金额大写转换

USE [toy]
GO
/****** Object:  StoredProcedure [dbo].[L2U]    Script Date: 03/19/2014 13:15:27 ******/
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
ALTER PROCEDURE [dbo].[L2U] 
(
@n_LowerMoney numeric(15,2),
@v_TransType int,
@RET VARCHAR(200) output
)
AS 

Declare @v_LowerStr VARCHAR(200) -- 小写金额 
Declare @v_UpperPart VARCHAR(200) 
Declare @v_UpperStr VARCHAR(200) -- 大写金额
Declare @i_I int

set nocount On

select @v_LowerStr = LTRIM(RTRIM(STR(@n_LowerMoney,20,2))) --四舍五入为指定的精度并删除数据左右空格

select @i_I = 1
select @v_UpperStr = ''''

while ( @i_I <= len(@v_LowerStr))
begin
select @v_UpperPart = case substring(@v_LowerStr,len(@v_LowerStr) - @i_I + 1,1)
WHEN '.' THEN ''
WHEN '0' THEN ''
WHEN '1' THEN ''
WHEN '2' THEN ''
WHEN '3' THEN ''
WHEN '4' THEN ''
WHEN '5' THEN ''
WHEN '6' THEN ''
WHEN '7' THEN ''
WHEN '8' THEN ''
WHEN '9' THEN ''
END
+ 
case @i_I
WHEN 1 THEN ''
WHEN 2 THEN ''
WHEN 3 THEN ''
WHEN 4 THEN ''
WHEN 5 THEN ''
WHEN 6 THEN ''
WHEN 7 THEN ''
WHEN 8 THEN ''
WHEN 9 THEN ''
WHEN 10 THEN ''
WHEN 11 THEN ''
WHEN 12 THEN '亿'
WHEN 13 THEN ''
WHEN 14 THEN ''
WHEN 15 THEN ''
WHEN 16 THEN ''
ELSE''''
END
select @v_UpperStr = @v_UpperPart + @v_UpperStr
select @i_I = @i_I + 1
end

--------print '//v_UpperStr ='+@v_UpperStr +'//'

if ( @v_TransType=0 )
begin
select @v_UpperStr = REPLACE(@v_UpperStr,'零拾','') 
select @v_UpperStr = REPLACE(@v_UpperStr,'零佰','') 
select @v_UpperStr = REPLACE(@v_UpperStr,'零仟','') 
select @v_UpperStr = REPLACE(@v_UpperStr,'零零零','')
select @v_UpperStr = REPLACE(@v_UpperStr,'零零','')
select @v_UpperStr = REPLACE(@v_UpperStr,'零角零分','')
select @v_UpperStr = REPLACE(@v_UpperStr,'零分','')
select @v_UpperStr = REPLACE(@v_UpperStr,'零角','')
select @v_UpperStr = REPLACE(@v_UpperStr,'零亿零万零元','亿元')
select @v_UpperStr = REPLACE(@v_UpperStr,'亿零万零元','亿元')
select @v_UpperStr = REPLACE(@v_UpperStr,'零亿零万','亿')
select @v_UpperStr = REPLACE(@v_UpperStr,'零万零元','万元')
select @v_UpperStr = REPLACE(@v_UpperStr,'万零元','万元')
select @v_UpperStr = REPLACE(@v_UpperStr,'零亿','亿')
select @v_UpperStr = REPLACE(@v_UpperStr,'零万','')
select @v_UpperStr = REPLACE(@v_UpperStr,'零元','')
select @v_UpperStr = REPLACE(@v_UpperStr,'零零','')
end

-- 对壹元以下的金额的处理 
if ( substring(@v_UpperStr,1,1)='' )
begin
select @v_UpperStr = substring(@v_UpperStr,2,(len(@v_UpperStr) - 1))
end

if (substring(@v_UpperStr,1,1)= '')
begin
select @v_UpperStr = substring(@v_UpperStr,2,(len(@v_UpperStr) - 1))
end

if (substring(@v_UpperStr,1,1)='')
begin
select @v_UpperStr = substring(@v_UpperStr,2,(len(@v_UpperStr) - 1))
end

if ( substring(@v_UpperStr,1,1)='')
begin
select @v_UpperStr = substring(@v_UpperStr,2,(len(@v_UpperStr) - 1))
end

if (substring(@v_UpperStr,1,1)='')
begin
select @v_UpperStr = '零元整'
end

select @ret=@v_UpperStr

 

posted @ 2014-03-19 13:17  美丽的矩阵  阅读(249)  评论(0编辑  收藏  举报