不能在PHP的另一个命名空间中使用命名空间类

Question

我仍然遇到PHP5命名空间的问题。

我有一个命名空间称为Project，我试图访问此Project命名空间有Library一个命名空间，以便在该文件的顶部是一个Project命名空间我使用这行内称为registry类use Library\Registry;

Registry类是Library命名空间内

这应该工作，但它并没有，而不是唯一的方式来访问我的registry类这里面Project命名空间是使用此

$this->registry = new \Library\Registry; 

我希望能够改为使用此...

$this->registry = new Registry; 

这是使用

use Library\Registry; 

在Project空间文件的顶部的全部原因

---

下面我有3个小考试在这样的文件夹结构中运行脚本。
Library/registry.class.php在我的库文件夹
Controller/controller.class.php和类在我的控制器目录
Controller/testing.php测试文件来运行该脚本的类。

E：\图书馆\ Registry.class.php文件

<?php 
namespace Library 
{ 
    class Registry 
    { 
     function __construct() 
     { 
      echo 'Registry.class.php Constructor was ran'; 
     } 
    } 
} 
?> 

E：\控制器\ Controller.class.php文件

<?php 
use Library\Registry; 

namespace Project 
{ 
    class Controller 
    { 
     public $registry; 

     function __construct() 
     { 
      include('E:\Library\Registry.class.php'); 

      // This is where my trouble is 
      // to make it work currently I have to use 
      // $this->registry = new /Library/Registry; 
      // But I want to be able to use it like below and that is why 
      // I have the `use Library\Registry;` at the top 
      $this->registry = new Registry; 
     } 

     function show() 
     { 
      $this->registry; 
      echo '<br>Registry was ran inside testcontroller.php<br>'; 
     } 
    } 
} 
?> 

E：\控制器\ testing.php文件

<?php 
use Project\Controller; 

include('testcontroller.php'); 

$controller = new Controller(); 
$controller->show(); 

?> 

我得到这个错误...

Fatal error: Class 'Project\Registry' not found in PATH to file 

，除非我因为该文件在顶部使用下面这个就controller.class.php文件

$this->registry = new \MyLibrary\Registry; 

我有use Library\Registry;我应该能够像这样访问它...

$this->registry = new Registry; 

请帮我得到它，我可以用它这样的，而不是

Answer 1

use Library\Registry; 

namespace Project 
{ 

我相信这一轮是错误的方式：你是use荷兰国际集团Library\Registry在全局命名空间，然后打开Project命名空间。

将use语句放入要导入的命名空间中。

namespace Project 
{ 
    use Library\Registry; 

Answer 2

只需添加： 使用\库\注册表;

在命名空间声明在你的脚本的顶部

然后，你可以说：

$注册表=新登记;

类

里面顺便说你的类声明是完全错误的。你不应该把你的名字空间包裹在大括号中，名字空间不是一个函数。

这是应该的。还要确保已使用include（'/ path/to/registry.php'）包含了Library \ Registry的类声明。或使用自动加载器

namespace Project; 

include（'E：\ Library \ Registry.class.php'）;

use \Library\Registry; 

    class Controller 
    { 
     public $registry; 

     function __construct() 
     { 


      // This is where my trouble is 
      // to make it work currently I have to use 
      // $this->registry = new /Library/Registry; 
      // But I want to be able to use it like below and that is why 
      // I have the `use Library\Registry;` at the top 
      $this->registry = new Registry; 
     } 

     function show() 
     { 
      $this->registry; 
      echo '<br>Registry was ran inside testcontroller.php<br>'; 
     } 
    } 

享受

Answer 3

您需要导入内部Project命名空间的Registry类，因为你需要它们存在，而不是在全球范围内。

<?php 
namespace Project 
{ 
    use Library\Registry; 

    class Controller 
    { 
     public $registry; 

     function __construct() 
     { 
      include('E:\Library\Registry.class.php'); 

      // This is where my trouble is 
      // to make it work currently I have to use 
      // $this->registry = new /Library/Registry; 
      // But I want to be able to use it like below and that is why 
      // I have the `use Library\Registry;` at the top 
      $this->registry = new Registry; 
     } 

     function show() 
     { 
      $this->registry; 
      echo '<br>Registry was ran inside testcontroller.php<br>'; 
     } 
    } 
} 
?> 

Answer 4

<?php namespace Project; 
    require_once 'your registry class file'; 
    use \Library\Registry as Registry; 

现在，您将能够使用...

$this->registry = new Registry;