![](/img/trans.png)
[英]Multiple select from one row from A table and insert result as multiple rows in B table and repeat in one query
[英]For each row from select insert multiple rows into one table
我想做这样的事情
SELECT ID, X, Y, Z FROM TABLE_1
这给我带来了类似的东西:
| IDTABLE_1 | X | Y | Z |
| 1 | a | b | c |
| 2 | d | e | f |
| 3 | g | h | i |
我有另一个表TABLE_2,其中包含一个列,其中包含上一个表中的列名(TABLE_1):
| IDTABLE_2 | COLUMN |
| 101 | X |
| 102 | Y |
| 103 | Z |
现在我需要将select中的所有数据插入到这个新表TABLE_3中,它必须是这样的:
| IDTABLE_3 | IDTABLE_1 | IDTABLE_2 | VALUE |
| 201 | 1 | 101 | a |
| 202 | 1 | 102 | b |
| 203 | 1 | 103 | c |
| 204 | 2 | 101 | d |
| 205 | 2 | 102 | e |
| 206 | 2 | 103 | f |
| 207 | 3 | 101 | g |
| 208 | 3 | 102 | h |
| 209 | 3 | 103 | i |
有关简单方法的任何建议吗?
SELECT
IDENTITY(int, 200, 1) IDTABLE_3,
T1.IDTABLE_1,
T2.IDTABLE_2,
CASE T2.COLUMN WHEN 'X' THEN T1.X
WHEN 'Y' THEN T1.Y
WHEN 'Z' THEN T1.Z END VALUE
INTO TABLE_3
FROM TABLE_1 T1
CROSS JOIN TABLE_2 T2
您可以使用UNPIVOT
表运算符执行此操作,如下所示:
WITH CTE
AS
(
SELECT IDTABLE_1, ID, value
FROM Table_1
UNPIVOT
(
ID
FOR value IN(x, y,z)
) u
)
SELECT
ROW_NUMBER() OVER(ORDER BY t1.ID) + 200 AS IDTABLE_3 ,
t1.IDTABLE_1,
t2.[IDTABLE_2],
t1.ID AS VALUE
FROM CTE t1
INNER JOIN TABLE_2 t2 ON t1.value = t2."Column";
这会给你:
| IDTABLE_3 | IDTABLE_1 | IDTABLE_2 | VALUE |
---------------------------------------------
| 201 | 1 | 101 | a |
| 202 | 1 | 102 | b |
| 203 | 1 | 103 | c |
| 204 | 2 | 101 | d |
| 205 | 2 | 102 | e |
| 206 | 2 | 103 | f |
| 207 | 3 | 101 | g |
| 208 | 3 | 102 | h |
| 209 | 3 | 103 | i |
根据您的SQL Server版本,您可以在第一个表上使用UNPIVOT
函数,将row_number()
应用于每个表,然后连接行号。 UNPIVOT
函数获取X
, Y
和Z
列并将它们转换为行。 完成此转换后,您将为每条记录提供由IDTABLE_1
分区的row_number()
。
select row_number() over(order by t1.IDTABLE_1) + 200 AS IDTABLE_3,
t1.IDTABLE_1, t2.IDTABLE_2, t1.value
from
(
select IDTABLE_1, value, col,
row_number() over(partition by IDTABLE_1 order by col) rn
from table1
unpivot
(
value
for col in (X, Y, Z)
) unpiv
) t1
inner join
(
select IDTABLE_2, [COLUMN],
row_number() over(order by IDTABLE_2) rn
from Table2
) t2
on t1.rn = t2.rn
order by t1.IDTABLE_1;
结果是:
| IDTABLE_3 | IDTABLE_1 | IDTABLE_2 | VALUE |
---------------------------------------------
| 201 | 1 | 101 | a |
| 202 | 1 | 102 | b |
| 203 | 1 | 103 | c |
| 204 | 2 | 101 | d |
| 205 | 2 | 102 | e |
| 206 | 2 | 103 | f |
| 207 | 3 | 101 | g |
| 208 | 3 | 102 | h |
| 209 | 3 | 103 | i |
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.