在PHP中查看生成和保留名称

问题描述：

这有点特别;我认为这实际上是不可能的，但 SO 社区一次又一次让我感到惊讶;所以这里。

This is a bit of a peculiar one; I don't think this is in fact possible, however the SO community has surprised me time and time again; so here goes.

在PHP中给出;我有以下代码段：

Given in PHP; I have the following snippet:

$path = 'path/to/file.php';
$buffer = call_user_func(function() use($path){
    ob_start();
    require $path;
    return ob_get_clean();
});

如果包含， path / to / file.php 在其范围内将有 $ path 。 有没有办法阻止此变量在包含文件的上下文中可用？

When included, path/to/file.php will have $path in it's scope. Is there any way to prevent this variable from being available in the context of the included file?

例如，给定 unset（）返回它未设置的变量的值，我可以这样做：

For instance, given unset() returned the value of the variable it was unsetting, I could do:

require unset($path);

但当然不起作用。

对于那些好奇的人，我试图阻止 $ path 继承include- 中的值呃。

For those curious, I'm trying to prevent $path from inheriting a value from the include-er.

通过混淆安全是我所做的考虑;传递类似 $ thisIsThePathToTheFileAndNobodyBetterUseThisName 之类的东西，但这看起来有点傻，但仍然不是万无一失。

"Security-by-obfuscation" is a consideration I made; passing something like $thisIsThePathToTheFileAndNobodyBetterUseThisName, but that seems a bit silly and still isn't foolproof.

其他保留应继承的变量，我已经使用 extract（）和 unset（）：

For other "reserved" variables that should be inherited, I've already went with extract() and unset():

$buffer = call_user_func(function() use($path, $collection){
    extract($collection);
    unset($collection);
    ob_start();
    // ...

编辑：

我最终选择的是：

What I finally went with:

$buffer = call_user_func(function() use(&$data, $registry){
    extract($registry, EXTR_SKIP);
    unset($registry);
    ob_start();
    // only $data and anything in $registry (but not $registry) are available
    require func_get_arg(0);
    return ob_get_clean();
}, $viewPath);

也许我的问题有点儿误导，通过我使用 use（）将变量传递到匿名函数范围;传递参数是我忽略的选项。

Perhaps my question was a bit misleading, through my use of use() to pass variables into the anonymous function scope; passing arguments was an option I neglected to mention.

关于@hakre和 use（） + func_get_args（）：

Regarding @hakre and use() + func_get_args():

$var = 'foo';
$func = function() use($var){
    var_dump(func_get_args());
};
$func(1, 2, 3);

/* produces
array(3) {
  [0]=>
  int(1)
  [1]=>
  int(2)
  [2]=>
  int(3)
}

答

使用 func_get_arg（） 而不是使用传统的函数参数：

Use func_get_arg() instead of using traditional function arguments:

$buffer = call_user_func(function() {
    ob_start();
    require func_get_arg(0);
    return ob_get_clean();
}, $path);

在PHP中查看生成和保留名称

编辑：

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