Oracle10g unwrap技术分析
Oracle为PL/SQL开发者提供的一种对他们所写的代码(oracle下的对象Package、Procedure、Function、Java Source等)进行加密的工具。当PL/SQL代码被加密以后,它就被描述为被“包装过”。使用wrap工具对代码进行包装,wrap工具会取出含有要包装的代码的文件名并输出下面的文件:
wrap iname=plain.sql oname=encrypted.plb
因为代码被加密了,所以其细节被隐藏了,而且Oracle不提供解包装设备。但是我们可以编写我们自己的unwrap程序。
Oracle为了防止自己的对象程序源码泄露,也对其做了wrap处理。因为许多PACKAGE、PROCEDURE等程序本身可能存在SQL注入漏洞,但是如果得不到源码,对其进行审计是很困难的。这样wrap在一定程度上也起到了保护作用。
这里随便找一个Oracle的PACKAGE为实例,这里我们找到dbms_hs包程序,其代码如图所示:
很明显代码被加密了。
0x03 Oracle Wrap的算法机制Oracle加密的原理就是先对源码进行lz压缩lzstr,然后对压缩数据进行SHA-1运算得到40位的加密串shstr,然后将加密串与压缩串拼接得到shstr+lzstr,然后对拼接后的字符串进行Oracle双字符转换(转换表)。最后将转换后的字符串进行base64编码,最终得到wrap的加密串。
关于LZ压缩,这里用老外提供的一个JAVA包,代码如下:
create or replace java source named MY_COMPRESSas import java.io.*; import java.util.zip.*; public class MY_COMPRESS { public static String Inflate( byte[] src ) { try { ByteArrayInputStream bis = new ByteArrayInputStream( src ); InflaterInputStream iis = new InflaterInputStream( bis ); StringBuffer sb = new StringBuffer(); for( int c = iis.read(); c != -1; c = iis.read() ) { sb.append( (char) c ); } return sb.toString(); } catch ( Exception e ) { } return null; } public static byte[] Deflate( String src, int quality ) { try { byte[] tmp = new byte[ src.length() + 100 ]; Deflater defl = new Deflater( quality ); defl.setInput( src.getBytes( "UTF-8" ) ); defl.finish(); int cnt = defl.deflate( tmp ); byte[] res = new byte[ cnt ]; for( int i = 0; i < cnt; i++ ) res[i] = tmp[i]; return res; } catch ( Exception e ) { } return null; } } / alter java source MY_COMPRESS compile / create or replace package mycompress is function deflate( src in varchar2 ) return raw; -- function deflate( src in varchar2, quality in number ) return raw; -- function inflate( src in raw ) return varchar2; -- end; / create or replace package body mycompress is function deflate( src in varchar2 ) return raw is begin return deflate( src, 6 ); end; -- function deflate( src in varchar2, quality in number ) return raw as language java name 'MY_COMPRESS.Deflate( java.lang.String, int ) return byte[]'; -- function inflate( src in raw ) return varchar2 as language java name 'MY_COMPRESS.Inflate( byte[] ) return java.lang.String'; -- end; /
这里deflate函数实现LZ压缩,而inflate则实现解压。
首先,我们以一小段PL/SQL代码来测试分先加密字符串的结构,这里我先讲程序加密处理。如图所示,这里将create PACKAGE a代码wrap之后变成了如下代码
为了进一步测试,分析其加密字符串中的结构,我们利用如下代码:
