0

例如，给定表中的以下内容alpha：

Field1 | Field2 | Field3
------------------------
Foo    | Bar    | ABCD

我怎么能把这些数据分解成：

Field1 | Field2 | Field3
------------------------
Foo    | Bar    | A
Foo    | Bar    | B
Foo    | Bar    | C
Foo    | Bar    | D

我敢肯定有一个花哨的join技巧可以做到这一点，但我无法弄清楚。速度优化不是优先事项 - 此查询仅用于一次性报告，所以我不介意它是否像糖蜜一样慢（让我有机会煮咖啡！）

4

5 回答 5

1

您可以尝试以下操作：

WITH CTE_LenF3 AS 
(
  -- find the length of each field3
  SELECT Field1, Field2, LEN(Field3) as Len_F3
  FROM alpha
)
,CTE_Numbers AS 
(
 --generate numbers from 1 to LEN(Filed3) for each field1,field2 combination
  SELECT Field1, Field2, 1 AS Nmb FROM CTE_LenF3
  UNION ALL
  SELECT c.Field1, c.Field2, Nmb + 1 FROM CTE_Numbers n
  INNER JOIN CTE_LenF3 c ON c.Field1 = n.Field1 AND c.Field2 = n.Field2
  WHERE Nmb + 1 <= LEN_F3
)
--join generated numbers with actual table and use substring to get the characters
SELECT a.Field1, a.Field2, SUBSTRING(a.Field3, n.Nmb, 1)
FROM CTE_Numbers n
INNER JOIN alpha a ON a.Field1 = n.Field1 AND a.Field2 = n.Field2
ORDER BY a.Field1, a.Field2, n.Nmb

SQLFiddle 演示

于 2013-07-08T09:33:56.220 回答
1

您可以通过以下步骤轻松完成：

  1. Step1：创建一个可以将单词拆分为字符的sql表值函数。您可以通过运行以下脚本来做到这一点。

    CREATE FUNCTION [dbo].[SPLITWORD](
@WORD VARCHAR(MAX) 
) RETURNS @words TABLE (item VARCHAR(8000))

 BEGIN
 declare @count int, @total int
 select @total = len(@WORD), @count = 0

 while @count <= @total
 begin
   insert into @words select substring(@WORD, @count, 1)
   select @count = @count + 1
 end

 RETURN
END

2.Steps 运行以下查询，将返回您想要的结果。

    SELECT A.FIELD1 , A.Field2 , B.ITEM
    FROM alpha AS A
    CROSS APPLY
    (
    SELECT * FROM SPLITWORD(A.Field3) WHERE ITEM != ''
    ) AS B
于 2013-07-08T09:34:30.593 回答
1

就像是：

declare @alpha table (Field1 varchar(20), Field2 varchar(20), Field3 varchar(6))
insert into @alpha(Field1, Field2, Field3) values
('Foo','Bar','ABCD')

;With Numbers(n) as (
    select 1 union all select 2 union all select 3 union all
    select 4 union all select 5 union all select 6
)
select Field1,Field2,SUBSTRING(Field3,n,1)
from
    @alpha
        inner join
    Numbers
        on
            n <= LEN(Field3)

（除非您已经有一个Numbers您在问题中没有提到的方便表，在这种情况下它更简单）

结果：

Field1               Field2               
-------------------- -------------------- ----
Foo                  Bar                  A
Foo                  Bar                  B
Foo                  Bar                  C
Foo                  Bar                  D
于 2013-07-08T09:35:15.727 回答
0

以下查询以任何方式实现您的功能

DECLARE @temp as table(newFiled3 varchar(1))
DECLARE @str_Value varchar(50),@count int,@i int=1
SET @str_Value=(SELECT Field3 FROM alpha)
SET @count=LEN(@str_Value)
WHILE(@i<=@count)
BEGIN
    INSERT INTO  @temp VALUES (SUBSTRING ( @str_Value ,@i , 1 ))
SET @i=@i+1
END


SELECT Field1,Field2,b.newFiled3 
FROM tblStudent a inner join @temp b ON a.Field1='Foo'

无论如何looping is not a good method，但我们仍然需要这样做，因为你Filed3 ABCD是dynamic

于 2013-07-08T09:38:28.677 回答
0 
DECLARE @xml as xml,@str as varchar(100),@delimiter as varchar(10)
SET @str='A,B,C,D,E'
SET @delimiter =','
SET @xml = cast(('<X>'+replace(@str,@delimiter ,'</X><X>')+'</X>') as xml)
SELECT N.value('.', 'varchar(10)') as value FROM @xml.nodes('X') as T(N)

替换你的分隔符

于 2013-07-08T09:35:30.807 回答