在 PHP 中使用来自另一个条件的变量 [英] Using a variable from another condition in PHP

问题描述

我有这个代码:

if(!isset($_GET["act"]))
{
    $display->display("templates/install_main.html");
    if(isset($_POST["proceed"]))
    {
        $prefix = $_POST["prefix"];
    }
}

if($_GET["act"] == "act")
{
    echo $prefix;
}

基本上我之前提出过类似的问题，问题是，如何使变量可访问?请提及是否有任何方法可以做到这一点，即使改变了它的完成方式(有人告诉我可以使用课程，但不太确定如何完成)，或任何其他使其可访问的方式.

Basically I've made a similar question before, thing is, HOW can I make the variable accessible? please mention if there is any way to do so, even with changing the way it's done (someone told me it's possible with a class but not quite sure how it can be done), or any other way to make it accessible.

谢谢！

推荐答案

PHP 的变量作用域是函数级的.$prefix 将在您的第二个 if() IF 中可用，另一个 if() 评估为真并实际执行了 $prefix = ... 代码.

PHP's variable scope is function-level. $prefix would be available in your second if() IF the other if()'s evaluated to true and actually executed that $prefix = ... code.

例如

if (true) {
    $foo = 'bar'; // always executes
}
if (false) {
    $baz = 'qux'; // never executes
}
echo $foo; // works just fine
echo $baz; // undefined variable, because $baz='qux' never executed.

另请注意，PHP 不能进行时间​​旅行:

Also note that PHP is not capable of time travel:

echo $x; // undefined variable;
$x = 'y';
echo $y; // spits out 'y'

较早"的代码不会有较晚的"变量可用，因为实际创建/分配值给这些变量的代码还没有执行.

"earlier" code will not have "later" variables available, because the code that actually creates/assigns values to those variables won't have executed yet.

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