php - 我可以使用同一张表以不同方式/以更 Cake 的方式重构我的多个用户类型的设置吗？

Question

我有一个 my_users 表，现在我需要通过删除 role 列来重构它，以支持每个用户支持多个角色。例如，我正在处理的 3 种用户类型是:

1. 斗士
2. 裁判
3. 经理

所以表格设置如下所示:

Users.php

use CakeDC\Users\Model\Table\UsersTable as BaseUsersTable;

class UsersTable extends BaseUsersTable
{
    public function initialize(array $config)
    {   
        parent::initialize($config);

        $this->setEntityClass('Users\Model\Entity\User');

        $this->belongsToMany('UserRoles', [
            'through' => 'users_user_roles'
        ]); 

        $this->hasMany('UsersUserRoles', [
            'className' => 'Users.UsersUserRoles',
            'foreignKey' => 'user_id',
            'saveStrategy' => 'replace',
        ]);
    }

    public function findRole(Query $query, array $options)
    {
        return $query
            ->innerJoinWith('UserRoles', function($q) use ($options) {
                return $q->where(['role_key' => $options['role_key']]);
            });
    }
}

MyUsersTable.php

class MyUsersTable extends Table
{   
    public function initialize(array $config)
    {
        parent::initialize($config);

        $this->setTable('my_users');

        $this->setPrimaryKey('user_id');

        $this->belongsTo('Users', [
            'className' => 'Users.Users',
            'foreignKey' => 'user_id',
        ]);
    }

    /**
     * @param Cake\Event\Event $event
     * @param Cake\ORM\Query $query
     * @param ArrayObject $options
     */
    public function beforeFind(Event $event, Query $query, \ArrayObject $options)
{
        // set the role
        if (defined(static::class.'::ROLE') && mb_strlen(static::ROLE) > 0) {
            $role = static::ROLE;

            // set user conditions
            $query->innerJoinWith('Users', function($query) use ($role) {
                return $query->find('role', [
                    'role_key' => $role,
                ]);
            });
        }
    }

    /**
     * @param \Cake\ORM\Query $query
     * @param array $options
     */
    public function findByRegistrationCode(Query $query, array $options): Query
    {
        $query->where([
            $this->aliasField('registration_no') => $options['registration_no']
        ]);

        return $query;
    }
}

FightersTable.php

use MyUsers\Model\Table\MyUsers;

class FightersTable extends MyUsersTable
{       
    const ROLE = 'fighter';

    public function initialize(array $config)
    {   
        parent::initialize($config);

        $this->setEntityClass('Fighters\Model\Entity\Fighter');
    }       

    /**
     * @param Validator $validator
     * @return Validator $validator
     */
    public function validationDefault(Validator $validator): Validator
    {
        $validator = parent::validationDefault($validator);

        $validator->allowEmpty('field');
    }
}

RefereesTable.php 和 ManagersTable.php 类似于 FightersTable 但有自己的验证规则，可能有自己特殊的实体虚拟属性什么不是。

问题:有没有更简单的方法来构建它，或者更具体地说，有一种替代方法来执行 beforeFind 来区分角色？如果对 role 的要求与用户保持 1:1 的比例，我可能会做这样的事情:

$this->belongsTo('Fighters', [
    'conditions' => [
        'role' => 'fighter'
    ],
    'foreignKey' => 'user_id',
    'className' => 'MyUsers',
]);

我将不胜感激任何关于重构它的见解。

Answer 1

您可以定义belongsTo 关联以使用findByRole finder :

 $this->belongsTo('Fighters', [
    'foreignKey' => 'user_id',
    'className' => 'MyUsers',
    'finder' => ['byRole' => ['role' => 'fighter']]
]);

当然，您必须在 MyUsers 中定义查找器:

public function findByRole(Query $query, \ArrayObject $options)
{
    $role = $options['role'];

    // set user conditions
    $query->innerJoinWith('Users', function($query) use ($role) {
       return $query->find('role', [
          'role_key' => $role,
           ]);
        });
    }
}

我还会将 beforeFind 逻辑提取到 RoleBehavior 或 MultiRoleBehavior behavior ， IE。: src/Model/Behavior/RoleBehavior.php:

<?php
namespace App\Model\Behavior;

use Cake\ORM\Behavior;
use Cake\ORM\Query;
use Cake\Event\Event;

/**
 * Role-specific behavior
 */
class RoleBehavior extends Behavior
{

    /**
     * @var array multiple roles support
     */
    protected $_defaultConfig = [
        'roles' => []
    ];


    public function initialize(array $config)
    {
        parent::initialize($config);
        if (isset($config['roles'])) {
            $this->config('roles', $config['roles'], false /* override, not merge*/);
        }
    }

    public function beforeFind(Event $event, Query $query, \ArrayObject $options, $primary) {
        // set user conditions
        $query->innerJoinWith('Users', function($query) use ($roles) {
            return $query->find('role', [
                'role_key' => $roles,
            ]);
        });
    }
}

它使用多个角色，但如果需要，您可以简单地将其更改为单个角色。这里还有一些其他要做的事情——可能将 $query->innerJoinWith('Users'... 提取到行为中的一个单独的方法中，并在 MyUsers::findByRole(. ..)...

接下来，您可以将此行为直接附加到扩展类中，只需将静态 ROLE 替换为行为配置即可:

use MyUsers\Model\Table\MyUsers;

class FightersTable extends MyUsersTable
{       
    public function initialize(array $config)
    {   
        parent::initialize($config);

        $this->setEntityClass('Fighters\Model\Entity\Fighter');

        $this->addBehavior('Role', [ 'roles' => ['fighter']]);
    }
}

或者您可以通过将行为附加到 MyUsers 表(即来自 Controller )来管理特定于角色的逻辑:

$this->MyUsers->addBehavior('Role', ['roles' => ['manager']])

You also can change the behavior setup on the fly :

$this->MyUsers->behaviors()->get('Role')->config([
   'roles' => ['manager'], 
]);

希望对你有帮助。