Php 为什么实现接口的类不是该接口的实例？

我想这样做依赖注入

<?php
    class UserModel
    {
        public function __construct(DatabaseInterface $db)
        {
            $this->db = $db;
        }
    }

我不知道为什么，因为PdoConnection实现了DatabaseInterface，所以我认为UserModel应该接受它

下面的调试

$class = new \Manager\Database\PdoConnection();
if ($class instanceof DatabaseInterface) {
    echo "variable class is a instance of DatabaseInterface";
} else {
    echo "variable class is not a instance of DatabaseInterface";
}

var_dump(class_implements($class));

结果

variable class is not a instance of DatabaseInterface

array(1) {
  'Manager\Database\DatabaseInterface' =>
  string(34) "Manager\Database\DatabaseInterface"
}

---

您需要到处编写

Manager\Database\DatabaseInterface

，而不是简单地编写

DatabaseInterface

我想你需要到处写

Manager\Database\DatabaseInterface

，而不是简单地写

DatabaseInterface

@Adder，谢谢！我忘了导入带有“use”的接口。请随意发布答案，我会接受的

$class = new \Manager\Database\PdoConnection();
if ($class instanceof DatabaseInterface) {
    echo "variable class is a instance of DatabaseInterface";
} else {
    echo "variable class is not a instance of DatabaseInterface";
}

var_dump(class_implements($class));

variable class is not a instance of DatabaseInterface

array(1) {
  'Manager\Database\DatabaseInterface' =>
  string(34) "Manager\Database\DatabaseInterface"
}