将同一表中的两个查询连接到不同的月份和总和值

Question

我想获取具有MONTH(NOW())的member_total_point列和MONTH(NOW() INTERVAL 1 MONTH)列的总和的表。 如何选择表格？

编辑

我不应该在选择上使用[..]

这是我的新查询：

SELECT la.member_created, la.member_api_domain, la.member_api_campaign, 
      la.total, lb.total_now
FROM (SELECT member_created, member_api_domain, member_api_campaign, SUM(member_total_point) AS 'total' 
     FROM lalights_member WHERE YEAR(member_created)=YEAR(NOW()) 
     AND MONTH(member_created)= MONTH(NOW()-INTERVAL 1 MONTH) 
     GROUP BY member_api_campaign ) AS la 
LEFT JOIN (SELECT member_created,SUM(member_total_point) 
     AS 'total_now' 
     FROM lalights_member WHERE YEAR(member_created)=YEAR(NOW()) 
     AND MONTH(member_created)= MONTH(NOW()) 
     ORDER BY member_api_campaign DESC) AS lb 
ON la.member_created = lb.member_created

但是错误出现在SQLyog上：

'[total]，lb. [total_now] FROM（选择第1行的member_created，member_api_domain，member_a'

即时通讯使用MySQL 5.1 ..

样本数据：

member_created       member_api_domain  member_api_campaign member_total_point
--------------       -----------------  ------------------- ------------------
2015-09-22 18:30:20  domain             campaign             4 (september)
2015-09-22 18:30:20  domain             campaign             7 (september)
2015-08-22 18:30:20  domain             campaign             2 (august)
2015-08-22 18:30:20  domain             campaign             7 (august)
2015-09-22 18:30:20  asdf               jujuju               4 (september)
2015-08-22 18:30:20  asdf               jujuju               9 (august)

样本输出：

member_api_campaign this_month last_1month last_2month
------------------- ---------- ----------- -----------
domain              campaign   11(september) 9 (august)
asdf                jujuju     4 (september) 9 (august)

提醒一下，现在是October 。

Answer 1

GROUP BY必须包括您选择的所有列。 在SQL Server 2005+ ，可以使用OVER（）子句 ，如下所示：

SELECT member_created, member_api_domain, member_api_campaign,
     SUM(CASE WHEN MONTH(member_created)= MONTH(NOW()) THEN member_total_point) OVER(PARTITION BY member_api_campaign) AS 'total',
     SUM(CASE WHEN MONTH(member_created)= MONTH(NOW()-INTERVAL 1 MONTH) THEN member_total_point) OVER(PARTITION BY member_api_campaign) AS 'total now'
FROM lalights_member 
WHERE YEAR(member_created)=2014 
AND (MONTH(member_created)= MONTH(NOW()) OR MONTH(member_created)= MONTH(NOW()-INTERVAL 1 MONTH))
ORDER BY member_api_campaign DESC

Answer 2

• 更改单引号（平均字符串）作为边引号，或仅删除字段别名的引号

   'total' for `total` 'total_now' for `total_now` 

• 删除ORDER BY member_api_campaign DESC

。

SELECT la.member_created
    ,la.member_api_domain
    ,la.member_api_campaign
    ,la.total
    ,lb.total_now
FROM (
    SELECT member_created
          ,member_api_domain
          ,member_api_campaign
          ,SUM(member_total_point) AS 'total'
    FROM lalights_member
    WHERE YEAR(member_created) = YEAR(NOW())
      AND MONTH(member_created) = MONTH(NOW() - INTERVAL 1 MONTH)
    GROUP BY member_api_campaign
    ) AS la
LEFT JOIN (
    SELECT member_created
          ,SUM(member_total_point) AS 'total_now'
    FROM lalights_member
    WHERE YEAR(member_created) = YEAR(NOW())
      AND MONTH(member_created) = MONTH(NOW())
    ) AS lb ON la.member_created = lb.member_created