Oracle中位运算函数试验

2014-11-24 18:13:41 · 作者: · 浏览: 1

1.rawtohex,hextoraw


rawtohex表示将raw类型数据转换为16进制字符串(nvarchar类型,Typ=1),hextoraw表示将16进制字符串转换为raw类型。
注意:hextoraw的参数如果是字符串,会当作16进制数字对待;如果是数字,也会认为是16进制而不是10进制;


SQL> select hextoraw('13'),hextoraw(13),hextoraw('D') from dual;


HEXTORAW('13') HEXTORAW(13) HEXTORAW('D')
-------------- ------------ -------------
13 13 0D


以上前两列返回结果是相同的(十六进制13即十进制19的字符串表示形式);
第三列返回'0D'是因为这样才是一个完整字节;
实际上hextoraw返回的值总是偶数位的,每两位表示一个字节;


dump一下可以证明(Typ=23表示raw类型):


SQL> select dump(hextoraw('13')),dump(hextoraw(13)),dump(hextoraw('D')) from dual;


DUMP(HEXTORAW('13')) DUMP(HEXTORAW(13)) DUMP(HEXTORAW('D'))
-------------------- ------------------ -------------------
Typ=23 Len=1: 19 Typ=23 Len=1: 19 Typ=23 Len=1: 13



注意以下例子:


SQL> select to_number('AB','xx') from dual;


TO_NUMBER('AB','XX')
--------------------
171


SQL> select hextoraw('AB') from dual;


HEXTORAW('AB')
--------------
AB


SQL> select hextoraw(to_number('AB','xx')) from dual;


HEXTORAW(TO_NUMBER('AB','XX'))
----------------------------------------
0171


虽然to_number('AB','xx')是将十六进制字符串'AB'转换为10进制数171,但是作为hextoraw参数时又被认为是16进制数了


2。utl_raw.bit_and,bitand


bitand的参数是十进制数字,将输入参数转化为二进制后求与,返回值是数值型;


SQL> select bitand(10,25) from dual;


BITAND(10,25)
-------------
8


SQL> select dump(bitand(10,25)) from dual;


DUMP(BITAND(10,25))
-------------------
Typ=2 Len=2: 193,9


10=1100b,25=11001b,按位与结果是01000b=8(Typ=2表示number类型)


utl_raw.bit_and的参数和返回值都是raw类型,


SQL> select utl_raw.bit_and(hextoraw('a'),hextoraw('19')) from dual;


UTL_RAW.BIT_AND(HEXTORAW('A'),
--------------------------------------------------------------------------------
08


SQL> select dump(utl_raw.bit_and(hextoraw('a'),hextoraw('19'))) from dual;


DUMP(UTL_RAW.BIT_AND(HEXTORAW(
--------------------------------------------------------------------------------
Typ=23 Len=1: 8