如何解析一个类而不执行它

I need to parse my classes (which they are in separate files) and search for a specified variable or method, then if there was that variable so have to execute that.

我需要解析我的类（它们在单独的文件中）并搜索指定的变量或方法，然后如果有那个变量则必须执行它。

I tried some ways:

我试过一些方法：

I used to parse my files with preg_match seeking for some pattern.(not a good idea)

我曾经用preg_match解析我的文件寻找一些模式。（不是一个好主意）
after that I try some code parser extension that was not good enough.

之后我尝试了一些不够好的代码解析器扩展。
finally I tried ArrayObject class that it was not fine at all.(because it was parsing objects not classes)

最后我尝试了ArrayObject类，它根本就不好。（因为它解析对象而不是类）

is there any way to do that?

有没有办法做到这一点？

Pseudo-code:

伪代码：

$file = file_get_content("dir/myClassFile.php");
$classname='foo';
if(there was '$variable=x' in $classname)
    include "dir/myClassFile.php";

edit

编辑

file content:

文件内容：

class foo {
    var include = true; //1st

    function bar()
    {
        $this->include = true; //2nd 
    }
}

I need to check if include is ture then I have to include then file.

我需要检查包含是否真的然后我必须包括然后文件。

1 个解决方案

#1

For simple syntax checking, using php on the command line goes a long way. Simply call:

对于简单的语法检查，在命令行上使用php有很长的路要走。只需致电：

$ php -l <filename>

For more complex tools, check out Sebastian Bergmann's github. He is the author of phpunit, and has developed some really useful tools for detecting copy-pasted code, dead code, etc.

有关更复杂的工具，请查看Sebastian Bergmann的github。他是phpunit的作者，并开发了一些非常有用的工具来检测复制粘贴的代码，死代码等。

Github Account

Github帐户

It would be a good idea to test your classes with PHPUnit, to make sure they function/work correctly.

使用PHPUnit测试您的类是一个好主意，以确保它们正常运行/工作。

There is also this PHP Code Sniffer that might come in use too.

这个PHP Code Sniffer也可能会被使用。

#1