Question

我有5个表tb1，tb2，tb3，tb4和tb5。

这是选择查询

SELECT c.COL1, f.COL2, SUM(a.colx1* l.clox2) AS COL3 
FROM tb1 c
  JOIN tb2 b ON (c.id = b.idC)
  JOIN tb3 l ON (b.id = l.idB)
  JOIN tb4 a ON (a.id = l.idA)
  JOIN tb5 f ON (f.id = a.idF) 
GROUP BY c.COL1, f.COL2

这是选择查询的结果

+-----------+-------+-----------+
| COL1      | COL2  | COL3      |
+-----------+-------+-----------+
| project1  | FAM1  | 1250.251  |
| project1  | FAM2  | 2145.325  |
| project2  | FAM1  | 2146.152  |
| project3  | FAM2  | 1248.002  |
| project4  | FAM1  | 3201.684  |
| project4  | FAM2  | 7325.981  |
| project5  | FAM1  | 4657.684  |

但是如果projectx没有FAMx

，我想用这种格式显示结果

+-----------+-------+-----------+
| COL1      | COL2  | COL3      |
+-----------+-------+-----------+
| project1  | FAM1  | 1250.251  |
| project1  | FAM2  | 2145.325  |
| project2  | FAM1  | 2146.152  |
| project2  | FAM2  | NULL      |
| project3  | FAM2  | NULL      |
| project3  | FAM2  | 1248.002  |
| project4  | FAM1  | 3201.684  |
| project4  | FAM2  | 7325.981  |
| project5  | FAM1  | 4657.684  |
| project5  | FAM1  | NULL      |

这是我对RIGHT JOIN的测试它不起作用！我错过了什么？

CREATE TEMPORARY TABLE tempt AS
   SELECT c.COL1, f.COL2, SUM(a.colx1* l.clox2) AS COL3 
    FROM tb1 c
      JOIN tb2 b ON (c.id = b.idC)
      JOIN tb3 l ON (b.id = l.idB)
      JOIN tb4 a ON (a.id = l.idA)
      JOIN tb5 f ON (f.id = a.idF) 
    GROUP BY c.COL1, f.COL2;
SELECT t.* FROM tempt t RIGHT JOIN tb5 f ON (f.COL2 = t.COL2)

tab5表的结构

+----+--------+------+
| id | COL1   | COL2 |
+----+--------+------+
|1   | F1     | FAM1 |
|2   | F2     | FAM2 |
+----+--------+------+

Answer 1

左边或右边的连接本身不能在所有＆＃34;项目＆＃34;之间创建你想要的额外关系。和＆＃34; FAM＆＃34;值。

您需要构建交叉连接以获取所需的所有行。

SELECT c.COL1, f.COL2, t.COL3 
FROM tb1 c
CROSS JOIN tb5 f
LEFT JOIN tempt t ON t.COL1=c.COL1 and t.COL2=f.COL2

如果tbl1.COL1或tbl5.COL2不是唯一的，则需要添加SELECT DISTINCT

也许整个事情可以归结为：

SELECT c.COL1, f.COL2, SUM(a.colx1* l.clox2) AS COL3 
FROM tb1 c
  CROSS JOIN tb5 f
  LEFT JOIN tb2 b ON (c.id = b.idC)
  LEFT JOIN tb3 l ON (b.id = l.idB)
  LEFT JOIN tb4 a ON (a.id = l.idA and f.id = a.idF)
GROUP BY c.COL1, f.COL2

右边选择结果

1 个答案: