7.20 ctf题目练习
1.php代码审计
if (isset($_GET['name']) and isset($_GET['password'])) {
if ($_GET['name'] == $_GET['password'])
echo 'Your password can not be your name!
';
else if (sha1($_GET['name']) === sha1($_GET['password']))
die('Flag: '.$flag);
else
echo 'Invalid password.
';
}
else{
echo 'Login first!
';
?>
- $_GET：URL 方式传递数据
（1）说明：其实就是浏览器的地址栏传递方式；
（2）方式：表单的get方式和超链接方式；
（3）$_GET 方式是在 URL 地址栏里面直接对接传输。
(4)传递方式为：传递方式：URL 地址?参数 1=值 1&参数 2=值2……………………
2.php中sah1函数
（1）定义和用法：sha1() 函数计算字符串的 SHA-1 散列；
sha1() 函数使用美国 Secure Hash 算法 1；
（2）语法：sha1(string,raw)
string ：必需，规定要计算的字符串。
raw ：可选，规定十六进制或二进制输出格式。
本题思路：分析代码逻辑，发现得到了两个字段name和password，获得flag要求的条件是：name != password & sha1(name) == sha1(password)，乍看起来这是不可能的，其实可以利用sha1()函数的漏洞来绕过。如果把这两个字段构造为数组，如：?name[]=a&password[]=b，这样在第一处判断时两数组确实是不同的，但在第二处判断时由于sha1()函数无法处理数组类型，将报错并返回false，if 条件成立，获得flag。
2.md5加密
$md51 = md5('QNKCDZO');
$a = @$_GET['a'];
$md52 = @md5($a); #$md52等于输入的a经过md5加密后的数
if(isset($a))
{ if ($a != 'QNKCDZO' && $md51 == $md52#a不等于QNKCDZO，但是等于QNKCDZO加密后的数
) { echo "nctf{*****************}"; }
else { echo "false!!!"; }}
else{echo "please input a";} ?>补充： ! a和isset( a 和 i s s e t ( a)和empty($a)的区别
（1）当 a没定义的时候，isset( a 没 定 义 的 时 候 ， i s s e t ( a)不会提示Notice错误并返回0，empty( a)也不会提示notice错误并返回1，而! a ) 也 不 会 提 示 n o t i c e 错 误 并 返 回 1 ， 而 ! a会提示notice错误。
（2）当 a=null时，isset( a = n u l l 时 ， i s s e t ( a)会返回0，empty( a)会返回1，! a ) 会 返 回 1 ， ! a会返回1
（3）当 a=′′时，isset( a = ″ 时 ， i s s e t ( a)会返回1，empty( a)会返回1，! a ) 会 返 回 1 ， ! a会返回1
（4）当 a=0时，isset( a = 0 时 ， i s s e t ( a)会返回1，empty( a)会返回1，! a ) 会 返 回 1 ， ! a会返回1
（5）当 a=false时，isset( a = f a l s e 时 ， i s s e t ( a)会返回1，empty( a)会返回1，! a ) 会 返 回 1 ， ! a会返回1
建议：使用empty( a)替换! a ) 替 换 ! a判断，isset和empty接收未定义变量时，都不会提示Notice错误。
注意：PHP函数isset()、empty()只能用于变量，传递任何其它参数都将造成解析错误
解题思路：由题目所给源代码分析，输入a ，要使a等于字符串 QNKCDZO 经过md5加密后的数且不能等于该字符串，该字符串经过md5加密后等于0，将QNKCDZO进行md5加密，密文为0e830400451993494058024219903391，发现密文为0e开头，PHP在进行比较运算时，如果遇到了0e\d+这种字符串，就会将这种字符串解析为科学计数法。因为0exx都等于0，所以让两者相等故只需要寻找一个经过md5加密后数值等于0的字符串提交，即可得到flag。（以oe开头md5汇总：https://www.bbsmax.com/A/kmzLxxDYzG/）
.
加index.php含义：index.php,是用上述方法制作好网业文件的格式后缀名。
3.密码爆破
解题思路：因为验证一次，验证码就失效，所以我们要绕过验证码这一项，目前绕过验证码的方法就是不输入或删除验证码。
（使验证码vcode后面为空白，解题步骤还是爆破就好）
4.上传也不一定得到flag
上传个试试，得到：“真乖，您上传了一张jpeg的图片，上传成功！”并没有得到flag。
所以查看了页面源代码
<html>
<head>
<meta http-equiv=Content-Type content="text/html;charset=utf-8">
<title>upload 1title>
<script>
function check(){
var filename=document.getElementById("file");
var str=filename.value.split(".");#意思是图片上传的后缀以“.”为分界且以“.”分界的第一个格式为jpg就返回为True。
var ext=str[1];
if(ext==='jpg'){
return true;#意思是如果上传的格式是jpg格式，返回False。
}else{
alert("请上传一张JPG格式的图片！");
return false;
}
return false;
}
script>
head>
<body>
请上传一张JPG格式的图片！
<form action="upload_file.php" method="post" enctype="multipart/form-data" onsubmit="return check()">
<label for="file">文件名label>
<input type="file" name="file" id="file" />
<br />
<input type="submit" name="submit" value="上传" />
form>
body>
html>
解题思路：如上代码分析，抓个包然后我们可以把传的文件名改为XX.jpg.txt或XX.jpg.php。
这个题也可以直接上传一张图片名称为第一个.后为jpg但不能是jpg格式的（上传jpg.php格式即可得到flag）
5.robots了解一下
首先robots文件指的是带有.txt的文件，如果站内没有robots.txt文件，则表明蜘蛛可以任意抓取网站的文件。对于网站优化来说并不是所有文件都需要让蜘蛛抓取的。例如一些不重要的文件，或者网站的死链接，我们就需要使用robots.txt来屏蔽掉，不让他抓取。
robots.txt文件存放在网站的什么地方：robots.txt文件存放在我们网站的根目录下，我们如何检测站点里面有没有robots.txt文件，我们只需要在网站的根域名后面输入robots.txt。
故该题的解题方法为：该题是为了让我们了解robots.txt要加在什么地方，要加在域的根目录后，不能加在子目录后。
故将http://www.wechall.net/challenge/training/www/robots/index.php改为：
http://www.wechall.net/robots.txt
6.自己试一下该提交什么
解题思路：首先抓个包看看，刚开始的key随便填（1111）得到如下图所示：
然后将其发送到 Intruder，执行go得到如下
显示check Failed!
Content-Row是一个用base64加密过的一串字符，然后将其解码便得到了pass key，我们把Content-Row后面的字符串填到右面的pass_key处再次执行go就可以得到key。
7.根据提示来，密码已经告诉你
需要用到爆破以及base解密，将解密的东西访问然后继续操作，最后得到flag。
解题思路：
首先我们点开链接是这样的一个页面
我们再根据题目要求是知道要进行爆破的，但是在这个登录页面我们并不知道密码的组成结构是什么样的，我们可以查看页面源代码（或按Fn加F12），出现下面代码
我们看到了password.txt，然后我们在浏览器里访问这个页面
出现了许多字典，我们把它们复制在一个txt文档中，然后开始进行爆破，得到，
然后将MjkwYmNhNzBjN2RhZTkzZGI2NjQ0ZmEwMGI5ZDgzYjkucGhw进行base64解密（set-cookie是一种保存方式，就是把静态数据存入到 cookie中 ），解密后得到290bca70c7dae93db6644fa00b9d83b9.php，然后访问http://39.107.92.230/web/web5/290bca70c7dae93db6644fa00b9d83b9.php得到如下页面
然后再进行一次抓包，
然后将IsLogin=0改为IsLogin=1（设为允许），然后将guest该问root（将访客权限改为管理员权限），然后点击Forward，然后出现留言成功页面
然后再进行一次抓包，出现flag。
提交这个flag显示失败，所以观察这个Flag=flag%7BC0ngratulati0n%7D我们对它进行了url解密，得到了flag{C0ngratulati0n}。
8./X00
view-source:
if (isset ($_GET['nctf'])) {
if (@ereg ("^[1-9]+$", $_GET['nctf']) === FALSE)
echo '必须输入数字才行';
else if (strpos ($_GET['nctf'], '#biubiubiu') !== FALSE)
die('Flag: '.$flag);
else
echo '骚年，继续努力吧啊~';
}
解题思路： GET[′nctf′]全为数字，并且 G E T [ ′ n c t f ′ ] 全 为 数 字 ， 并 且 _GET[‘nctf’]含有’#biubiubiu’字符串，对于ereg()这个函数存在NULL截断漏洞，所以构造payload:nctf=11%00%23biubiubiu，文件名后缀就一个%00字节，可以截断某些函数对文件名的判断。在许多语言函数中，处理字符串的函数中0x00被认为是终止符。
例如：网站上传函数处理xxx.php%00.jpg时，首先后缀名是合格的jpg格式，可以上传，在保存文件时，遇到%00字符丢弃后面的jpg，文件后缀最终保存的后缀名为xxx.php。
故该题为
即可得到Flag: flag:nctf{use_00_to_jieduan}
9.web5
解题思路：查看页面源代码
<html>
<body>
<div 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<form action="index.php" method="post" >
JSPFUCK??????答案格式CTF{******}<br>
<br>
<input type="input" name="flag" id="flag" />
<input type="submit" name="submit" value="Submit" />
form>
body>
html>
在控制台打开（可以按Fn加F12），然后将
这些在控制台输出就可以得到 ctf{whatfk}
10.矛盾
$num=$_GET['num'];
if(!is_numeric($num))#如果输入的是数字返回0
{
echo $num;
if($num==1)
echo 'flag{**********}';
}解题思路：我们要输入的num要使其满足值为1且不能只是数字，所以我们可以在地址栏加:
?num=1a
(1)is_numeric – 检测变量是否为数字或数字字符串
(2)php中==与===的区别
==（等于）：只会比较左右两端的值，不比较它们的类型；
===（全等于）：不仅比较左右两端的值，也比较它们的类型。
11.文件上传比想象的难一点
解题思路：我们可以先通过抓包，得到
这里有uploads（上传），所以在后面加上time.php timg.jp，然后点开Hex
将20改为00，在点到Raw可以看到
然后点击Forward，就可以得到flag:nctf{welcome_to_hacks_world}
白名单检测绕过：白名单就是只只能上传规定后缀的文件
主要利用截断上传绕过，0x00截断与%00截断
0x00截断是将上传文件名或路径名中使用ascii码值为0的字符（也就是null）来进行截断。多用在路径和文件名截断中，在上传的时候，当文件系统读到【0x00】时，会认为文件已经结束。利用00截断就是利用程序员在写程序时对文件的上传路径过滤不严格，产生0x00上传截断漏洞。
通过抓包截断将【evil.php.jpg】后面的一个【.】换成【0x00】。在上传的时候，当文件系统读到【0x00】时，会认为文件已经结束，从而将【evil.php.jpg】的内容写入到【evil.php】中，从而达到攻击的目的。