suctf2019

easy_sql

应该是非预期解，输入abs(1)就有flag

不过也可以这样

后期在线维护的时候泄露了源码

<?php
    session_start();

    include_once "config.php";

    $post = array();
    $get = array();
    global $MysqlLink;

    //GetPara();
    $MysqlLink = mysqli_connect("localhost",$datauser,$datapass);
    if(!$MysqlLink){
        die("Mysql Connect Error!");
    }
    $selectDB = mysqli_select_db($MysqlLink,$dataName);
    if(!$selectDB){
        die("Choose Database Error!");
    }

    foreach ($_POST as $k=>$v){
        if(!empty($v)&&is_string($v)){
            $post[$k] = trim(addslashes($v));
        }
    }
    foreach ($_GET as $k=>$v){
        }
    }
    //die();
    ?>

<html>
<head>
</head>

<body>

<a> Give me your flag, I will tell you if the flag is right. </a>
<form action="" method="post">
<input type="text" name="query">
<input type="submit">
</form>
</body>
</html>

<?php

    if(isset($post['query'])){
        $BlackList = "prepare|flag|unhex|xml|drop|create|insert|like|regexp|outfile|readfile|where|from|union|update|delete|if|sleep|extractvalue|updatexml|or|and|&|\"";
        //var_dump(preg_match("/{$BlackList}/is",$post['query']));
        if(preg_match("/{$BlackList}/is",$post['query'])){
            //echo $post['query'];
            die("Nonono.");
        }
        if(strlen($post['query'])>40){
            die("Too long.");
        }
        $sql = "select ".$post['query']."||flag from Flag";
        mysqli_multi_query($MysqlLink,$sql);
        do{
            if($res = mysqli_store_result($MysqlLink)){
                while($row = mysqli_fetch_row($res)){
                    print_r($row);
                }
            }
        }while(@mysqli_next_result($MysqlLink));

    }

    ?>

看源码不太明白这个题目的用意，后来参考师傅们的wp时提到了||拼接

mysql 修改sql_mode 实现字符串管道‘||’连接

一种思路是直接使用*,1去拿到flag，另一种方式就是设置sql_mode来实现字符串的拼接

python nginx

首页有源码

        @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"
    </code>
    <!-- Dont worry about the suctf.cc. Go on! -->
    <!-- Do you know the nginx? -->

这个题目似乎有一个姿势

黑帽大会的ppt

urlparse函数

zz :: ~ » python
Python 3.5.4 (v3.5.4:3f56838, Aug  8 2017, 02:17:05) [MSC v.1900 64 bit (AMD64)] on win32
Type "help", "copyright", "credits" or "license" for more information.
>>> from urllib.parse import urlparse
>>> r = "https://prontosil.club"
>>> urlparse_result = urlparse(r)
>>> urlparse_result
ParseResult(scheme='https', netloc='prontosil.club', path='', params='', query='', fragment='')
>>> urlparse_result.hostname
'prontosil.club'
>>>