Question

问题：

我无法找到使用QueryBuilder构建查询的解决方案（可能首先使用常规SQL查询完成它会有所帮助）：

尝试检索用户的所有客户（在其中一个商店用户链接的店铺积分），需要总积分（属于该用户的商店的积分总和）作为虚拟列（能够订购） on），使用paginate（）。

数据库结构：

表客户

  id           email       other_fields
   1        1@email.com        f
   2        2@email.com        o
   3        3@email.com        o

表用户

  id           email       other_fields
   1        1@user.com        b
   2        2@user.com        a
   3        3@user.com        r

餐馆

  id           name       other_fields
   1          Shop 1        m
   2          Shop 1        o
   3          Shop 1        o

表user_shops

  user_id     shop_id
   1             1       
   1             2       
   3             3

表customer_shop_credits

  customer_id        shop_id       credits
   1                   1             55
   1                   2             45
   2                   2             3
   3                   3             44

预期结果：

为用户1检索客户时，我希望获得100个积分的客户1和3个积分的客户2

我最近了：

$credits_query = CustomerShopCreditQuery::create()
        ->useShopQuery()
            ->useUserShopQuery()
                ->filterByUserId($user->getId())
            ->endUse()
        ->endUse()
        ;


    $customers = CustomerQuery::create()
        ->addSelectQuery($credits_query, 'credits_alias', false)
        ->useCustomerShopCreditQuery()
            ->useShopQuery()
                ->useUserShopQuery()
                    ->filterByUserId($user->getId())
                ->endUse()
            ->endUse()
        ->endUse()
        ->withColumn('sum(credits_alias.credits)', 'credits')
        ->groupById()
        ->orderBy($order_by_column, $direction)
        ->paginate($page, $page_size);

导致以下查询：

SELECT customers.id, customers.email, sum(credits_alias.credits) AS credits 
FROM customers 
CROSS JOIN (
    SELECT customer_shop_credits.id, customer_shop_credits.customer_id, customer_shop_credits.shop_id, customer_shop_credits.credits 
    FROM customer_shop_credits 
    INNER JOIN shops ON (customer_shop_credits.shop_id=shops.id) 
    INNER JOIN user_shops ON (shops.id=user_shops.shop_id) 
    WHERE user_shops.user_id=159
    ) AS credits_alias 
INNER JOIN customer_shop_credits ON (customers.id=customer_shop_credits.customer_id) 
INNER JOIN shops ON (customer_shop_credits.shop_id=shops.id) 
INNER JOIN user_shops ON (shops.id=user_shops.shop_id) 
WHERE user_shops.user_id=159 
GROUP BY customers.id 
ORDER BY customers.id DESC 
LIMIT 25

但是给我的结果有错误的学分。不确定CROSS JOIN。当我编辑这个查询并使其成为一个JOIN并使用ON (credits_alias.customer_id = customers.id)作为条件时，学分的总和会更好，但似乎有加倍总和的经典连接问题

Propel2;如何使用querybuilder进行子选择查询

0 个答案: