本文介绍了使用 via 或 viaTable 关系从关系生成 mysql 查询的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
限时送ChatGPT账号..考虑下面的例子...
class Customers extends ActiveRecord
{
public function getOrders()
{
return $this->hasMany(Orders::className(), ['customer_id' => 'id']);
}
public function getOrderItems()
{
return $this->hasMany(OrderItems::className(), ['order_id' => 'id'])
->via('orders');
}
}
我如何从 getOrderItems()
关系生成任何一个以下查询
how i can generate any one of the follwing query from the getOrderItems()
relation
SELECT * FROM `order-items`
LEFT JOIN `orders` ON `orders`.`id` = `order-items`.`order_id`
LEFT JOIN `customers` ON `customer`.`id` = `orders`.`customer_id`
或
SELECT `order-items`.* FROM `order-items`,`orders`,`customers`
WHERE `customer`.`id` = `orders`.`customer_id` AND `orders`.`id` = `order-items`.`order_id`
或
SELECT * FROM `order-items` WHERE `order_id` IN(
SELECT * FROM `orders` WHERE `customer_id` IN(
SELECT * FROM `customers`
)
)
我使用以下代码来做到这一点.
i use the following code to do this.
$customers = Customers::findAll();
$query = $customers[0]->getOrderItems()->createCommand()->rawSql;
但它只生成
SELECT * FROM `order-items`
怎么办……???
推荐答案
使用这个:
$q = Customers::findAll()->innerJoinWith('orderItems')->createCommand()->rawSql;
你必须像这样使用关系名称而不是函数
you have to use relation name like this not as function
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