php – 我可以使用相同的表以不同的方式重构我的多个用户类型的
我有一个my_users表,我现在需要通过删除角色列来重构它,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:
>战士 因此表设置如下所示: Users.php use CakeDCUsersModelTableUsersTable as BaseUsersTable; class UsersTable extends BaseUsersTable { public function initialize(array $config) { parent::initialize($config); $this->setEntityClass('UsersModelEntityUser'); $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',]); } /** * @param CakeEventEvent $event * @param CakeORMQuery $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 CakeORMQuery $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 MyUsersModelTableMyUsers; class FightersTable extends MyUsersTable { const ROLE = 'fighter'; public function initialize(array $config) { parent::initialize($config); $this->setEntityClass('FightersModelEntityFighter'); } /** * @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,可能有自己的特殊实体虚拟属性,而不是. 问题:是否有一种更好的结构方式,或者更具体地说是另一种方法来进行beforeFind以区分角色?如果角色的要求与用户保持1:1,我可能会做这样的事情: $this->belongsTo('Fighters',[ 'conditions' => [ 'role' => 'fighter' ],'className' => 'MyUsers',]); 我很欣赏任何有关重组的见解.
您可以定义belongsTo关联以使用findByRole
finder:
$this->belongsTo('Fighters',[ 'foreignKey' => 'user_id','finder' => ['byRole' => ['role' => 'fighter']] ]); 当然,你必须在MyUsers中定义finder: 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,]); }); } } 我还将beforeFind逻辑提取到RoleBehavior或MultiRoleBehavior behavior中,即: <?php namespace AppModelBehavior; use CakeORMBehavior; use CakeORMQuery; use CakeEventEvent; /** * 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,ArrayObject $options,$primary) { // set user conditions $query->innerJoinWith('Users',function($query) use ($roles) { return $query->find('role',[ 'role_key' => $roles,]); }); } } 它使用多个角色,但如果需要,您只需将其更改为单个角色即可.以下是其他一些事情要做 – 可能将$query-> innerJoinWith(‘Users’…提取到行为中的单独方法中,并在MyUsers :: findByRole(…)中调用它… 接下来,您可以将此行为直接附加到扩展类,只需使用行为配置替换静态ROLE: use MyUsersModelTableMyUsers; class FightersTable extends MyUsersTable { public function initialize(array $config) { parent::initialize($config); $this->setEntityClass('FightersModelEntityFighter'); $this->addBehavior('Role',[ 'roles' => ['fighter']]); } } 或者,您可以通过将行为附加到MyUsers表(即从控制器)来管理您的特定于角色的逻辑: $this->MyUsers->addBehavior('Role',['roles' => ['manager']]) You also can change the behavior setup on the fly: $this->MyUsers->behaviors()->get('Role')->config([ 'roles' => ['manager'],]); 我希望它有所帮助. (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |