Question

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：A＃2没有两个D通过B和C关联，它只有D＃2通过两个中间表关联，但在上面的结果中它仍然被列出。我如何限制这一点，因此两个关联必须属于同一模型？

Answer 1

对每个关系使用while应该有效，例如：

D::whereHas('B', function ($query) {
    $query->whereExists(function ($query) { 
        $query->select(DB::raw(1))
                    ->from('B')
                    ->whereRaw('B.A_id = A.id');
    });
})->whereHas('C', function ($query) {
    $query->whereExists(function ($query) { 
        $query->select(DB::raw(1))
                    ->from('C')
                    ->whereRaw('C.A_id = A.id');
    });
})->get();

完全未经测试，祝你好运。

Answer 2

如果您希望通过A和D获得至少有A个与B相关联的所有C，那么原始SQL查询看起来像这样：

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
)

我还为此准备了SQL Fiddle。

我的建议是在A上创建一个新关系，找到所有D到B，我们发现C也连接回A {1}}。听起来很奇怪，但很容易看起来像这样：

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

顺便说一下，这需要你设置适当的关系。但就我的问题而言，这应该没问题。

如果我们定义了关系，我们应该能够执行以下非常简单的查询：

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

Laravel Eloquent相关模型必须匹配两个外键

2 个答案: