CakePHP的3：多级联想

Question

我有3个表，categories，products和seller_products及其关联就像 CategoriesTable.php

$this->hasMany('Products', [ 
    'foreignKey' => 'category_id' 
]); 

ProductsTable.php

$this->hasMany('SellerProducts', [ 
    'foreignKey' => 'product_id' 
]); 

$this->belongsTo('Categories', [ 
    'foreignKey' => 'category_id', 
    'joinType' => 'INNER' 
]); 

SellerProductsTable.php

$this->belongsTo('Products', [ 
    'foreignKey' => 'product_id', 
    'joinType' => 'INNER' 
]); 

现在，在categories的view（site.com/categories/view/2），我不得不从products和sellerProducts选择所有产品数据，其中产品所属类别ID，也存在于sellerProducts。 即，

products.category_id = $id AND sellerProducts.product_id = Products.id 

有CakePHP中3任何简单的方法来获得结果呢？

编辑2

这就是我正在努力。在CategoriesController.php

$this->loadModel('Products'); 
     $sellerProducts = $this->Products->find('all', [ 
      'conditions' => [ 
      'category_id' => $id 
      ], 
      'joins' => [ 
       'table' => 'SellerProducts', 
       'alias' => 'SellerProducts', 
       'type' => 'INNER', 
       'conditions' => [ 
       'SellerProducts.product_id' => 'Products.id', 
       'stock >' => 0 
       ] 
      ] 
     ]); 
     debug($sellerProducts); 
     foreach($sellerProducts as $a) { 
      debug($a); 
     } 

上debugview()行动也只给出从Products表，但没有从SellerProducts

Answer 1

，而不是加入你们可以只包含SellerProducts

$this->Products->find() 
    ->where(['category_id' => $id]) 
    ->contain(['SellerProducts' => function($q) { 
     return $q->where(['stock >' => 0]) 
    }]); 

，如果你想查询数据您可以做的卖家产品的数据

$this->Products->SellerProducts->find() 
    ->where([ 
     'Products.category_id' => $id 
     'SellerProducts.stock >' => 0]) 
    ->contain(['Products', 'Products.Categories']);