问题描述

MySQLi 在没有命名空间的类内部和类外部都可以正常工作.

MySQLi works fine inside a class with no namespace and outside a class.

我最近开始使用命名空间，现在我偶然发现了一个类似于以下的代码:

I recently started using namespace and now I have stumbled on a code much like the following:

 namespace Project;

 class ProjectClass{

      public static function ProjectClassFunction{
          $db = new mysql(data, data, data, data);
      }

 }

但是，它通过一条消息向我报告

However, it reports back to me with a message

致命错误:未找到类Projectmysqli""

"Fatal error: Class 'Projectmysqli' not found"

如何在使用命名空间的类中使用 mysqli?

How do I use mysqli inside a class which uses namespace?

推荐答案

默认情况下，PHP 将尝试从您当前的命名空间加载类.引用全局命名空间中的类:

By default, PHP will try to load classes from your current namespace. Refer to the class in the global namespace:

$db = new mysqli(/* ... */);

这与您在引用不同命名空间中的类时所做的相同:

This is the same thing you'd do when referring to a class in a different namespace:

$foo = new SomeNamespaceFoo();

请注意，如果您不使用开头的反斜杠，PHP 将尝试加载相对于您当前命名空间的类.以下代码将在名称空间 ProjectSomeNamespace 中查找名为 Foo 的类:

Note that if you left off the beginning backslash, PHP would try to load the class relative to your current namespace. The following code will look in the namespace ProjectSomeNamespace for a class named Foo:

namespace Project;
$foo = new SomeNamespaceFoo();

或者，您可以显式导入命名空间并避免歧义:

Alternatively, you can explicitly import namespaces and save yourself ambiguity:

namespace Project;

use Mysqli;

class ProjectClass
{
    public static function ProjectClassFunction()
    {
        $db = new Mysqli(/* ... */);
    }
}

这篇关于如何在命名空间中使用 MySQLi的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！