Question

我在varchar（max）列中有以下字符串：

PREV - FirstName: John / LAST - FirstName: Johan; PREV- LastName: Crescot / LAST - LastName: Crescott;

每个分号后可以获得无限量的PREV值和LAST值突变，具体取决于源系统中所做的更改量。

我需要编写一个只返回PREV值的查询。如果是上面的字符串，所需的结果将是：

FirstName: John; LastName: Crescot

所有斜杠（/）分隔符和破折号也需要删除，正如您在所需结果中所见。

有人可以帮我吗？谢谢你们！

Answer 1

如果对UDF开放，请考虑以下内容。

厌倦了提取字符串（charindindex，patindex，left，right ...），我修改了一个解析函数来接受两个不相似的参数。在这种情况下，'PREV'和'/'

示例

Declare @YourTable table (ID int,SomeCol varchar(max)) Insert Into @YourTable values (1,'PREV - FirstName: John / LAST - FirstName: Johan; PREV- LastName: Crescot / LAST - LastName: Crescott;') Select A.ID ,B.NewVal From @YourTable A Cross Apply ( Select NewVal = Stuff((Select '; '+ltrim(rtrim(replace(RetVal,'-',''))) From [dbo].[udf-Str-Extract](A.SomeCol,'PREV','/') For XML Path ('')),1,2,'') ) B

<强>返回

ID NewVal 1 FirstName: John; LastName: Crescot

感兴趣的UDF

CREATE FUNCTION [dbo].[udf-Str-Extract] (@String varchar(max),@Delimiter1 varchar(100),@Delimiter2 varchar(100)) 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 N1,cte1 N2,cte1 N3,cte1 N4,cte1 N5,cte1 N6) A ), cte3(N) As (Select 1 Union All Select t.N+DataLength(@Delimiter1) From cte2 t Where Substring(@String,t.N,DataLength(@Delimiter1)) = @Delimiter1), cte4(N,L) As (Select S.N,IsNull(NullIf(CharIndex(@Delimiter1,@String,s.N),0)-S.N,8000) From cte3 S) Select RetSeq = Row_Number() over (Order By N) ,RetPos = N ,RetVal = left(RetVal,charindex(@Delimiter2,RetVal)-1) From ( Select *,RetVal = Substring(@String, N, L) From cte4 ) A Where charindex(@Delimiter2,RetVal)>1 ) /* Max Length of String 1MM characters Declare @String varchar(max) = 'Dear [[FirstName]] [[LastName]], ...' Select * From [dbo].[udf-Str-Extract] (@String,'[[',']]') */

Answer 2

 create table #temp(val varchar(max))

 Insert into #temp values('PREV - FirstName: John / LAST - FirstName: Johan; PREV - LastName: Crescot / LAST - LastName')

Select stuff(
 (SELECT ';'+ 
  Replace(stuff(Tbl.Col.value('./text()[1]','varchar(50)'),charindex('/',Tbl.Col.value('./text()[1]','varchar(50)')),len(Tbl.Col.value('./text()[1]','varchar(50)')),''),'PREV -','')as ColName
FROM 
(Select cast('<a>'+ replace((SELECT val As [*] FOR XML PATH('')), ';', '</a><a>') + '</a>' as xml)as t 
 from #temp) tl
Cross apply 
tl.t.nodes('/a') AS Tbl(Col) for xml path(''),type).value('.','NVARCHAR(MAX)'),1,2,'')

此方法不需要任何其他UDF。分解以上查询以便于理解： 1.基于分号';'

将一行字符串转换为多行

SELECT 
 Tbl.Col.value('./text()[1]','varchar(50)')
FROM 
(Select cast('<a>'+ replace((SELECT val As [*] FOR XML PATH('')), ';', '</a><a>') + '</a>' as xml)as t 
  from #temp) tl
Cross apply 
tl.t.nodes('/a') AS Tbl(Col)

2.在上面提取的值中，使用replace和stuff命令删除不必要的字符

        SELECT 
   Replace(stuff(Tbl.Col.value('./text()[1]','varchar(50)'),charindex('/',Tbl.Col.value('./text()[1]','varchar(50)')),len(Tbl.Col.value('./text()[1]','varchar(50)')),''),'PREV -','')as ColName
FROM 
(Select cast('<a>'+ replace((SELECT val As [*] FOR XML PATH('')), ';', '</a><a>') + '</a>' as xml)as t 
  from #temp) tl
Cross apply 
tl.t.nodes('/a') AS Tbl(Col)

3。使用stuff和xml路径将多行返回到一行，根据需要用分号分隔

     Select stuff(
(SELECT ';'+ 
 Replace(stuff(Tbl.Col.value('./text()[1]','varchar(50)'),charindex('/',Tbl.Col.value('./text()[1]','varchar(50)')),len(Tbl.Col.value('./text()[1]','varchar(50)')),''),'PREV -','')as food_Name
FROM 
(Select cast('<a>'+ replace((SELECT val As [*] FOR XML PATH('')), ';', '</a><a>') + '</a>' as xml)as t 
  from #temp) tl
 Cross apply 
tl.t.nodes('/a') AS Tbl(Col) for xml path(''),type).value('.','NVARCHAR(MAX)'),1,2,'')

SQL查询：删除long varchar（max）列中的多个字符串

2 个答案: