Question

我正在研究一个由Symfony 3驱动的RESTful应用程序，使用FOSRestBundle，FOSUserBundle和FOSOAuthBundle。

我的问题是我想扩展FOS UserManager类，但这会引发致命错误：

类型错误：传递给FOS \ UserBundle \ Doctrine \ UserManager :: __ construct（）的参数1必须实现接口FOS \ UserBundle \ Util \ PasswordUpdaterInterface，没有给出，在/ var / www / html / var / cache / dev中调用第2538行的/appDevDebugProjectContainer.php

以下是我的应用程序文件的一部分：

# app/config/config.yml

fos_user:
    # [...]
    service:
        user_manager: foo.user_manager

# src/FooBundle/Resources/config/services.yml

services:
    # [...]
    foo.user_manager:
        class: FooBundle\Service\UserManager

    FooBundle\Service\UserManager:
        arguments:
            $class: '%fos_user.model.user.class%'

<?php

// src/FooBundle/Service/UserManager.php

namespace FooBundle\Service;

use FOS\UserBundle\Doctrine\UserManager as BaseUserManager;

class UserManager extends BaseUserManager implements UserManagerInterface
{
    /**
    * {@inheritDoc}
    * @see \FooBundle\Service\UserManagerInterface::findUsersAtPage()
    */
    public function findUsersAtPage($usersPerPage, $page) {
        $offset = $usersPerPage * $page;
        return $this->getRepository()->findBy(array(), null, $usersPerPage, $offset);
    }
}

<?php

// src/FooBundle/Service/UserManagerInterface.php

namespace FooBundle\Service;

use FOS\UserBundle\Model\UserManagerInterface as BaseUserManagerInterface;

interface UserManagerInterface extends BaseUserManagerInterface
{
    /**
     * Returns a collection with specified number of user instances at specified page.
     * @param int $usersPerPage The number of users per page.
     * @param int $page The page number.
     * @return \Traversable
     */
    public function findUsersAtPage($usersPerPage, $page);
}

我观察到的是，当我使用默认的FOS用户管理器时，appDevDebugProjectContainer.php会生成getter方法（带参数）：

<?php

// var/cache/dev/appDevDebugProjectContainer.php

class appDevDebugProjectContainer extends Container
{
    // [...]

    protected function getFosUser_UserManagerService()
    {
    return $this->services['fos_user.user_manager'] = new \FOS\UserBundle\Doctrine\UserManager(${($_ = isset($this->services['fos_user.util.password_updater']) ? $this->services['fos_user.util.password_updater'] : $this->getFosUser_Util_PasswordUpdaterService()) && false ?: '_'}, ${($_ = isset($this->services['fos_user.util.canonical_fields_updater']) ? $this->services['fos_user.util.canonical_fields_updater'] : $this->getFosUser_Util_CanonicalFieldsUpdaterService()) && false ?: '_'}, ${($_ = isset($this->services['fos_user.object_manager']) ? $this->services['fos_user.object_manager'] : $this->getFosUser_ObjectManagerService()) && false ?: '_'}, 'FooBundle\\Entity\\User');
    }

    // [...]
}

当我使用我的（没有参数）时：

<?php

// var/cache/dev/appDevDebugProjectContainer.php

class appDevDebugProjectContainer extends Container
{
    // [...]

    protected function getFosUser_UserManagerService()
    {
        return $this->services['fos_user.user_manager'] = new \FooBundle\Service\UserManager();
    }

    // [...]
}

拜托，有人可以帮助我吗？

Answer 1

好的家伙，这就是答案：

必须在服务定义中设置服务参数，而不是类服务定义。这不是很清楚，比word更好，有些代码：

# src/FooBundle/Resources/config/services.yml

services:
    # [...]
    foo.user_manager:
        class: FooBundle\Service\UserManager
        arguments:
            $class: '%fos_user.model.user.class%'

    FooBundle\Service\UserManager: '@foo.user_manager'

我以为我试过了，但我没有......

Answer 2

刚刚解决了类似的问题。也许对某人有帮助。 Check this answer

或者只是尝试一下

services:
    you_bundle.user_manager:
        class: YourBundle\Services\UserManager
        arguments: ["@fos_user.util.password_updater", "@fos_user.util.canonical_fields_updater", "@doctrine.orm.entity_manager", "%fos_user.model.user.class%"]

Symfony FOSUserBundle用户管理器扩展

2 个答案: