Laravel Eloquent hasMany condition in parent table

You may try this:

$users = \App::User::with('reviews' => function($query) {
    $query->where('as_anoniem', '!=', 1);
})->get();

This will require you to create a one-to-many relationship in App\User model, for example:

// App\User.php
public function reviews()
{
    // namespace: App\Review
    return $this->hasMany(Review::class);
}

Assumed that, the namespace for both User & Review is App, they are in the same directory.

Update After the original question was changed by OP:

$places = \App::Place::with('reviews' => function($query) {
    $query->with('user')->where('reviews.as_anoniem', '!=', 1);
})
->get();

Place Model:

public function reviews()
{
    // namespace: App\Review
    return $this->hasMany(Review::class);
}

Review Model:

public function user()
{
    // namespace: App\User
    return $this->belongsTo(User::class);
}

It is possible to use values of model in conditions:

class Content extends Model
{

    // ...

    /**
     * Get linked content
     * @return \Illuminate\Database\Eloquent\Relations\HasMany
     */
    public function linked()
    {
        return $this->hasMany(self::class, 'source_content_id', 'source_content_id')
            ->where('source_content_type_id', '=', $this->source_content_type_id)
            ->where('id', '<>', $this->id);
    }
}

You can check out this link https://stackoverflow.com/a/18600698/16237933.

class Game extends Eloquent {
    // many more stuff here

    // relation without any constraints ...works fine 
    public function videos() {
        return $this->hasMany('Video');
    }

    // results in a "problem", se examples below
    public function available_videos() {
        return $this->videos()->where('available','=', 1);
    }
}