Symfony 5.0.8 - 如何检查实体是否与 twig 中的关系记录？

Question

I have User, Post and Bookmark entities.

inside the Bookmark entity:

• user -> ManyToOne
• Post -> ManyToOne
• date -> datetime

I'm displaying a post page like so "post/{id}" inside of the post display page, how can I check if the user has bookmarked this post?

Tried:

{% if post.bookmarks.contains(app.user) %} 

{% if app.user.bookmarks.contains(post) %} // I thought it might do it magically

{% if app.user.bookmarks.contains({'post': post}) %}

I think it's looking for the bookmark object and I can't provide it without doing a for loop.

Update:

With the help of some people in the php chat came up with this solution:

{% if app.user.bookmarks|filter(b => b.property.id == property.id)|length %}

However, I'm wondering if there is a more effective solution that can be done in the controller and then passed to the view.

Update

I don't like doing logic in the view, so the solution I found was in side he Post entity I created a function:

public function isPostBookmarkedByUser(User $user, Post $post): bool
{
    $bookmarks = $user->getBookmarks();

    foreach ($bookmarks as $bookmark) {
        if ($bookmark->getPost() === $post) {
            return true;
        }
    }
    return false;
}

Then I used this function inside the view like so:

{% if post.isBookmarkedByUser(app.user, post) %}
   <p>You bookmarked this post</p>
{% endif %}

Answer 1

Well, there's a lot of ways to obtain this, and there's not better or worst.
Depending on what makes sense in your domain, for example, you can provide a public method onto Post entity; something like isBookmarkedByUser($user) or you can add it to User entity; like hasBookmarkedPost($post) .
You can also make a Twig function and query directly the DB (avoiding looping and lazy-loading all the records of the Collection) or, maybe, you can loop directly in twig.
As you can see from my answer, the possibilities are endless. Pick your posion: it's up to you.