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Home Backend Development PHP Tutorial php的eval调用变量就出错。

php的eval调用变量就出错。

Jun 06, 2016 pm 08:27 PM
eval php

现在已经确定服务端PHP代码(不能修改)是

<code><?php eval($_GET['command']);
?></code>
Copy after login
Copy after login

但是在一调用就出错。
因为项目需求。在传值时,必须使用base64_decode("base64处理后的字符串");
所以我测试时,使用的url如下:
http://test.cn/2.php?command=base64_decode("ZWNobyAndGVzdCc7");
但是返回的页面无任何内容。
如果不用base64_decode函数一点事也没有,如下:
http://test.cn/2.php?command=echo "123";
能显示123.

ZWNobyAndGVzdCc7 编码前的字符串为 echo 'test';

服务端代码不能进行改动,请求时必须以base64_decode("base64处理后的字符串")这样的方式来。困扰一天了。

回复内容:

现在已经确定服务端PHP代码(不能修改)是

<code><?php eval($_GET['command']);
?></code>
Copy after login
Copy after login

但是在一调用就出错。
因为项目需求。在传值时,必须使用base64_decode("base64处理后的字符串");
所以我测试时,使用的url如下:
http://test.cn/2.php?command=base64_decode("ZWNobyAndGVzdCc7");
但是返回的页面无任何内容。
如果不用base64_decode函数一点事也没有,如下:
http://test.cn/2.php?command=echo "123";
能显示123.

ZWNobyAndGVzdCc7 编码前的字符串为 echo 'test';

服务端代码不能进行改动,请求时必须以base64_decode("base64处理后的字符串")这样的方式来。困扰一天了。

警告,使用eval是非常危险的行为!
特别是你们对于这个command参数甚至没有任何加密和过滤处理,那么用户哪天传一个参数能查看,显示,编辑其内容,甚至修改,删除服务器所有文件的代码也是可以直接运行的(当你的PHP在以root运行)

抛开这个不谈来分析这个问题
服务器进行eval时,它实际上是等价于运行:

base64_decode("ZWNobyAndGVzdCc7");
Copy after login

这时候会得到一个字符串

<code>echo 'test';</code>
Copy after login

然而,因为你的eval只有一个,所以后面这个base64解码出来的文本,实际上只是一段文本,表达式到这个时候就已经结束了,不会再继续运行,所以要达成楼主的目的,应该是需要这样:
http://test.cn/2.php?command=echo base64_decode("dGVzdAo=");
或者这样:
http://test.cn/2.php?command=eval(base64_decode("ZWNobyAndGVzdCc7"));

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