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--前提,必须确保[]成对出现.
declare @s varchar(20)
declare @i int
declare @j int
declare @s1 varchar(20)
set @s = '某某[12345]'
set @i = charindex('[',@s)
set @j = charindex(']',@s,@i )
select @s1 = substring(@s,@i+1,@j- @i-1)
print @s1
declare @s varchar(20) = '某某[12345]';
select left (STUFF(@s,1,charindex('[',@s),''),CHARINDEX(']',STUFF(@s,1,charindex('[',@s),''))-1)
declare @s varchar(100)
set @s='某某[12345]'
select SUBSTRING(@s,charindex('[',@s)+1,CHARINDEX(']',@s)-charindex('[',@s)-1)
/*----------------------------------------------------------------------------------------------------
12345
(1 行受影响)
*/