with src as ( select 'PACKAGE a' txt from dual ) , wrap as ( select src.txt, dbms_ddl.wrap( 'create ' || src.txt ) wrap from src ) Select rtrim( substr( wrap.wrap, instr(wrap.wrap, chr(10), 1, 20 ) + 1 ),chr(10) ) from wrap;
这段代码获得加密串,如图所示
进一步对这段代码做base64解码,获得经过Oracle双字符转换后的字符串,如图所示:
因为字符串是由两部分组成,先经过LZ压缩,然后经过SHA-1处理,然后再将SHA-1字符串和压缩字符串拼接得到,因为SHA-1总共40位,所以40位以后的全为经过Oracle双字符转换的压缩串,也就是308399B8F5339FF5BF5CB891A6A6CBBFE1DC
0x04 计算转换表当wrap完成sha-1字符串和压缩字符串的拼接之后,紧接着会对照一个字符代替表进行Oracle双字符转换。这个表可能是Oracle的商业机密,所以官方没有给出该表的信息。
但是,既然我们已经得到了经过字符转换之后的LZ压缩串,同时我们可以通过前面提到的JAVA包得到纯净的LZ压缩串,通过对比这两个字符串,我们便可以推算出这个转换表。
在使用这个java包时,涉及到一个压缩级别参数,这个等级参数不一样,压缩得到的字符串完全一不样。有人可能要问,这样搞岂不是没法得到替换表了吗?是的,但也不完全正确。因为可供选择的等级参数有限,我们可以从0等级开始一个一个进行测试,经过测试发现,ORACLE用的是“9”等级。所以我们用以下代码对比两个字符串:
with src as ( select 'PACKAGE c' txt from dual ) , wrap as ( select src.txt, dbms_ddl.wrap( 'create ' || src.txt ) wrap from src ) , subst as ( select substr( utl_encode.base64_decode( utl_raw.cast_to_raw( rtrim( substr( wrap.wrap, instr(wrap.wrap, chr(10), 1, 20 ) + 1 ),chr(10) ) ) ), 41 ) x , mycompress.deflate( wrap.txt || chr(0),9 ) d from wrap ) select to_number( substr( x, r * 2 - 1, 2),'xx' ) wrapped , to_number( substr( d, r * 2 - 1, 2),'xx' ) zipped from subst , ( select rownum r from dual connect by rownum < 19 );
结果如下:
通过对结果的排序,没有出现同一个BASE64编码对应不同的十六进制的情况,因此我们知道了可以用这个SQL为基础,通过用不同的SOURCE串来产生替换表的内容。
根据上面的SQL首先建一个表来存储替换表的内容,
CREATE TABLE SYS.IDLTRANSLATE ( C_BASE64DECODE VARCHAR2(2) NOT NULL, C_LZDEFLATECODE VARCHAR2(2) NULL )
然后写一段PLSQL块来生成替换表的内容:
declare nCnt integer; nLoop integer; nSLoop integer; nCharmax integer; nCharmin integer; vChar Varchar2(3); cursor getchar is with src AS ( select 'PACKAGE '||vChar txt from dual ), wrap as ( select src.txt , dbms_ddl.wrap( 'create ' || src.txt ) wrap from src ), subst as (select substr( utl_encode.base64_decode( utl_raw.cast_to_raw(rtrim( substr( wrap.wrap, instr( wrap.wrap, chr( 10 ), 1, 20 ) + 1 ), chr(10) ) ) ), 41 ) x, mycompress.deflate( wrap.txt || chr(0), 9 ) d from wrap ) select substr( x, r * 2 - 1, 2 ) xr , substr( d, r * 2 - 1, 2 ) dr from subst , ( select rownum r from dual connect by rownum <= ( select length( x ) / 2 from subst ) ); begin nCharmax:=97; nCharmin:=122; For nLoop In 97..122 Loop For nSloop In 0..99 Loop vChar := chr(nLoop)||to_char(nSloop); For abc In getchar Loop Select Count(*) Into nCnt From sys.idltranslate WHERE c_base64decode = abc.xr; If nCnt < 1 Then Insert INTO sys.idltranslate VALUES (abc.xr,abc.dr); Commit; Else Select Count(*) Into ncnt From sys.idltranslate WHERE c_base64decode = abc.xr AND c_lzdeflatecode=abc.dr; If nCnt < 1 Then DBMS_OUTPUT.PUT_LINE('wrong orginal char:'||vchar||' hex base64:'||abc.xr); End If; End If; End Loop; End Loop; End Loop; end;
