例如,从表alpha
中可以看出以下内容:
Field1 | Field2 | Field3
------------------------
Foo | Bar | ABCD
我怎样才能将这些数据分解为:
Field1 | Field2 | Field3
------------------------
Foo | Bar | A
Foo | Bar | B
Foo | Bar | C
Foo | Bar | D
我确信有一种奇特的join
技巧可以做到,但我无法弄清楚。速度优化不是优先考虑的问题 - 这个查询只用于一次性报告,所以我不介意它是否像糖蜜一样慢(让我有机会喝咖啡!)
答案 0 :(得分: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
<强> SQLFiddle DEMO 强>
答案 1 :(得分:1)
您可以通过以下步骤轻松完成:
步骤1:创建一个可以将单词拆分为字符的sql表值函数。你可以通过运行以下脚本来完成它。
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,它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
答案 2 :(得分:1)
类似的东西:
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)
(除非你已经有一个方便的Numbers
表,你在问题中没有提到,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
答案 3 :(得分: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
答案 4 :(得分: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'
任何方式looping is not a good method
但我们仍然需要这样做,因为Filed3 ABCD
是dynamic