代码之家 › 专栏 › 技术社区 › Calcimicium

Symfony FOSUserBundle用户管理器扩展

symfony-3.3 fosuserbundle symfony php

Calcimicium · 技术社区 · 8 年前

我正在开发一个由Symfony 3支持的RESTful应用程序,使用FOSRestBundle、FOSUserBundle和FOSOAuthBundle。

类型错误:传递给FOS\UserBundle\Doctrine\UserManager的参数1:\uu construct()必须实现接口FOS\UserBundle\Util\PasswordUpdaterInterface,未给定,在/var/www/html/var/cache/dev/appDevDebugProjectContainer中调用。php在线2538

以下是我的部分应用程序文件:

# 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);
}

<?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');
    }

    // [...]
}

当我使用我的时(没有参数):

<?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,有人能帮我吗?

2 回复 | 直到 8 年前

Calcimicium 8 年前

好了,伙计们,这就是答案:

服务参数必须在服务定义中设置,而不是在类服务定义中设置。这不是很清楚,所以比word更好,一些代码:

# 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'

Anton G 8 年前

Check this answer

或者试试这个

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%"]