运行上面这段SQL大概会产生100多条记录,还未达到00-FF总共256条记录的要求,建议替换select 'PACKAGE '||vChar txt from dual中的PACKAGE关健字为procedure或者function类似的,继续运行直到替换表中有不重复的256条记录为止。有了替换表的内容,unwrap出明文也就不再困难了。
这里给出替换表的内容,我将其写到一个变量里,如下所示:
int charmap[] = {0x3d, 0x65, 0x85, 0xb3, 0x18, 0xdb, 0xe2, 0x87, 0xf1, 0x52, 0xab, 0x63, 0x4b, 0xb5, 0xa0, 0x5f, 0x7d, 0x68, 0x7b, 0x9b, 0x24, 0xc2, 0x28, 0x67, 0x8a, 0xde, 0xa4, 0x26, 0x1e, 0x03, 0xeb, 0x17, 0x6f, 0x34, 0x3e, 0x7a, 0x3f, 0xd2, 0xa9, 0x6a, 0x0f, 0xe9, 0x35, 0x56, 0x1f, 0xb1, 0x4d, 0x10, 0x78, 0xd9, 0x75, 0xf6, 0xbc, 0x41, 0x04, 0x81, 0x61, 0x06, 0xf9, 0xad, 0xd6, 0xd5, 0x29, 0x7e, 0x86, 0x9e, 0x79, 0xe5, 0x05, 0xba, 0x84, 0xcc, 0x6e, 0x27, 0x8e, 0xb0, 0x5d, 0xa8, 0xf3, 0x9f, 0xd0, 0xa2, 0x71, 0xb8, 0x58, 0xdd, 0x2c, 0x38, 0x99, 0x4c, 0x48, 0x07, 0x55, 0xe4, 0x53, 0x8c, 0x46, 0xb6, 0x2d, 0xa5, 0xaf, 0x32, 0x22, 0x40, 0xdc, 0x50, 0xc3, 0xa1, 0x25, 0x8b, 0x9c, 0x16, 0x60, 0x5c, 0xcf, 0xfd, 0x0c, 0x98, 0x1c, 0xd4, 0x37, 0x6d, 0x3c, 0x3a, 0x30, 0xe8, 0x6c, 0x31, 0x47, 0xf5, 0x33, 0xda, 0x43, 0xc8, 0xe3, 0x5e, 0x19, 0x94, 0xec, 0xe6, 0xa3, 0x95, 0x14, 0xe0, 0x9d, 0x64, 0xfa, 0x59, 0x15, 0xc5, 0x2f, 0xca, 0xbb, 0x0b, 0xdf, 0xf2, 0x97, 0xbf, 0x0a, 0x76, 0xb4, 0x49, 0x44, 0x5a, 0x1d, 0xf0, 0x00, 0x96, 0x21, 0x80, 0x7f, 0x1a, 0x82, 0x39, 0x4f, 0xc1, 0xa7, 0xd7, 0x0d, 0xd1, 0xd8, 0xff, 0x13, 0x93, 0x70, 0xee, 0x5b, 0xef, 0xbe, 0x09, 0xb9, 0x77, 0x72, 0xe7, 0xb2, 0x54, 0xb7, 0x2a, 0xc7, 0x73, 0x90, 0x66, 0x20, 0x0e, 0x51, 0xed, 0xf8, 0x7c, 0x8f, 0x2e, 0xf4, 0x12, 0xc6, 0x2b, 0x83, 0xcd, 0xac, 0xcb, 0x3b, 0xc4, 0x4e, 0xc0, 0x69, 0x36, 0x62, 0x02, 0xae, 0x88, 0xfc, 0xaa, 0x42, 0x08, 0xa6, 0x45, 0x57, 0xd3, 0x9a, 0xbd, 0xe1, 0x23, 0x8d, 0x92, 0x4a, 0x11, 0x89, 0x74, 0x6b, 0x91, 0xfb, 0xfe, 0xc9, 0x01, 0xea, 0x1b, 0xf7, 0xce}
这里为256条记录,按顺序00~FF对应数组每一个元素。
0x05 破解程序利用前面得出的JAVA包以及转换表,结合下面这段代码unwrap出明文,如下
set serveroutput on; create or replace procedure unwrap(o in varchar,n in varchar, t in varchar) as vWrappedtext Varchar2(32767); vtrimtext Varchar2(32767); vChar Varchar2(2); vRepchar Varchar2(2); vLZinflatestr Varchar2(32767); nLen Integer; nLoop Integer; nCnt Integer; code varchar(512); Begin code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sys.idltranslate表内容存到字符数组 vtrimtext:=''; select count(*) into ncnt from DBA_SOURCE Where owner=o And Name = n And Type=t ; if ncnt >0 and ncnt <=5 then for i in 1..ncnt loop if i=1 then select rtrim(substr(TEXT,instr(TEXT,chr(10),1,20)+1),chr(10)) --保存去掉前边20行的BASE64码正文 into vLZinflatestr from DBA_SOURCE Where owner=o And Name = n And Type=t and line=i; else select text into vLZinflatestr from DBA_SOURCE Where owner=o And Name = n And Type=t and line=i; end if; vtrimtext:=vtrimtext||vLZinflatestr; end loop; end if; vtrimtext:=replace(vtrimtext,chr(10),''); nLen := Length(vtrimtext)/256 ; vWrappedtext :=''; for i in 0..nLen loop --if i< nLen then vWrappedtext:=vWrappedtext||utl_encode.base64_decode( utl_raw.cast_to_raw(substrb(vtrimtext,256*i+1 , 256 ))) ; -- else --vWrappedtext:=vWrappedtext||utl_encode.base64_decode( utl_raw.cast_to_raw(substrb(vtrimtext,64*i+1 ))) ; --end if; --DBMS_OUTPUT.PUT_LINE(vWrappedtext); End Loop; --vWrappedtext:=substr(vWrappedtext,41); nLen := Length(vWrappedtext)/2 - 1; vLZinflatestr :=''; For nLoop In 20..nLen Loop --从第41字节开始 vChar := Substrb(vWrappedtext,nLoop*2+1,2); vLZinflatestr := vLZinflatestr ||substr(code,to_number(vChar,'XX')*2+1,2); --从字符串变量匹配 --DBMS_OUTPUT.PUT_LINE(vLZinflatestr); End Loop; --DBMS_OUTPUT.PUT_LINE(vLZinflatestr); DBMS_OUTPUT.PUT_LINE(mycompress.inflate(vLZinflatestr)); End; / exec unwrap('SYS','AA','PROCEDURE');
这里将转换表IDLTRANSLATE的内容存储变量code当中,这样可以在unwrap的过程中不必再去创建IDLTRANSLATE这个表。
执行以上代码查看一下输出的内容,至此Oracle所有PACKAGE程序没有密码可言了,当然其他对象程序存储过程、函数等也可以得到明文了。
这里执行一段unwrap以后的程序,然后保存以上代码为code.sql,解码得到明文。
同时还收集了老外提供的一个python脚本,代码如下:
#!/usr/bin/python # # This script unwraps Oracle wrapped plb packages, does not support 9g # Contact: niels at teusink net / blog.teusink.net # # License: Public domain # import re import base64 import zlib import sys # simple substitution table charmap = [0x3d, 0x65, 0x85, 0xb3, 0x18, 0xdb, 0xe2, 0x87, 0xf1, 0x52, 0xab, 0x63, 0x4b, 0xb5, 0xa0, 0x5f, 0x7d, 0x68, 0x7b, 0x9b, 0x24, 0xc2, 0x28, 0x67, 0x8a, 0xde, 0xa4, 0x26, 0x1e, 0x03, 0xeb, 0x17, 0x6f, 0x34, 0x3e, 0x7a, 0x3f, 0xd2, 0xa9, 0x6a, 0x0f, 0xe9, 0x35, 0x56, 0x1f, 0xb1, 0x4d, 0x10, 0x78, 0xd9, 0x75, 0xf6, 0xbc, 0x41, 0x04, 0x81, 0x61, 0x06, 0xf9, 0xad, 0xd6, 0xd5, 0x29, 0x7e, 0x86, 0x9e, 0x79, 0xe5, 0x05, 0xba, 0x84, 0xcc, 0x6e, 0x27, 0x8e, 0xb0, 0x5d, 0xa8, 0xf3, 0x9f, 0xd0, 0xa2, 0x71, 0xb8, 0x58, 0xdd, 0x2c, 0x38, 0x99, 0x4c, 0x48, 0x07, 0x55, 0xe4, 0x53, 0x8c, 0x46, 0xb6, 0x2d, 0xa5, 0xaf, 0x32, 0x22, 0x40, 0xdc, 0x50, 0xc3, 0xa1, 0x25, 0x8b, 0x9c, 0x16, 