Question

我想返回记录用户只能查看的网站。当用户访问其他人的网站但您不在该网站内时，您应该无法查看该网站。这应该只返回与您关联的站点。我有一个单元测试，它通过但似乎有一个更好的方法来做到这一点。 TIA

$user = $this->userRepo->findUserById($userId);

$userRepo = new UserRepository($user);

$sites = $userRepo->findSites();

$loggedUser = app('request')->user();

$loggedUserSites = $loggedUser->sites()->get()->all();

// Return only the sites of the user being access that is the same with the currently logged user
$sites = $sites->filter(function (Site $site) use ($loggedUserSites) {
    foreach ($loggedUserSites as $userSite) {
        if($site->id === $userSite->id) {
            return $site;
        };
    }
});

// user 1: [1,2,3] - `/users/2/sites` - should return [1,2] (default since user 2 is only associated with this 2 sites)
// user 2: [1,2] - `/users/1/sites` - should return [1,2] (no 3 since user has no site #3)

Answer 1

您可以使用类似whereIn()的内容：

$sites = $sites->whereIn('id', $loggedUserSites->pluck('id')->toArray())->all();

如果你使用＆gt; = 5.3，你也应该能够删除->toArray()方法。

使用laravel中的数组过滤集合

1 个答案: