Question

我试图加入两张桌子，并且还得到一个SUM并且非常糟糕。我需要获得每个会员的总佣金金额，其中affiliate.approved = 1 AND order.status = 3.

//affiliate table
affiliate_id | firstname | lastname | approved |
     1            joe        shmoe       1
     2           frank       dimag       0
     3            bob        roosky      1

这里是订单表

//order
affiliate_id | order_status_id | commission
    1                3              0.20
    1                0              0.30
    2                3              0.10
    3                3              0.25
    1                3              0.25
    2                3              0.15
    2                0              0.20

以及我希望返回的查询：

affiliate_id | commission
    1             0.45
    3             0.25

这是我的尝试，但不起作用。它只输出一行。

 SELECT order.affiliate_id, SUM(order.commission) AS total, affiliate.firstname, affiliate.lastname FROM `order`, `affiliate` WHERE order.order_status_id=3 AND affiliate.approved=1 AND order.affiliate_id = affiliate.affiliate_id ORDER BY total;

感谢您的帮助。

Answer 1

您可以针对您的解决方案尝试此查询： -

   SELECT order.affiliate_id, SUM(order.commission) AS total,affiliate.firstname,
  affiliate.lastname 
    FROM `order`, `affiliate`
    WHERE order.order_status_id=3 
    AND affiliate.approved=1 
    AND order.affiliate_id = affiliate.affiliate_id 
    GROUP BY order.affiliate_id
    ORDER BY total;

Answer 2

你错过了GROUP BY，试试这个：

SELECT 
      `order`.affiliate_id,
      SUM(`order`.commission) AS total,
      affiliate.firstname,
      affiliate.lastname
FROM `order` 
JOIN `affiliate` 
ON `order`.order_status_id = 3 AND affiliate.approved = 1 AND `order`.affiliate_id = affiliate.affiliate_id
GROUP BY `order`.affiliate_id
ORDER BY total;

Demo Here

Answer 3

首先：删除隐式join语法。这令人困惑。

第二：您需要按affiliate_id进行分组。通过将结果集折叠为单行，使用不带组的聚合函数。

以下是使用INNER JOIN的查询：

SELECT
    `order`.affiliate_id,
    SUM(`order`.commission) AS total,
    affiliate.firstname,
    affiliate.lastname
FROM `order`
INNER JOIN`affiliate` ON  `order`.affiliate_id = affiliate.affiliate_id
WHERE `order`.order_status_id = 3
AND affiliate.approved = 1
GROUP BY affiliate.affiliate_id
ORDER BY total;

WORKING DEMO

警告：您选择了一个MySQL的保留字作为表名（order）。请注意始终用（`）反引号将其括起来。

_{只是一个温和的提醒}

Answer 4

以下是解决方案：

select affiliate.affiliate_id,sum(`order`.commission) as total from affiliate left join `order` on affiliate.affiliate_id=`order`.affiliate_id
where affiliate.approved=1 and `order`.order_status_id=3 group by affiliate.affiliate_id

此外，＆＃34;订单＆＃34;是SQL的关键词，我建议你不要将它用作表/列名。

mysql：两个表连接sum

4 个答案: