PHP:通过字符串值获取静态类的实例 [英] PHP: Get instance of static class by string value

问题描述

我正在开发一个PHP Web API，该Web API已交付给我，其中包含许多需要重构的代码.编写该代码的人希望将静态配置类包含到api资源中，然后获取该类的实例，如下所示:

I'm working on a php web api that was handed to me with a lot of code that needs to be refactored. The ones that wrote the code wanted to include a static configuration class to an api resource and then get an instance of that class something like this:

<?php
$obj = "User";
$confObjectSuffix = "_conf";
$confObject = $obj.$confObjectSuffix;
if ($confObject::inst()->checkMethod($method)) {
.....

由于$ confObject是字符串而不是对象，因此出现错误解析错误:语法错误，在.....中出现意外的T_PAAMAYIM_NEKUDOTAYYIM".

This gives the error "Parse error: syntax error, unexpected T_PAAMAYIM_NEKUDOTAYIM in ....." since $confObject is a string and not a object.

我写了一些测试代码:

<?php

$class = "User_conf";
echo "<pre>";
print_r($$class::Inst());
echo "</pre>";

class User_conf {
    private static $INSTANCE = null;

    public static function Inst() {
        if(User_conf::$INSTANCE === null) {
            User_conf::$INSTANCE = new User_conf();
        }

        return User_conf::$INSTANCE;
    }
}

但是也不能使其与$$一起使用，是否还有其他解决方法?我不想重写过多的内容.

But can't get it to work with $$ either, is there some other way around this? I don't want to rewrite more than necessary.

推荐答案

您可以使用call_user_func捕获实例，然后根据需要对其进行处理:

You can use call_user_func to capture the instance, then process it as needed:

$instance = call_user_func(array($confObject, 'inst'));

if($instance->checkMethod($method)) {
    ...

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