "releases"
inner join "content_game" on "releases"."id" = "content_game"."release_id"
where "content_game"."game_id" in (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)
and (
select count(*)
from "content"
inner join "content_game" on "content"."id" = "content_game"."content_id"
where "content_game"."release_id" = "releases"."id"
) >= 1
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谁能建议一种更有效的方法来实现相同的输出?
谢谢。
是的,这个问题有点过时了,但我前一段时间对此感到困惑,正确的答案可能对其他人有用:)
设置数据库表和关系的方式存在一些问题,因此让我们先避免它们:
如果我理解正确,内容与游戏之间存在多对多关系,游戏与发行之间存在多对多关系。因此,您将为内容和游戏之间的关系创建数据透视表,并在发布表上使用外键。这意味着重命名枢轴(为清楚起见)并删除release_id列。
完成后,我们必须更新模型关系。“内容属于ToMany游戏”,“游戏属于ToMany内容”和“游戏hasMany发布”。
现在只需要使用嵌套的雄辩关系,此后可能已经在文档中添加了一些内容。
Content::with(['games', 'games.releases'])->get()->toArray();
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