Oracle实现金额小写转大写函数
今天工作的时候遇到了，然后上网百度了一下，发现了有个大佬写了一个很牛的函数，在此记录下来。
原文：http://www.itpub.net/thread-240281-1-1.html
函数：
[code]create or replace function F_upper_money(p_num in number default null)return nvarchar2 is/*Ver:1.0 Created By xsb on 2003-8-18 For: 将金额数字(单位元)转换为大写(采用从低至高算法) 数字整数部分不得超过16位,可以是负数。Ver：1.1 Modified By xsb on 2003-8-20 For:个位数处理也放在For循环中。 Ver：1.2 Modified By xsb on 2003-8-22 For:分后不带整字。Ver：1.3 Modified By xsb on 2003-8-28 For:完善测试用例。测试用例:SET HEAD OFFSET FEED OFFselect '无参数时='||f_upper_money() from dual;select 'null='||f_upper_money(null) from dual;select '0='||f_upper_money(0) from dual;select '0.01='||f_upper_money(0.01) from dual;select '0.126='||f_upper_money(0.126) from dual;select '01.234='||f_upper_money(01.234) from dual;select '10='||f_upper_money(10) from dual;select '100.1='||f_upper_money(100.1) from dual;select '100.01='||f_upper_money(100.01) from dual;select '10000='||f_upper_money(10000) from dual;select '10012.12='||f_upper_money(10012.12) from dual; select '20000020.01='||f_upper_money(20000020.01) from dual;select '3040506708.901='||f_upper_money(3040506708.901) from dual;select '40005006078.001='||f_upper_money(40005006078.001) from dual;select '-123456789.98='||f_upper_money(-123456789.98) from dual;select '123456789123456789.89='||f_upper_money(123456789123456789.89) from dual;*/Result nvarchar2(100);--返回字符串num_round nvarchar2(100) :=to_char(abs(round(p_num,2)));--转换数字为小数点后2位的字符(正数)num_left nvarchar2(100);--小数点左边的数字num_right nvarchar2(2);--小数点右边的数字str1 nchar(10) :='零壹贰参肆伍陆柒捌玖';--数字大写str2 nchar(16) :='元拾佰仟万拾佰仟亿拾佰仟万拾佰仟';--数字位数(从低至高)num_pre number(1):=1;--前一位上的数字num_current number(1);--当前位上的数字num_count number:=0;--当前数字位数beginif p_num is null then return null;end if;--转换数字为null时返回nullselect to_char(nvl(substr(to_char(num_round),1,decode(instr(to_char(num_round),'.'),0,length(num_round),instr(to_char(num_round),'.')-1)),0)) into num_left from dual;--取得小数点左边的数字select substr(to_char(num_round),decode(instr(to_char(num_round),'.'),0,length(num_round)+1,instr(to_char(num_round),'.')+1),2)into num_right from dual;--取得小数点右边的数字if length(num_left)>16 then return '**********'; end if;--数字整数部分超过16位时 --采用从低至高的算法，先处理小数点右边的数字 if length(num_right)=2 then if to_number(substr(num_right,1,1))=0 then result:='零'||substr(str1,to_number(substr(num_right,2,1))+1,1)||'分'; else result:=substr(str1,to_number(substr(num_right,1,1))+1,1)||'角'|| substr(str1,to_number(substr(num_right,2,1))+1,1)||'分'; end if; elsif length(num_right)=1 then result:=substr(str1,to_number(substr(num_right,1,1))+1,1)||'角整'; else result :='整'; end if; --再处理小数点左边的数字 for i in reverse 1..length(num_left) loop --(从低至高) num_count:=num_count+1;--当前数字位数 num_current:=to_number(substr(num_left,i,1));--当前位上的数字 if num_current>0 then --当前位上数字不为0按正常处理 result:=substr(str1,num_current+1,1)||substr(str2,num_count,1)||result; else --当前位上数字为0时 if mod(num_count-1,4)=0 then --当前位是元、万或亿时 result:=substr(str2,num_count,1)||result; num_pre:=0;--元、万,亿前不准加零 end if; if num_pre>0 or length(num_left)=1 then --上一位数字不为0或只有个位时 result:=substr(str1,num_current+1,1)||result; end if; end if; num_pre:=num_current; end loop; if p_num |
