如何为嵌套关系构建适当的CakePHP查询

我是CakePHP新手，我正在尝试为结果构建复杂的查询。我快死了。也许有人能帮我我用的是蛋糕2.7

我有两个具有3个关系（多对多）的表<代码>邻域和

多边形

。 案例类似于此

邻域

有许多

邻域

，它也属于许多

邻域

。而且

邻域

有许多

多边形

，

多边形

属于许多

邻域

neighborary

表包含两个字段

name

和

zip

。我从用户那里得到的是

邮政编码

现在我想要的是： 我想从

neighbourth

获取所有

Polygons

，它是

neighbourth

。其中

neighbourth.zip=由用户定义

我该怎么做？我应该编写自定义查询还是将过程划分为更小的步骤？我整天都在为这个而斗争

以下是模型关系的样子：

class Neighbourhood extends AppModel
{
    var $hasAndBelongsToMany = array(
        'Neighbourhoods' => array(
            'className' => 'Neighbourhood',
            'joinTable' => 'neighbourhoods_neighbours',
            'foreignKey' => 'neighbourhood_id',
            'associationForeignKey' => 'neighbour_id',
            'unique' => false
        ),
        'Polygon' => array(
            'className' => 'Polygon',
            'joinTable' => 'neighbourhoods_polygons',
            'foreignKey' => 'neighbourhood_id',
            'associationForeignKey' => 'polygon_id',
            'unique' => false
        ),
    );
}


class Polygon extends AppModel
{
    var $hasAndBelongsToMany = array(
        'Neighbours' => array(
            'className' => 'Neighbourhood',
            'joinTable' => 'neighbourhoods_polygons',
            'foreignKey' => 'polygon_id',
            'associationForeignKey' => 'neighbourhood_id',
            'unique' => false,
        )
    );
}

---

您需要在您的模型中启用可控制行为，或者在应用程序模型中设置该行为后更好

public $actsAs = array('Containable');

然后开始建立您的查询，例如

$conditions = array(
    'conditions' => array('id' => '123'),
    'contain' => array(
        'Neighbourhood'=>array(
              'conditions' => array('Neighbourhood.id' => '123')
         )
     ),
     // or even joins
     'joins' => array(
          array(
                    'table' => $this->getTableName('default', 'neighbourhoods'),
                    'alias' => 'Neighbourhood',
                    'type' => 'RIGHT', // OR LEFT
                    'conditions' => array(
                        'Neighbourhood.id = Polygon.neighbourhood_id',
                        'Neighbourhood.deleted' => 0,
                    )
          )
     )
 );

$polygons = $this->Polygon->find('all', $conditions);

如果您认为这还不够（为了进行更复杂的查询），那么您需要构建查询语句。e、 g.从多边形模型运行查询：

    $dbo = $this->getDataSource();
    $query = $dbo->buildStatement(
            array(
                'fields' => array(
                    'Polygon.name AS polygon_name', 'Polygon.size',
                    'Neighbourhood.name AS neighbourhood_name', 'Neighbourhood.lat', 
                    'IF( Neighbourhood.created > DATE_SUB(NOW(), INTERVAL 1 DAY) , 1 , 0) AS new_neighbourhood'  
                ), 
                'table' => $dbo->fullTableName($this),
                'alias' => 'Polygon',
                'limit' => null,
                'offset' => null,
                'joins' => array(
                    array(
                        'table' => $this->getTableName('default', 'neighbourhoods'),
                        'alias' => 'Neighbourhood',
                        'type' => 'LEFT',
                        'conditions' => array(
                            'Neighbourhood.id = Polygon.neighbourhood_id',
                        ), 
                        'order' => 'Neighbourhood.name ASC',
                    ),
                    ....
                 ), 
                'conditions' => $conditions,
                'group' => 'Polygon.name',
                'order' => 'Polygon.name ASC',
           ),
           $this
     );

 $polygons = $this->query($query);

如果你觉得这还不够，你就得像这样唱出美妙的优雅

$polygons = $this->query("Here your sql query");

 debug($polygons);

---

我设法用你的方式创建了一个查询，谢谢你给我展示了一种方式，我很高兴。请记住，这是针对CakePHP2.x的，您没有太多的灵活性。但在CakePHP3.x中，也就是ORM，你可以自由地做几乎每件事，而且要简单得多。框架和版本不是我的选择，而是客户的选择。我照吩咐的去做。我不是说Cakephp不好。Cakephp是最好的顶级框架之一。你可以用它做很多事。祝你好运