我正在为 Laravel 应用程序编写测试。在我的 AuthServiceProvider->boot()
中,我根据数据库中的权限表使用 $gate->define()
定义了许多用户能力。
基本上是这样的:
foreach ($this->getPermissions() as $permission) {
$gate->define($permission->name, function ($user) use ($permission) {
return $user->hasPermission($permission->name);
});
}
在我的测试中,我正在动态创建权限,但是 AuthServiceProvider
已经启动,这意味着我无法使用 @can
验证用户权限, 门
等
有没有合适的方法来处理这个问题?
最佳答案
我知道我在这个问题上迟到了一点,但我自己也遇到了同样的问题,因此这个问题没有一个全面的答案,这是我对同一问题的解决方案(在Laravel 5.3):
我在我的 app\Providers\AuthServiceProvider
中有这个:
/**
* Register any authentication / authorization services.
*
* @param Gate $gate
*/
public function boot(Gate $gate)
{
$this->registerPolicies();
if (!app()->runningInConsole()) {
$this->definePermissions($gate);
}
}
/**
* @param Gate $gate
*/
private function definePermissions(Gate $gate)
{
$permissions = Permission::with('roles')->get();
foreach($permissions as $permission) {
$gate->define($permission->key, function($user) use ($permission) {
return $user->hasRole($permission->roles);
});
}
}
这会在不测试时处理正常的应用程序流程,并在测试时禁用过早的策略注册。
在我的 tests/TestCase.php
文件中,我定义了以下方法(注意 Gate
指向 Illuminate\Contracts\Auth\Access\Gate
):
/**
* Logs a user in with specified permission(s).
*
* @param $permissions
* @return mixed|null
*/
public function loginWithPermission($permissions)
{
$user = $this->userWithPermissions($permissions);
$this->definePermissions();
$this->actingAs($user);
return $user;
}
/**
* Create user with permissions.
*
* @param $permissions
* @param null $user
* @return mixed|null
*/
private function userWithPermissions($permissions, $user = null)
{
if(is_string($permissions)) {
$permission = factory(Permission::class)->create(['key'=>$permissions, 'label'=>ucwords(str_replace('_', ' ', $permissions))]);
if (!$user) {
$role = factory(Role::class)->create(['key'=>'role', 'label'=>'Site Role']);
$user = factory(User::class)->create();
$user->assignRole($role);
} else {
$role = $user->roles->first();
}
$role->givePermissionTo($permission);
} else {
foreach($permissions as $permission) {
$user = $this->userWithPermissions($permission, $user);
}
}
return $user;
}
/**
* Registers defined permissions.
*/
private function definePermissions()
{
$gate = $this->app->make(Gate::class);
$permissions = Permission::with('roles')->get();
foreach($permissions as $permission) {
$gate->define($permission->key, function($user) use ($permission) {
return $user->hasRole($permission->roles);
});
}
}
这使我能够以多种方式在测试中使用它。考虑我的 tests/integration/PermissionsTest.php
文件中的用例:
/** @test */
public function resource_is_only_visible_for_those_with_view_permission()
{
$this->loginWithPermission('view_users');
$this->visit(route('dashboard'))->seeLink('Users', route('users.index'));
$this->visit(route('users.index'))->assertResponseOk();
$this->actingAs(factory(User::class)->create());
$this->visit(route('dashboard'))->dontSeeLink('Users', route('users.index'));
$this->get(route('users.index'))->assertResponseStatus(403);
}
/** @test */
public function resource_action_is_only_visible_for_those_with_relevant_permissions()
{
$this->loginWithPermission(['view_users', 'edit_users']);
$this->visit(route('users.index'))->seeLink('Edit', route('users.edit', User::first()->id));
$this->loginWithPermission('view_users');
$this->visit(route('users.index'))->dontSeeLink('Edit', route('users.edit', User::first()->id));
}
这在我的所有测试中都运行良好。希望对您有所帮助。
关于php - 在测试 Laravel 应用程序时,一个加载门如何定义?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/33110073/