0x60, 0x5c, 0xcf, 0xfd, 0x0c, 0x98, 0x1c, 0xd4, 0x37, 0x6d, 0x3c, 0x3a, 0x30, 0xe8, 0x6c, 0x31, 0x47, 0xf5, 0x33, 0xda, 0x43, 0xc8, 0xe3, 0x5e, 0x19, 0x94, 0xec, 0xe6, 0xa3, 0x95, 0x14, 0xe0, 0x9d, 0x64, 0xfa, 0x59, 0x15, 0xc5, 0x2f, 0xca, 0xbb, 0x0b, 0xdf, 0xf2, 0x97, 0xbf, 0x0a, 0x76, 0xb4, 0x49, 0x44, 0x5a, 0x1d, 0xf0, 0x00, 0x96, 0x21, 0x80, 0x7f, 0x1a, 0x82, 0x39, 0x4f, 0xc1, 0xa7, 0xd7, 0x0d, 0xd1, 0xd8, 0xff, 0x13, 0x93, 0x70, 0xee, 0x5b, 0xef, 0xbe, 0x09, 0xb9, 0x77, 0x72, 0xe7, 0xb2, 0x54, 0xb7, 0x2a, 0xc7, 0x73, 0x90, 0x66, 0x20, 0x0e, 0x51, 0xed, 0xf8, 0x7c, 0x8f, 0x2e, 0xf4, 0x12, 0xc6, 0x2b, 0x83, 0xcd, 0xac, 0xcb, 0x3b, 0xc4, 0x4e, 0xc0, 0x69, 0x36, 0x62, 0x02, 0xae, 0x88, 0xfc, 0xaa, 0x42, 0x08, 0xa6, 0x45, 0x57, 0xd3, 0x9a, 0xbd, 0xe1, 0x23, 0x8d, 0x92, 0x4a, 0x11, 0x89, 0x74, 0x6b, 0x91, 0xfb, 0xfe, 0xc9, 0x01, 0xea, 0x1b, 0xf7, 0xce] #print charmap[124] def decode_base64_package(base64str): base64dec = base64.decodestring(base64str)[20:] # we strip the first 20 chars (SHA1 hash, I don't bother checking it at the moment) decoded = '' for byte in range(0, len(base64dec)): decoded += chr(charmap[ord(base64dec[byte])]) return zlib.decompress(decoded) sys.stderr.write("=== Oracle 10g/11g PL/SQL unwrapper 0.2 - by Niels Teusink - blog.teusink.net ===\n\n" ) if len(sys.argv) < 2: sys.stderr.write("Usage: %s infile.plb [outfile]\n" % sys.argv[0]) sys.exit(1) infile = open(sys.argv[1]) outfile = None if len(sys.argv) == 3: outfile = open(sys.argv[2], 'w') lines = infile.readlines() for i in range(0, len(lines)): # this is really naive parsing, but works on every package I've thrown at it matches = re.compile(r"^[0-9a-f]+ ([0-9a-f]+)$").match(lines[i]) #print matches if matches: base64len = int(matches.groups()[0], 16) base64str = '' j = 0 while len(base64str) < base64len: j+=1 base64str += lines[i+j] base64str = base64str.replace("\n","") if outfile: outfile.write(decode_base64_package(base64str) + "\n") else: print decode_base64_package(base64str)
执行结果,如图所示:
0x06 总结Oracle 10g PL/SQL的wrap过程是对源码先进行lz压缩lzstr,然后对压缩数据进行SHA-1运算得到40位的加密串shstr,然后将加密串与压缩串拼接得到shstr+lzstr,然后对拼接后的字符串进行Oracle双字符转换(转换表),最后将转换后的字符串进行base64编码,最终得到wrap的加密串。
而unwrap的过程正好相反,破解的关键是得到ORACLE的双字符转换表,因为在10g以后Oracle中无法查看转换表的内容。这里捎带提一下oracle 9i的wrap技术,它使用一种完全不同于10g的方法。9i wrap的时候,代码被转换为DIANA(ADA的中间描述语言),这是一个相当复杂的过程,具体细节请阅读Pete Finnegan的论文“How to Unwrap PL/SQL”,大家可以去http://www.blackhat.com/presentations/bh-usa-06/BH-US-06-Finnigan.pdf下载