T-SQL 在分隔符上拆分

Question

我正在使用以下格式的员工层次结构字符串。这些数字代表employeeID 数字以及它们在公司内部的结构，从而能够遵循管理链。

123|456|789|012|345|320

我正在尝试获取此数据字符串并将其转换为临时表，以便我可以将每个 ID 用作它们自己的值。

我尝试制作一个函数来拆分字符串：

ALTER FUNCTION [dbo].[SplitString]
    (@String NVARCHAR(4000),
     @Delimiter NCHAR(1))
RETURNS TABLE
AS
    RETURN
        (WITH Split(stpos, endpos) AS
         (
             SELECT 0 AS stpos, CHARINDEX(@Delimiter, @String) AS endpos
             UNION ALL
             SELECT endpos + 1, CHARINDEX(@Delimiter, @String, endpos+1)
             FROM Split
             WHERE endpos > 0
         )
         SELECT 
             'Id' = ROW_NUMBER() OVER (ORDER BY (SELECT 1)),
             'Data' = SUBSTRING(@String, stpos, COALESCE(NULLIF(endpos, 0), LEN(@String) + 1))
         FROM 
             Split
)

然而，这导致了以下结果：

Id  Data 
-------------------
1   123
2   456|7893
3   7893|012|345|
4   012|345|320
5   345|320
6   320

有没有更好的方法来解决这个问题，可能根本不需要函数或者需要它来实现这个？

Answer 1

我使用这个版本的拆分功能。

CREATE FUNCTION [dbo].[Split]
(
  @delimited nvarchar(max),
  @delimiter nvarchar(100)
) RETURNS @t TABLE
(
-- Id column can be commented out, not required for sql splitting string
  id int identity(1,1), -- I use this column for numbering splitted parts
  val nvarchar(max)
)
AS
BEGIN
  declare @xml xml
  set @xml = N'<root><r>' + replace(@delimited,@delimiter,'</r><r>') + '</r></root>'

  insert into @t(val)
  select
    r.value('.','varchar(max)') as item
  from @xml.nodes('//root/r') as records(r)

  RETURN
END

你的查询看起来像....

SELECT * 
FROM TableName t 
  CROSS APPLY [dbo].[Split](t.EmpIDs, '|')

Answer 2

没有解析函数

Declare @YourTable table (ID int,IDList varchar(Max))
Insert Into @YourTable values
(1,'123|456|789|012|345|320'),
(2,'123|456')

Select A.ID
      ,B.*
 From @YourTable A
 Cross Apply (
                Select RetSeq = Row_Number() over (Order By (Select null))
                      ,RetVal = LTrim(RTrim(B.i.value('(./text())[1]', 'varchar(max)')))
                From (Select x = Cast('<x>'+ replace((Select A.IDList as [*] For XML Path('')),'|','</x><x>')+'</x>' as xml).query('.')) as A  
                Cross Apply x.nodes('x') AS B(i)
             ) B

返回

ID  RetSeq  RetVal
1   1       123
1   2       456
1   3       789
1   4       012
1   5       345
1   6       320
2   1       123
2   2       456

或使用 SUPER DUPER Parse（下面列出的原始来源/一些调整）

Select A.ID
      ,B.*
 From @YourTable A
 Cross Apply [dbo].[udf-Str-Parse-8K](A.IDList,'|') B

返回的结果和上面一样

CREATE FUNCTION [dbo].[udf-Str-Parse-8K] (@String varchar(max),@Delimiter varchar(10))
Returns Table 
As
Return (  
    with   cte1(N)   As (Select 1 From (Values(1),(1),(1),(1),(1),(1),(1),(1),(1),(1)) N(N)),
           cte2(N)   As (Select Top (IsNull(DataLength(@String),0)) Row_Number() over (Order By (Select NULL)) From (Select N=1 From cte1 a,cte1 b,cte1 c,cte1 d) A ),
           cte3(N)   As (Select 1 Union All Select t.N+DataLength(@Delimiter) From cte2 t Where Substring(@String,t.N,DataLength(@Delimiter)) = @Delimiter),
           cte4(N,L) As (Select S.N,IsNull(NullIf(CharIndex(@Delimiter,@String,s.N),0)-S.N,8000) From cte3 S)

    Select RetSeq = Row_Number() over (Order By A.N)
          ,RetVal = Substring(@String, A.N, A.L) 
    From   cte4 A
);
--Orginal Source http://www.sqlservercentral.com/articles/Tally+Table/72993/
--Much faster than str-Parse, but limited to 8K
--Select * from [dbo].[udf-Str-Parse-8K]('Dog,Cat,House,Car',',')
--Select * from [dbo].[udf-Str-Parse-8K]('John||Cappelletti||was||here','||')

编辑 - 独立

Declare @String varchar(max) = '123|456|789|012|345|320'
Declare @Delim  varchar(10)  = '|'

Select RetSeq = Row_Number() over (Order By (Select null))
      ,RetVal = LTrim(RTrim(B.i.value('(./text())[1]', 'varchar(max)')))
From (Select x = Cast('<x>'+ replace((Select @String as [*] For XML Path('')),@Delim,'</x><x>')+'</x>' as xml).query('.')) as A 
Cross Apply x.nodes('x') AS B(i)

Answer 3

如果您需要一个字符串“拆分器”，可以找到 2012 年（2016 年之前）可用的最快的字符串 here。这将把迄今为止发布的任何内容都吹走。如果您的物品/令牌大小相同，那么更快的方法是：

DECLARE @yourstring varchar(8000) = '123|456|789|012|345|320';

WITH E(N) AS (SELECT 1 FROM (VALUES (1),(1),(1),(1),(1),(1),(1),(1),(1),(1))t(v)),
iTally(N) AS (SELECT TOP ((LEN(@yourstring)/4)+1) ROW_NUMBER() OVER (ORDER BY (SELECT 1)) 
              FROM e a, e b, e c, e d)
SELECT itemNumber = ROW_NUMBER() OVER (ORDER BY N), item = SUBSTRING(@yourstring, ((N*4)-3), 3) 
FROM iTally;

结果：

itemNumber           item
-------------------- ----
1                    123
2                    456
3                    789
4                    012
5                    345
6                    320

我写了更多关于这方面的文章，并提供了如何将此逻辑放入函数here 的示例。