Question

请参阅以下SQL：

SELECT M.username, count(*) as Total, date(status_date) as Date FROM com_result
   LEFT JOIN members as M on M.member_id = com_result.member_id 
GROUP BY date(status_date), com_result.member_id  Order By status_date DESC

这将显示基于DATE和member_id

的记录总数

示例结果：

+----------+-------+------------+
| username | Total | DATE       |
+----------+-------+------------+
| bx7      |     3 | 2012-09-10 |
| bx2      |    25 | 2012-09-04 |
| bx2      |   401 | 2012-09-03 |
| bx1      |   703 | 2012-09-02 |
| bx4      |  1075 | 2012-09-02 |
+----------+-------+------------+

它运行良好，现在我想合并/联合到另一个相同的表结构。 第二个SQL：

SELECT M.username, count(*) as Total, date(status_date) as Date FROM com_result_b
   LEFT JOIN members as M on M.member_id = com_result_b.member_id 
GROUP BY date(status_date), com_result_b.member_id  Order By status_date DESC

假设com_result的结果为：

| bx2   |    25 | 2012-09-04 |

com_result_b的结果是：

 | bx2   |    50 | 2012-09-04 |

所以当它合并/联合时 - 我希望结果如下：

 | bx2   |    75 | 2012-09-04 |

正如您所看到的，Total的值从两个表中的相同日期开始增加。怎么办？

Answer 1

SELECT
  username,
  IF(q1.Total IS NULL,0,q1.Total)+IF(q2.Total IS NULL,0,q2.Total) AS Total,
  q1.Date
FROM (
  SELECT member_id, M.username, count(*) as Total, date(status_date) as Date 
  FROM com_result
  LEFT JOIN members as M on M.member_id = com_result.member_id 
  GROUP BY date(status_date), com_result.member_id  
) AS q1
LEFT JOIN (
  SELECT member_id, M.username, count(*) as Total, date(status_date) as Date 
  FROM com_result_b
  LEFT JOIN members as M on M.member_id = com_result_b.member_id 
  GROUP BY date(status_date), com_result_b.member_id
) AS q2 ON q1.member_id=q2.member_id AND q1.Date=q2.Date
ORDER BY q1.Date DESC

Answer 2

使用SUM

select username,SUM(Total) , DATE
(
    SELECT M.username, count(*) as Total, date(status_date) as Date FROM com_result_b 
       LEFT JOIN members as M on M.member_id = com_result.member_id  
    GROUP BY date(status_date), com_result.member_id  
   UNION ALL
   SELECT M.username, count(*) as Total, date(status_date) as Date FROM com_result_b   
      LEFT JOIN members as M on M.member_id = com_result_b.member_id    
   GROUP BY date(status_date), com_result_b.member_id 
) v

group by username, date

Answer 3

SELECT X.username, SUM(X.Total)  AS TOTAL , X.Date
FROM( SELECT ... FROM com_result Union Select .. From com_result_b)X
GROUP BY X.username,  X.Date

Answer 4

试试这个：

最简单的方法是：

您只需要将两个结果放在带有union all运算符的派生表中

 SELECT username,SUM(Total),Date
FROM   (
        SELECT M.username, count(*) as Total, date(status_date) as Date 
        FROM com_result
        LEFT JOIN members as M on M.member_id = com_result.member_id 
        GROUP BY date(status_date), com_result.member_id   
        union all
        SELECT M.username, count(*) as Total, date(status_date) as Date 
        FROM com_result_b
        LEFT JOIN members as M on M.member_id = com_result_b.member_id 
        GROUP BY date(status_date), com_result_b.member_id )a
GROUP BY username,Date
ORDER BY Date DESC

合并2个或更多结果？

4 个答案: