CakePHP模型连接问题,连接表中缺少字段.

ste*_* mc 0 cakephp join model

我可能错过了一些非常明显的东西.我试图在Cake中将多个表连接在一起,但我只从第一个表中获取字段.请考虑以下代码..

$joins = array();
$joins[] =  array(
            'table' => 'products', 
            'alias' => 'Product', 
            'type' => 'LEFT',
            'conditions' => array(
                'Device.id = Product.device_id'
            )
    );

$joins[] =  array(
            'table' => 'pricepoints', 
            'alias' => 'Pricepoints', 
            'type' => 'LEFT',
            'conditions' => array(
                'Pricepoints.product_id = Product.id'
            )
    ); 
$all_products = $this->Device->find('all', array("joins" => $joins); 
Run Code Online (Sandbox Code Playgroud)

这将返回以下SQL

SELECT `Device`.`id`, `Device`.`manufacturer_id`, `Device`.`name`, `Device`.`type_id`, `Manufacturer`.`id`, `Manufacturer`.`name` FROM `devices` AS `Device` LEFT JOIN products AS `Product` ON (`Device`.`id` = `Product`.`device_id`) LEFT JOIN pricepoints AS `Pricepoints` ON (`Pricepoints`.`product_id` = `Product`.`id`) LEFT JOIN `manufacturers` AS `Manufacturer` ON (`Device`.`manufacturer_id` = `Manufacturer`.`id`) 
Run Code Online (Sandbox Code Playgroud)

即.它只返回父模型中的字段ie.设备.如何从连接中选择所有字段?我认为它与我建立模型关系的方式有关,但我认为我已经正确设置了这些.

有人建议吗?

ple*_*ong 7

您可以在查找查询中指定字段:

$all_products = $this->Device->find('all', array("fields" => array('Device.*','Product.*','Pricepoints.*')
                                                 "joins" => $joins
); 
Run Code Online (Sandbox Code Playgroud)

希望这可以帮助