可以通过to_number函数实现
| SQL> select to_number(’19f’,’xxx’) from dual; TO_NUMBER(’19F’,’XXX’) ---------------------- 415 SQL> select to_number(’f’,’xx’) from dual; TO_NUMBER(’F’,’XX’) ------------------- 15 |
二 10进制转换为16进制
可以通过to_char函数转换
| SQL> select to_char(123,’xxx’) from dual; TO_C ---- 7b SQL> select to_char(4567,’xxxx’) from dual; TO_CH ----- 11d7 |
三 2进制转换为10进制
从Oracle9i开始,提供函数bin_to_num进行2进制到10进制的转换
| SQL> select bin_to_num(1,1,0,1) a,bin_to_num(1,0) b from dual; A B ----- ---------- 13 2 SQL> select bin_to_num(1,1,1,0,1) from dual; BIN_TO_NUM(1,1,1,0,1) --------------------- 29 |
四 通过自定义函数实现进制转换
以下函数来自AskTom网站,是Tom给出的例子,供参考:
| create or replace function to_base( p_dec in number, p_base in number ) return varchar2 is l_str varchar2(255) default NULL; l_num number default p_dec; l_hex varchar2(16) default ’0123456789ABCDEF’; begin if ( trunc(p_dec) <> p_dec OR p_dec < 0 ) then raise PROGRAM_ERROR; end if; loop l_str := substr( l_hex, mod(l_num,p_base)+1, 1 ) || l_str; l_num := trunc( l_num/p_base ); exit when ( l_num = 0 ); end loop; return l_str; end to_base; / create or replace function to_dec ( p_str in varchar2, p_from_base in number default 16 ) return number is l_num number default 0; l_hex varchar2(16) default ’0123456789ABCDEF’; begin for i in 1 .. length(p_str) loop l_num := l_num * p_from_base + instr(l_hex,upper(substr(p_str,i,1)))-1; end loop; return l_num; end to_dec; / show errors create or replace function to_hex( p_dec in number ) return varchar2 is begin return to_base( p_dec, 16 ); end to_hex; / create or replace function to_bin( p_dec in number ) return varchar2 is begin return to_base( p_dec, 2 ); end to_bin; / create or replace function to_oct( p_dec in number ) return varchar2 is begin return to_base( p_dec, 8 ); end to_oct; / |