SQL - 将字段的数据分解为单独的行

时间:2013-07-08 09:18:30

标签: sql sql-server

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!)

5 个答案:

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

答案 1 :(得分:1)

您可以通过以下步骤轻松完成:

  1. 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. 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 ABCDdynamic