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小写转大写金额存储过程

  【IT168技术文档】

  下面是个SQL Server的小写转大写金额存储过程:

CREATE 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 GO

  调用过程:

declare @ret varchar(200) exec L2U 567983.897,1,@ret output select @ret
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