Question

- 我在sql server中需要像F11.20，F13.20，F14.10这样的数据

declare @S varchar(200) = ',F11.20:,F13.20:Sedative, hypnotic o,F14.10:Cocaine abuse, uncom';
select left(@S, charindex(':', @S, charindex(':', @S)+2)-2);

Answer 1

任何分割/解析函数都可以解决问题，但您必须执行辅助逻辑来清理已解析的字符串。也就是说，我修改了一个解析函数来接受任何两个非类似的分隔符（开始/结束）。在这种情况下，a和：

此外，作为一个表格值函数，很容易合并到一个交叉应用程序中或作为一个独立程序，如下所示。

示例

Select NewString = Stuff((Select ',' +RetVal From [dbo].[udf-Str-Extract](@S,',',':') For XML Path ('')),1,1,'')

<强>返回

F11.20,F13.20,F14.10

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,'[[',']]') */

编辑只是为了帮助实现可视化

如果您单独执行TVF：

declare @S varchar(200) = ',F11.20:,F13.20:Sedative, hypnotic o,F14.10:Cocaine abuse, uncom'; Select * From [dbo].[udf-Str-Extract](@S,',',':')

<强>返回

RetSeq RetPos RetVal 1 2 F11.20 2 10 F13.20 3 38 F14.10

编辑2 - 通过交叉申请执行

Declare @YourTable table (ID int,SomeString varchar(200)) Insert Into @YourTable values (1,',F11.20:,F13.20:Sedative, hypnotic o,F14.10:Cocaine abuse, uncom'), (2,',Z99.55:,Z25.10:Someother text') Select A.ID ,B.* From @YourTable A Cross Apply ( Select NewString = Stuff((Select ',' +RetVal From [dbo].[udf-Str-Extract](A.SomeString,',',':') For XML Path ('')),1,1,'') ) B

<强>返回

ID NewString 1 F11.20,F13.20,F14.10 2 Z99.55,Z25.10

Answer 2

如果您提取的文本总是以（字母，数字，数字，。，数字，数字）的形式出现，并且该文本总是有3个实例，那么您可以这样做：

WITH 
s1(string, ci) AS (SELECT @S, CHARINDEX(':', @S)),
s2(ci) AS (SELECT CHARINDEX(':', @S, ci+1) FROM s1),
s3(ci) AS (SELECT CHARINDEX(':', @S, ci+1) FROM s2)
SELECT 
  SUBSTRING(string, s1.ci-6, 6)+','+
  SUBSTRING(string, s2.ci-6, 6)+','+
  SUBSTRING(string, s3.ci-6, 6)
FROM s1, s2, s3;

执行计划：

它没有那么高效。

如果它总是在任何“：”实例之前的6个字符，你可以抓取NGrams8K的副本并执行此操作：

declare @S varchar(200) = ',F11.20:,F13.20:Sedative, hypnotic o,F14.10:Cocaine abuse, uncom';

SELECT NewString = STUFF
  ((SELECT ','+SUBSTRING(@S, position-6, 6)
    FROM dbo.NGrams8k(@S, 1)
    WHERE token = ':'
    FOR XML PATH('')),1,1,'');

使用NGrams8K和变量的另一种方式：

declare @S varchar(200) = ',F11.20:,F13.20:Sedative, hypnotic o,F14.10:Cocaine abuse, uncom';
declare @newstring varchar(100)='';

declare @S varchar(200) = ',F11.20:,F13.20:Sedative, hypnotic o,F14.10:Cocaine abuse, uncom';
declare @newstring varchar(100)='';

SELECT 
  @newstring += 
    CASE @newstring WHEN '' THEN '' ELSE ',' END +SUBSTRING(@S, position-6, 6)
FROM dbo.NGrams8k(@S, 1)
WHERE token = ':';

SELECT @newstring;

我想在sql server中删除冒号后的字符串

2 个答案: