Question

我在下面有这个功能我正在打破一个字符串，例如 string =＆＃39; 100001 | 125000,50001 - 75000＆＃39;下面的结果表有一个更简单的方法吗？

     low     high
     100001  125000
     50001   75000


ALTER FUNCTION [dbo].[GetRange] (@String varchar(max))
returns @temptable TABLE (low int, high int)  
as       
begin       

declare @separated table(items varchar(max))

insert into @separated select * from dbo.Split(@String, '|')

update @separated
set items = items + ',2147483647'
where charindex(',', items) = 0

insert into @temptable 
select left(items, charindex(',', items) - 1)
     , right(items, len(items) - charindex(',', items)) 
       from @separated
return       
end

Answer 1

嗯......您可以删除拆分功能并将其替换为UNION，并且可能更快。这将消除您的功能和嵌套功能。我想这取决于你的用例。它至少会移除您的dbo.Split功能，该功能希望基于注意事项from this article.，但如果不是，则可能会非常慢。

declare @table table (a varchar(4000))

insert into @table (a) values
('100001|125000,50001 - 75000'),
('101321320001|1250132100,52130001 - 755000')

select
    left(left(a,charindex(',',a) - 1),charindex('|',left(a,charindex(',',a) - 1)) - 1) as Low
    ,right(left(a,charindex(',',a) - 1),len(left(a,charindex(',',a) - 1)) - charindex('|',left(a,charindex(',',a) - 1))) as High
from @table
union all
select
    left(right(a,len(a) - charindex(',',a)),CHARINDEX('-',right(a,len(a) - charindex(',',a))) - 1)
    ,right(right(a,len(a) - charindex(',',a)),len(right(a,len(a) - charindex(',',a))) - charindex('-',right(a,len(a) - charindex(',',a))))
from @table

Answer 2

注意：这是假设分隔符是逗号并且范围可以用|分割或破折号。您可能会注意到破折号被替换为|为了一致性。

CREATE FUNCTION [dbo].[TestFunction] (@String varchar(max))
Returns Table 
As
Return (  
    Select Low  = try_convert(int,Left(RetVal,charindex('|',RetVal)-1))
          ,High = try_convert(int,Right(RetVal,charindex('|',reverse(RetVal))-1))
    From  (
            Select RetVal = LTrim(RTrim(B.i.value('(./text())[1]', 'varchar(500)')))
            From  (Select x = Cast('<x>' + replace((Select replace(replace(@String,'-','|'),',','§§Split§§') as [*] For XML Path('')),'§§Split§§','</x><x>')+'</x>' as xml).query('.')) as X
            Cross Apply x.nodes('x') AS B(i)
          ) B
)


Select * from [dbo].[TestFunction]('100001|125000,50001 - 75000')

返回

Low     High
100001  125000
50001   75000

更简单的方法来分解字符串sql server

2 个答案: