日期:2014-05-17  浏览次数:20609 次

一个16进制转10进制函数运行错误?
CREATE FUNCTION dbo.f_hex_dec(@s varchar(32))
RETURNS bigint
AS
BEGIN
  DECLARE @i bigint,@result bigint
  SELECT @i=0,@result=0,@s=RTRIM(LTRIM(UPPER(REVERSE(@s))))
  WHILE @i<LEN(@s)
  BEGIN
  IF SUBSTRING(@s,@i+1,1) not between '0' and '9' and SUBSTRING(@s,@i+1,1) not between 'A' and 'F'
  BEGIN
  SELECT @result=0
  break
  END
  SELECT @result=@result+(CHARINDEX(SUBSTRING(@s,@i+1,1),'0123456789ABCDEF')-1)*POWER(16,@i),@i=@i+1
  END
  RETURN @result
END
GO

调用函数运行:select oid from lists where oid=dbo.f_hex_dec('7a7a003500bc0001') and state=0 and tid=13出现错误为:类型 int 发生算术溢出错误,值 = 4294967296.000000。
为什么呢?RETURNS为bigint,@i bigint,@result bigint都为bigint,这些来存储16个16进制数应该没问题啊?请知道者帮忙修改下代码?谢谢了

------解决方案--------------------
修改后

SQL code
create FUNCTION dbo.f_hex_dec(@s varchar(max))  
RETURNS bigint  
AS  
BEGIN  
    DECLARE @i BIGINT ,  
        @result BIGINT 
         
    DECLARE @data BIGINT  --添加了这个
    SET @data =16
    
    SELECT  @i = 0 ,  
            @result = 0 ,  
            @s = RTRIM(LTRIM(UPPER(REVERSE(@s))))  
    WHILE @i < DATALENGTH(@s)   
        BEGIN  
            IF SUBSTRING(@s, @i + 1, 1) NOT BETWEEN '0' AND '9'  
                AND SUBSTRING(@s, @i + 1, 1) NOT BETWEEN 'A' AND 'F'   
                BEGIN  
                    SELECT  @result = 0  
                    BREAK  
                END  
            SELECT  @result = @result + ( CHARINDEX(SUBSTRING(@s, @i + 1, 1),  
                                                    '0123456789ABCDEF') - 1 )  
                    * POWER(@data, @i) ,  
                    @i = @i + 1  
        END  
    RETURN @result  
END  

SELECT dbo.f_hex_dec('7a7a003500bc0001')

--8825366647431495681