PHP评估问题与PHP + HTML代码 [英] PHP eval issue with PHP + HTML code

问题描述

我已经将PHP和HTML代码存储在数据库表中.当我得到这些数据时，我需要回显HTML并处理PHP.我以为可以使用eval()来工作，如果执行eval("echo 'dlsj'; ?> EVALED ");，我会打印出"dlsjEVALED".

I've got PHP and HTML code stored in a database table. When I get this data, I need to echo the HTML and process the PHP. I thought I could use eval() for this, which works, if I do this eval("echo 'dlsj'; ?> EVALED "); I get "dlsjEVALED" printed out.

问题是，当我运行更长的脚本时，出现致命错误.像这样的东西:

The problem is, I get a fatal error when I run longer scripts. Things like:

解析错误:语法错误，意外的'<'在/home/content.php(18)中:第1行上的eval()代码

Parse error: syntax error, unexpected '<' in /home/content.php(18) : eval()'d code on line 1

推荐答案

我猜您正在尝试eval()包含开头<?php标记的内容.这会导致手头的错误.

I would guess that you're trying to eval() something that contains an opening <?php tag. And that leads to the error at hand.

这篇关于PHP评估问题与PHP + HTML代码的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！