Php Laravel雄辩的相关模型必须匹配两个外键

我在laravel框架中使用雄辩。我想通过表B和表C从表D中获得与表A相关的模型

有办法做到这一点吗

以下是数据库中的情况：

这是数据库的示例数据：

INSERT INTO A (idA) VALUES(1);
INSERT INTO A (idA) VALUES(2);
INSERT INTO A (idA) VALUES(3);

INSERT INTO B (idB, A_idA) VALUES(1, 1);
INSERT INTO B (idB, A_idA) VALUES(2, 2);

INSERT INTO C (idC, A_idA) VALUES(1, 1);
INSERT INTO C (idC, A_idA) VALUES(2, 2);
INSERT INTO C (idC, A_idA) VALUES(3, 3);

INSERT INTO D (idD, B_idB, C_idC) VALUES(1, 1, 1);
INSERT INTO D (idD, B_idB, C_idC) VALUES(2, 2, 2);
INSERT INTO D (idD, B_idB, C_idC) VALUES(3, 2, 3);

以下是模型：

class A extends Model
{
    protected $table = "A";
    protected $primaryKey = "idA";
    public $timestamps = false;

    public function B()
    {
        return $this->hasMany("App\B", "A_idA");
    }

    public function C()
    {
        return $this->hasMany("App\C", "A_idA");
    }
}

class B extends Model
{
    protected $table = "B";
    protected $primaryKey = "idB";
    public $timestamps = false;

    public function D()
    {
        return $this->hasMany("App\D", "B_idB");
    }

    public function A()
    {
        return $this->belongsTo("App\A", "A_idA");
    }
}

class C extends Model
{
    protected $table = "C";
    protected $primaryKey = "idC";
    public $timestamps = false;

    public function D()
    {
        return $this->hasMany("App\D", "C_idC");
    }

    public function A()
    {
        return $this->belongsTo("App\A", "A_idA");
    }
}

class D extends Model
{
    protected $table = "D";
    protected $primaryKey = "idD";
    public $timestamps = false;

    public function B()
    {
        return $this->belongsTo("App\B", "B_idB");
    }

    public function C()
    {
        return $this->belongsTo("App\C", "C_idC");
    }
}

这是控制器

class TestController extends Controller
{
    public function test()
    {
        $aWithRelatedDs = A::with(['B.D'])->whereHas("B.D")->whereHas("C.D")->get();
        dd($aWithRelatedDs);
    }
}

这是输出：

Collection {#380 ▼
  #items: array:2 [▼
    0 => A {#390 ▶}
    1 => A {#388 ▼
      #table: "A"
      ... snip ...
      #relations: array:1 [▼
        "B" => Collection {#387 ▼
          #items: array:1 [▼
            0 => B {#398 ▼
              #table: "B"
              ... snip ...
              #relations: array:1 [▼
                "D" => Collection {#395 ▼
                  #items: array:2 [▼
                    0 => D {#407 ▼
                      #table: "D"
                      ... snip ...
                      #attributes: array:3 [▼
                        "idD" => 2
                        "B_idB" => 2
                        "C_idC" => 2
                      ]
                      ... snip ...
                    }
                    1 => D {#408 ▼
                      #table: "D"
                      ... snip ...
                      #attributes: array:3 [▼
                        "idD" => 3
                        "B_idB" => 2
                        "C_idC" => 3
                      ]
                      ... snip ...
                    }]}]
... snip ...

---

问题是：A#2没有通过B和C关联的两个D，它只有通过两个中间表关联的D#2，但在上面的结果中它仍然列出。我如何限制这一点，使两个关联必须是同一个模型？

如果您想获得所有

A

，其中至少有一个

D

，通过

B

和

C

连接到

A

，那么原始SQL查询将如下所示：

SELECT *
FROM A
WHERE EXISTS (
  SELECT D.id
  FROM D
    INNER JOIN B ON D.b_id = B.id
    INNER JOIN C ON D.c_id = C.id
  WHERE B.a_id = A.id
    AND C.a_id = A.id
)

public function doubleLinkedDs(): HasManyThrough
{
    return $this->hasManyThrough(D::class, B::class)
        ->whereHas('C', function ($q) {
            $q->where('a_id', $this->id);
        });
}

我也为此准备了一份报告

我的建议是在

a

上创建一个新的关系，通过

B

找到所有

D

，我们找到一个

C

，它也连接回

a

。听起来很奇怪，但是很简单，看起来是这样的：

SELECT *
FROM A
WHERE EXISTS (
  SELECT D.id
  FROM D
    INNER JOIN B ON D.b_id = B.id
    INNER JOIN C ON D.c_id = C.id
  WHERE B.a_id = A.id
    AND C.a_id = A.id
)

public function doubleLinkedDs(): HasManyThrough
{
    return $this->hasManyThrough(D::class, B::class)
        ->whereHas('C', function ($q) {
            $q->where('a_id', $this->id);
        });
}

顺便说一句，这要求您建立了适当的关系。但就我从问题中所见，这应该是好的

如果定义了关系，我们应该能够执行以下非常简单的查询：

A::has('doubleLinkedDs')
    ->with('doubleLinkedDs')
    ->get();

---

请张贴您已经尝试过的代码。正确的方法是使用两个whereHas。你在做wrong@Namoshek我添加了代码。谢谢。@devk肯定是我看不到的东西。哦，它们一定和同一个模型D关联。这改变了它。我不确定还有什么路要走。像

A:：whereHas（'b.d.c'）->get（）这样的东西怎么样。注意，我不确定点表示法是否适用于
whereHas
，只是根据您的示例假设它适用