php 第 xx 行的 eval() 代码
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eval()'d code on line xx
提问by Hawkar Oghiz
i have this code , i want to convert the get(title) to variable , its correctly working , for example i have variable 'news , sport, program ' , my problem is that when someone is changing the title="news" to title=new , and i dont have the new variable how i can control it when the user change the title ; this will appear an error that the variable undefined and eval()'d code on line xx error
我有这个代码，我想将 get(title) 转换为变量，它可以正常工作，例如我有变量“新闻、体育、节目”，我的问题是当有人将标题 =“新闻”更改为标题时=new ，而且我没有新变量，当用户更改标题时如何控制它；这将出现一个错误，变量 undefined 和 eval()'d 代码在第 xx 行错误
if(isset($_GET['title']))
$title=stripslashes($_GET['title']);
else
$title="news";
$trimed=trim(strtolower($title));
$variable="$".str_replace(" ","_",$trimed);
$title=eval('return '.$variable.';');
if(!isset($title))
$title=$$title;
thank you for help
谢谢你的帮助
采纳答案by dynamic
Just do like this:
只是这样做：
$nameOfVar = 'return' . $variable;
$title = $$nameOfVar;
回答by Sarfraz
You can use ${}syntax and replace:
您可以使用${}语法并替换：
$title = eval('return '. $variable.';');
With:
和：
$title = ${'return '. $variable};
