雄辩的嵌套关系与一些约束 [英] Eloquent Nested Relation with Some Constraint

问题描述

我有以下三个表：

A
-------------
| id | name |
-------------

B
--------------------
| id | A_id | name |
--------------------

C
--------------------
| id | B_id | name |
--------------------

因此，表 C 中的数据属于表 B 中的数据，属于表中的数据 A 。现在，我想查询 C ，同时还从 B 和 A 和以下代码做的伎俩很好。

So, the data in table C belongs to the data in table B which belongs to the data in table A. Now, I want to query C, while also retrieving data from B and A and the following code does the trick just fine.

C::with('B.A')->get();

现在的问题是，我想查询 C 有一些约束。其中一个约束是 id A 。我尝试了以下内容：

The problem now, is that I want to query C with some constraints. One of these constraints is the id of A. I've tried the following:

C::with(array('B.A' => function ($query)
{
    $query->where('id', '=', $constraint);
}))->get();

但是似乎Eloquent会检索 C 甚至不考虑约束，除非它执行查询以检索表 A 中的数据。如何解决这个问题？我需要在 C 中添加另一个字段，即 A_id ，并匹配 $ constraint 对于该字段

But it seems that Eloquent will retrieve all the rows in C without even taking the constraint into account, except when it's executing the query to retrieve data in table A. How do I get around this problem? Do I need to add another field in C, that is A_id, and match $constraint against that field?

推荐答案

您将与（） / code>方法与SQL的 JOIN ，发生了很多。

当您使用 Foo :: with（'bar'） - > where_something（1）时，Laravel将首先加载 Foo 然后，根据 Foo.bar_id ，它将加载 Bar 。它的目的是告诉Laravel对您的模型的加载依赖关系的组合查询，大大提高了这些模型的迭代性能。

When you use Foo::with('bar')->where_something(1), Laravel will first load the Foo and then, based on Foo.bar_id, it will load the Bar. It serves the purpose of telling Laravel to eager load dependencies of your model on a combined query, greatly improving performance of iterations on those models.

如果不使用它应执行以下查询：

If you don't use it, the following queries should be executed:

SELECT * FROM foos WHERE foos.something = 1;
SELECT * FROM bars WHERE bars.id = 30;
SELECT * FROM bars WHERE bars.id = 57;
SELECT * FROM bars WHERE bars.id = 134;
SELECT * FROM bars WHERE bars.id = 1096;

另一方面，如果您使用它：

If you use it, on the other hand:

SELECT * FROM foos WHERE foos.something = 1;
SELECT * FROM bars WHERE bars.id IN (30, 57, 134, 1096); // Eager loading

当您使用（），你只是限制这些依赖关系的热切加载，而不是第一个查询。

When you add a condition to that with(), you are only constraining the eager loading of those dependencies, and not the first query.

为了达到你想要的目的，你需要使用 - > join（）。

To achieve what you want, you'll need to use ->join().

C::with(array('b', 'b.a'))
 ->join('b', 'b.id', '=', 'c.b_id')
 ->join('a', 'a.id', '=', 'b.a_id')
 ->where('a.id', '=', $ID)
 ->get('c.*');

我已经将与（） ，因为我不知道你是否需要访问 $ c-> b-> a 。如果不这样做，而您只需要 $ c 数据，则可以使用（）删除，因为它将不必要地查询B和A。

I've included the with(), because I didn't know if you would need to access $c->b->a. If you don't, and you just need $c data, you can remove the with() since it will query for B's and A's unnecessarily.

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