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cxl862002755 2020-07-04 11:42 原文

CTF-never give up 100

点击进入界面,F12看下有个提示

1.html,查看一下发现是官方…….

 

进行brupsuite抓包试试

发现有一大串java代码+注释

进行url解码

var Words ="<script>window.location.href%3D'http%3A//www.bugku.com'%3B</script>

<!--JTIyJTNCaWYlMjglMjElMjRfR0VUJTVCJTI3aWQlMjclNUQlMjklMEElN0IlMEElMDloZWFkZXIlMjglMjdMb2NhdGlvbiUzQSUyMGhlbGxvLnBocCUzRmlkJTNEMSUyNyUyOSUzQiUwQSUwOWV4aXQlMjglMjklM0IlMEElN0QlMEElMjRpZCUzRCUyNF9HRVQlNUIlMjdpZCUyNyU1RCUzQiUwQSUyNGElM0QlMjRfR0VUJTVCJTI3YSUyNyU1RCUzQiUwQSUyNGIlM0QlMjRfR0VUJTVCJTI3YiUyNyU1RCUzQiUwQWlmJTI4c3RyaXBvcyUyOCUyNGElMkMlMjcuJTI3JTI5JTI5JTBBJTdCJTBBJTA5ZWNobyUyMCUyN25vJTIwbm8lMjBubyUyMG5vJTIwbm8lMjBubyUyMG5vJTI3JTNCJTBBJTA5cmV0dXJuJTIwJTNCJTBBJTdEJTBBJTI0ZGF0YSUyMCUzRCUyMEBmaWxlX2dldF9jb250ZW50cyUyOCUyNGElMkMlMjdyJTI3JTI5JTNCJTBBaWYlMjglMjRkYXRhJTNEJTNEJTIyYnVna3UlMjBpcyUyMGElMjBuaWNlJTIwcGxhdGVmb3JtJTIxJTIyJTIwYW5kJTIwJTI0aWQlM0QlM0QwJTIwYW5kJTIwc3RybGVuJTI4JTI0YiUyOSUzRTUlMjBhbmQlMjBlcmVnaSUyOCUyMjExMSUyMi5zdWJzdHIlMjglMjRiJTJDMCUyQzElMjklMkMlMjIxMTE0JTIyJTI5JTIwYW5kJTIwc3Vic3RyJTI4JTI0YiUyQzAlMkMxJTI5JTIxJTNENCUyOSUwQSU3QiUwQSUwOXJlcXVpcmUlMjglMjJmNGwyYTNnLnR4dCUyMiUyOSUzQiUwQSU3RCUwQWVsc2UlMEElN0IlMEElMDlwcmludCUyMCUyMm5ldmVyJTIwbmV2ZXIlMjBuZXZlciUyMGdpdmUlMjB1cCUyMCUyMSUyMSUyMSUyMiUzQiUwQSU3RCUwQSUwQSUwQSUzRiUzRQ%3D%3D-->"

function OutWord()

{

var NewWords;

NewWords = unescape(Words);

document.write(NewWords);

}

OutWord();

将注释再进行base64解码

";if(!$_GET['id'])

{

    header('Location: hello.php?id=1');

    exit();

}

$id=$_GET['id'];

$a=$_GET['a'];

$b=$_GET['b'];

if(stripos($a,'.'))

{

    echo 'no no no no no no no';

    return ;

}

$data = @file_get_contents($a,'r');

if($data=="bugku is a nice plateform!" and $id==0 and strlen($b)>5 and eregi("111".substr($b,0,1),"1114") and substr($b,0,1)!=4)

{

    require("f4l2a3g.txt");

}

else

{

    print "never never never give up !!!";

}

 

 

?>

ÃÜ0

得到这段代码

发现有f4l2a3g.txt

得尝试绕过?

看的有点难,百度了一下发现直接查找这个txt就可以了如下:

得到flag

当然也可以构造payload

参看链接:https://blog.csdn.net/changer_WE/article/details/89001076

总结:这题莫非是考加密解密求出源码,再进行绕过获取flag?这作者给力啊

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