PHP：从变量中实例化一个类奇怪的失败 [英] PHP: Instantiating a class from a variable oddly fails

问题描述

我试图通过一个变量来实例化一个类（Laravel3雄辩模型），而且我收到一个错误，说没有找到该类。

当我对类名进行硬编码时，但是，它的工作原理很好。

（FYI，代码如下 $ contact_type 预计是电话，传真或电子邮件。）

I'm trying to instantiate a class (a Laravel3 Eloquent model) by a variable, and I'm getting an error saying that the class is not found.
When I hardcode the class name, though, it works just fine.
(FYI, in the code below $contact_type is expected to either be Phone, Fax, or Email.)

这是我正在玩的时刻：

    foreach( $input AS $contact_type => $contact_info )
    {
        foreach( $contact_info AS $data )
        {
            $obj = new $contact_type( (array)$data);
            echo'<pre>Obj: ',print_r($obj),'</pre>';  // <----- For Testing
        }
    }

当我运行上面的代码时，它会引发一个 Class'Phone'not found 错误。

当我替换新$ contact_type（）与新的电话（）（或传真或电子邮件），它的工作很好。

我打赌有一些简单的我只是看看:)我缺少什么？

请帮助！

When I run the code as above, it throws a "Class 'Phone' not found" error.
When I replace new $contact_type() with new Phone() (or Fax or Email), it works just fine.
I bet there's something simple I'm just looking over :) What am I missing?
Please help!

推荐答案

涵盖此的相关手册条目是此处和 here 。引用第二个链接：

The relevant manual entries that covers this are here and here. To quote the second link:

如果一个包含类的名称的字符串与new一起使用，则该类的新实例将被创建。如果类位于命名空间中，则在完成此操作时必须使用其完全限定名称。

If a string containing the name of a class is used with new, a new instance of that class will be created. If the class is in a namespace, its fully qualified name must be used when doing this.

这对我有用：

class Phone {}
$classtype = 'Phone';
$phone = new $classtype();
var_dump($phone);

生成输出：

object(Phone)#1 (0) {
}

查看以确保您不在命名空间中（如果您是字符串，请包括Phone类的名称空间）。或者，您也可以尝试使用反射：

Look to make sure you're not in a namespace (include the Phone class' namespace in the string if you are). Or, you can also try using reflection:

class Phone {}
$classtype = 'Phone';
$reflectionClass = new ReflectionClass($classtype);
$phone = $reflectionClass->newInstanceArgs();
var_dump($phone);

如果Phone类在命名空间中，这也可以：

If the Phone class is in a namespace, this also works:

Phone.php

<?php namespace Contact;

class Phone {}

test.php

<?php

include 'Phone.php';

$classtype = 'Contact\Phone';
$phone = new $classtype();
var_dump($phone);

我不是100％肯定为什么，虽然我怀疑当评估变量类名时命名空间映射不可见。考虑这个代码：

I'm not 100% sure why, although I suspect that when evaluating a variable class name the current namespace mappings aren't visible. Consider this code:

<?php namespace Foo;

use SomePackage\SomeClass as WeirdName;

$test = new WeirdName();

与以下项目进行比较：

<?php namespace Foo;

use SomePackage\SomeClass as WeirdName;

$class = 'WeirdName';
$test = new $class();

当PHP决定为新实例分配内存时，它将如何知道映射类名称别名 WeirdName to SomePackage\Someclass ？该别名仅对当前文件有效，在 userland 代码，更不用说相同的文件。

When PHP decides to allocate memory for a new instance, how will it know to map the class name alias of WeirdName to SomePackage\Someclass? That alias is only valid for the current file, and the code that actually performs that operation isn't even in userland code, much less the same file.

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