Question

您好我将我正在处理的mysql查询转换为laravel eloquent查询并需要一些帮助。我有一个保留表，链接到具有多对多关系的产品表。无论有多少产品与预订相关，我都想提取所有预订，而不是它找到的第一个产品。这是我的sql：

SELECT reservations.id,
       reservations.play_date,
       reservations.group_size,
       reservations.status,
       reservations.event_title,
       t4.product_id,
       t4.id AS link_id,
       p1.name,
       CONCAT_WS(" ", customers.first_name, customers.last_name, customers.group_name) AS customerName,
       reservations.event_type
FROM reservations
LEFT JOIN customers ON reservations.customer_id = customers.id
LEFT JOIN
  (SELECT *
   FROM product_reservation AS t3
   GROUP BY t3.reservation_id ) AS t4 ON t4.reservation_id = reservations.id
LEFT JOIN products AS p1 ON t4.product_id = p1.id

我可以将它作为原始查询放置，但是会产生一个带有结果的数组 - 我需要能够创建一个查询对象，这样我才能在结果上使用另一个模块有没有雄辩的方法来做到这一点 - 或者我怎样才能让这个查询在laravel中工作？谢谢

Answer 1

是的，你可以使用Eloquent关系。它们看起来像这样......

class Reservation extends Eloquent
{
    public function products()
    {
        return $this->belongsToMany('Product');
    }
}

class Product
{
    public function reservations()
    {
        return $this->belongsToMany('Reservation');
    }
}

$reservations = Reservation::with(array('products', function($q) {
    $q->take(1);
}))->get();

foreach($reservations as $reservation) {
    echo $reservation->name;
    foreach($reservation->products as $product) {
        echo $product->description;
    }
}

laravel：将多个（但只有一个相关项）mysql查询转换为laravel查询

1 个答案: