日期:2014-05-17 浏览次数:20706 次
Create function AtoC
(@ChangeMoney Money )
retuens varchar(100)
as
Set Nocount ON
Declare @String1 char(20)
Declare @String2 char(30)
Declare @String4 Varchar(100)
Declare @String3 Varchar(100) –从原A值中取出的值
Declare @i int –循环变量
Declare @J Int –A的值乘以100的字符串长度
Declare @Ch1 Varchar(100) –数字的汉语读法
Declare @Ch2 Varchar(100) –数字位的汉字读法
Declare @Zero Int –用来计算连续有几个零
Declare @ReturnValue VarChar(100)
Select @ReturnValue = ”
Select @String1 = ‘零壹贰叁肆伍陆柒捌玖’
Select @String2 = ‘万仟佰拾亿仟佰拾万仟佰拾元角分’
Select @String4 = Cast(@ChangeMoney*100 as int)
select @J=len(cast((@ChangeMoney*100) as int))
Select @String2=Right(@String2,@J)
Select @i = 1
while @i<= @j Begin
Select @String3 = Substring(@String4,@i,1)
if @String3<>‘0′ Begin
Select @Ch1 = Substring(@String1, Cast(@String3 as Int) + 1, 1)
Select @Ch2 = Substring(@String2, @i, 1)
Select @Zero = 0 –表示本位不为零
end
else Begin
If (@Zero = 0) Or (@i = @J – 9) Or (@i = @J – 5) Or (@i = @J – 1)
Select @Ch1 = ‘零’
Else
Select @Ch1 = ”
Select @Zero = @Zero + 1 –表示本位为0
–如果转换的数值需要扩大,那么需改动以下表达式 I 的值。
Select Ch2 = ”
If @i = @J – 10 Begin
Select @Ch2 = ‘亿’
Select @Zero = 0
end
If @i = @J – 6 Begin
Select @Ch2 = ‘万’
Select @Zero = 0
end
if @i = @J – 2 Begin
Select @Ch2 = ‘元’
Select @Zero = 0
end
If @i = @J
Select @Ch2 = ‘整’
end
Select @ReturnValue = @ReturnValue + @Ch1 + @Ch2
select @i = @i+1
end
–最后将多余的零去掉
If CharIndex(‘仟仟’,@ReturnValue) <> 0
Select @ReturnValue = Replace(@ReturnValue, ‘仟仟’, ‘仟’)
If CharIndex(‘佰佰’,@ReturnValue) <> 0
Select @ReturnValue = Replace(@ReturnValue, ‘佰佰’, ‘佰’)
If CharIndex(‘零元’,@ReturnValue) <> 0
Select @ReturnValue = Replace(@ReturnValue, ‘零元’, ‘元’)
If CharIndex(‘零万’,@ReturnValue) <> 0
Select @ReturnValue = Replace(@ReturnValue, ‘零万’, ‘万’)
If CharIndex(‘零亿’,@ReturnValue) <> 0
S