• 暴力求值


    1.0(3),2.0(4),3.5(5),3.5111(78),55.98(23345)
    12.0(2),3.5(5)
    4567.0(344),234.43(2323),3.5(5),3.5111(78),55.98(23345)
    
    
    
    1307155.46580
    41.50000
    3422773.35580
    
    
    
    create function toolkindvalues(@str varchar(8000) )
    returns decimal(18,5)
    as 
    begin
    declare @result decimal(18,5)
    set @result=0
    declare @charindex int 
    set  @charindex=0
    declare @replacestr varchar(500) 
    
    while  (charindex(',',@str,0)>0)
    begin
     
    select  @charindex=charindex(',',@str,0)
    select @replacestr=substring (@str,0,@charindex)
    
    select  @str= replace( @str,@replacestr+',','')
    
     --select   @replacestr
     --select   left(@replacestr, charindex( '(',@replacestr,0)-1),left( right(@replacestr,2),len(right(@replacestr,2))-1)
    select  @result=@result+ cast( left(@replacestr, charindex( '(',@replacestr,0)-1)  as decimal(18,5))  *  cast ( substring( replace( @replacestr,left(@replacestr, charindex( '(',@replacestr,0)-1),''),2,len(replace( @replacestr,left(@replacestr, charindex( '(',@replacestr,0)-1),''))-2)   as decimal(18,5) )
    end 
    
     --select   @str
    
    select   @result=@result+ cast(  left(@str, charindex( '(',@str,0)-1) as decimal(18,5))  *   cast ( substring( replace( @str,left(@str, charindex( '(',@str,0)-1),''),2,len(replace( @str,left(@str, charindex( '(',@str,0)-1),''))-2)  as decimal(18,5) )
    
    
    return @result
    
    end 
    
    
    
    
    --drop  table #ddd
    
    create table #ddd(str varchar(500))
    
    
    
    insert #ddd
    
    select '1.0(3),2.0(4),3.5(5),3.5111(78),55.98(23345)'   
    union 
    select '4567.0(344),234.43(2323),3.5(5),3.5111(78),55.98(23345)'   
    union 
    select '12.0(2),3.5(5)'   
    
    
    select  *  from  #ddd
    
    select dbo.toolkindvalues(str)   from #ddd
    
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  • 原文地址:https://www.cnblogs.com/qanholas/p/2081787.html
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