简体   繁体   English

PHP - 使用常量的值来引用数据成员

[英]PHP - Using a constant's value to reference a data member

I am trying to access one class object's data member by using a constant. 我试图通过使用常量来访问一个类对象的数据成员。 I was wondering if this is possible with a syntax similar to what I am using? 我想知道这是否可能使用类似于我正在使用的语法?

When I attempt to do this in the following script I get this error: Parse error: syntax error, unexpected T_PAAMAYIM_NEKUDOTAYIM 当我尝试在以下脚本中执行此操作时,我收到此错误: 解析错误:语法错误,意外T_PAAMAYIM_NEKUDOTAYIM

class Certificate {
    const BALANCE = 'cert_balance';

    public function __construct() {}

}

class Ticket {
    public $cert_balance = null;

    public function __construct()
    {
        $this->cert_balance = 'not a chance';
        echo $this->cert_balance."<br />";
    }
}

$cert = new Certificate();

$ticket = new Ticket();

// This next code line should be equal to: $ticket->cert_balance = 'nice'; 

$ticket->$cert::BALANCE = 'nice!';

You need to disambiguate the expression with braces. 您需要使用大括号消除表达式的歧义。 Also, prior to PHP 5.3, you need to refer to the constant via the class name, like this: 此外,在PHP 5.3之前,您需要通过类名引用常量,如下所示:

$ticket->{Certificate::BALANCE} = 'nice!';

The PHP manual section on class constants says this 关于类常量PHP手册部分说明了这一点

As of PHP 5.3.0, it's possible to reference the class using a variable 从PHP 5.3.0开始,可以使用变量引用该类

So in PHP 5.3.0 and higher, this will work: 所以在PHP 5.3.0及更高版本中,这将起作用:

$ticket->{$cert::BALANCE} = 'nice!';

Do: 做:

$ticket->{$cert::BALANCE} = 'nice';

So the parser knows it has to process $cert::BALANCE first. 所以解析器知道它必须首先处理$cert::BALANCE It seems you need PHP 5.3 for this to work. 看来你需要PHP 5.3才能工作。 Otherwise, use the classname instead of $cert . 否则,请使用classname而不是$cert

The point is that you have to put it into {} . 关键是你必须把它放进{}

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM