PHP 5.2中的语法错误

In the php 5.3 I can use class name as variable and I can call static variable.

在php 5.3中我可以使用类名作为变量，我可以调用静态变量。

$class_name = 'Test';
$class_name::$static_var;

How to call it in the php 5.2 version?

如何在php 5.2版本中调用它？

Parse error: syntax error, unexpected T_PAAMAYIM_NEKUDOTAYIM in ...

3 个解决方案

#1

T_PAAMAYIM_NEKUDOTAYIM is the double colon scope resolution thingy PHP uses - :: You can try this $class_name = 'Test'; $class_name->$static_var;

T_PAAMAYIM_NEKUDOTAYIM是PHP使用的双冒号范围解析 - ::你可以尝试这个$ class_name ='Test'; $ CLASS_NAME - > $ static_var;

#2

@user762799 here is the solution for what you want to do it in php 5.2

@ user762799这里是你想在php 5.2中做到的解决方案

class Sample{
    public static $name;

    public function __construct(){
        self::$name = "User 1";
    }
}

$sample = new Sample();
$class = 'Sample';
$name = 'name';
$val_name = "";
$str = '$class::$$name';
eval("\$val_name = \"$str\";");
//echo $val_name."<br>";
eval("\$name = $val_name;");
echo $name;

PAAMAYIM_NEKUDOTAYIM means scope resolution operator(::) actually in your code PHP is unable to identify $static_var in the scope of $class_name that is why the error occured.

PAAMAYIM_NEKUDOTAYIM意味着范围解析运算符（：:)实际上在您的代码中PHP无法识别$ class_name范围内的$ static_var，这就是发生错误的原因。

If you still not clear, let me know. Thank you :)

如果你还不清楚，请告诉我。谢谢：）

#3

You really should update your PHP version, 5.2 isn't supported anymore, but ...

你真的应该更新你的PHP版本，不再支持5.2，但是......

... in PHP 5.2 the only way to hack around this is to use eval:

...在PHP 5.2中，解决这个问题的唯一方法是使用eval：

$return = eval($class_name . '::\\$static_var;');

But be sure to validate $class_name before you use this, otherwise arbitrary code could be injected (e.g. $class_name = 'do_bad_things(); Class_Name).

但是在使用它之前一定要验证$ class_name，否则可能会注入任意代码（例如$ class_name ='do_bad_things（）; Class_Name）。

#1