Question

我有两张桌子。我需要将这两个表的每一行组合成table3中的一行。我设法获得table1 SUM金额但不是table2。

例如

表用户

 +---------+-----------+
 | user_id | user_name |
 +---------+-----------+
 | 001     |   JOHN    |
 | 002     |   ADAM    |
 +---------+-----------+

表1

+-----------+----------------+-------------------+---------------------+
| table1_id | table1_user_id |    table1_amount  |       table1_date   |
+-----------+----------------+-------------------+---------------------+
|  6        | 001            |    100            |  01/11/2014 10:55   |  
|  7        | 002            |   100             |  01/11/2014 10:55   | 
|  8        | 001            |    50             |  25/10/2014 10:55   |  
|  9        | 001            |   100             |  23/10/2014 11:00   |  
|  10       | 002            |    0              |  21/10/2014 11:00   | 
+-----------+----------------+-------------------+---------------------+

表2

 +-----------+----------------+----------------+--------------------+
 | table2_id | table2_user_id |  table2_amount |  table2_date       |   
 +-----------+----------------+----------------+--------------------+
 |   1       |    001         |   100          |   15/11/2014 10:55 |    
 |   2       |    001         |   100          |   15/10/2014 10:55 |     
 |   3       |    002         |   100          |   11/10/2014 10:55 |     
 |   4       |    001         |   50           |   11/10/2014 10:55 |    
 +-----------+----------------+----------------+--------------------+

预期结果：

表3

+-----+---------+---------------+---------------+----------+---------+
| id  | user_id | table1_amount | table2_amount |    Year  |  Month  |
+-----+---------+---------------+---------------+----------+---------+
| 1   |  001    |    100        |       100     |     2014 |    11   |
| 2   |  002    |    100        |        0      |     2014 |    11   |
| 3   |  001    |    150        |       150     |     2014 |    10   |
| 4   |  002    |     0         |       100     |     2014 |    10   |
+-----+---------+---------------+---------------+----------+---------+

我的尝试，但它没有显示预期的结果。每行中table2_amount的数量为NULL：

SQL=" INSERT INTO table3 
      SELECT user_id,SUM(table1_amount),t2.amount2,
      YEAR(table1_date),MONTH(table1_date) FROM table1 a 
      LEFT JOIN 
     (SELECT c.table2_user_id,SUM(c.table2_amount) as amount2,c.table2_date
              FROM   table2 c
       GROUP BY DATE_FORMAT(c.table2_date,'%Y-%m'),c.table2_user_id ASC
              ) t2
     on t2.table2_user_id = a.table1_user_id AND t2.table2_date = a.table1_date
     GROUP BY DATE_FORMAT(a.table1_date,'%Y-%m'),table1_user_id ASC ";
"

Answer 1

UNION

这是一项不错的任务

SELECT tx.uid,SUM(tx.a1),SUM(tx.a2),YEAR(tx.d),MONTH(tx.d)
FROM
(
SELECT t1.table1_user_id as uid,
  t1.table1_amount as a1,
  0 as a2,
  t1.table1_date as d 
  FROM table1 t1
UNION
SELECT t2.table2_user_id as uid,
  0 as a1,
  t2.table2_amount as a2,
  t2.table2_date as d 
  FROM table2 t2
) tx
GROUP BY DATE_FORMAT(d,'%Y-%m'),uid ASC

Answer 2

感谢David162795的启发性讨论。

当两个表中的日期不同时，错过的点是按日期和用户ID分组INNER QUERY。我们需要在Inner Query中按个别日期对它们进行分组，然后按时间变量对主SELECT查询进行分组。

这是我对这个案例的回答：

$SQL = "
 INSERT INTO table3 (user_id, table1_amount, table2_amount,Year,  Month)
 SELECT tx.uid, SUM(tx.sum1), SUM(tx.sum2),YEAR(tx.d) as year,MONTH(tx.d) as month 
  FROM 
  (SELECT  b.table1_user_id as uid,b.table1_amount as sum1,0 as sum2,
           b.table1_date as d  FROM   table1 b
    GROUP BY DATE_FORMAT(d,'%Y-%m'),uid ASC

    UNION

    SELECT c.table2_user_id as uid,0 as sum1,
           sum(c.table2_amount) as sum2,c.table2_date   as d1
              FROM   table2 c
    GROUP BY DATE_FORMAT(d1,'%Y-%m'),uid ASC
  ) tx 

 GROUP BY year,month,uid"

MYSQL总和来自两个不同的表组

2 个答案: