How to access model hasMany Relation with where condition?

I think that this is the correct way:

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);
    }
}

And then you'll have to

$game = Game::find(1);
var_dump( $game->available_videos()->get() );

I think this is what you're looking for (Laravel 4, see http://laravel.com/docs/eloquent#querying-relations)

$games = Game::whereHas('video', function($q)
{
    $q->where('available','=', 1);

})->get();

use getQuery() to add condition

public function videos() {
    $instance = $this->hasMany('Video');
    $instance->getQuery()->where('available','=', 1);
    return $instance
}

simply

public function videos() {
    return $this->hasMany('Video')->where('available','=', 1);
}

If you want to apply condition on the relational table you may use other solutions as well.. This solution is working from my end.

public static function getAllAvailableVideos() {
        $result = self::with(['videos' => function($q) {
                        $q->select('id', 'name');
                        $q->where('available', '=', 1);
                    }])                    
                ->get();
        return $result;
    }

 public function outletAmenities()
{
    return $this->hasMany(OutletAmenities::class,'outlet_id','id')
        ->join('amenity_master','amenity_icon_url','=','image_url')
        ->where('amenity_master.status',1)
        ->where('outlet_amenities.status',1);
}

I have fixed the similar issue by passing associative array as the first argument inside Builder::with method.

Imagine you want to include child relations by some dynamic parameters but don't want to filter parent results.

Model.php

public function child ()
{
    return $this->hasMany(ChildModel::class);
}

Then, in other place, when your logic is placed you can do something like filtering relation by HasMany class. For example (very similar to my case):

$search = 'Some search string';
$result = Model::query()->with(
    [
        'child' => function (HasMany $query) use ($search) {
            $query->where('name', 'like', "%{$search}%");
        }
    ]
);

Then you will filter all the child results but parent models will not filter. Thank you for attention.

Model (App\Post.php):

/**
 * Get all comments for this post.
 */
public function comments($published = false)
{
    $comments = $this->hasMany('App\Comment');
    if($published) $comments->where('published', 1);

    return $comments;
}

Controller (App\Http\Controllers\PostController.php):

/**
 * Display the specified resource.
 *
 * @param int $id
 * @return \Illuminate\Http\Response
 */
public function post($id)
{
    $post = Post::with('comments')
        ->find($id);

    return view('posts')->with('post', $post);
}

Blade template (posts.blade.php):

{{-- Get all comments--}}
@foreach ($post->comments as $comment)
    code...
@endforeach

{{-- Get only published comments--}}
@foreach ($post->comments(true)->get() as $comment)
    code...
@endforeach

To access a hasMany relationship with a where condition, you can use the whereHas method.

Here's an example:

$orders = Order::whereHas('items', function ($query) {
   $query->whereNotNull('merchant_referral_percentage');
})->get();

In this example, Order is the model with the hasMany relationship to items. The whereHas method is used to filter the orders based on a condition on the items relationship. The condition in this case is that the merchant_referral_percentage column should not be null.

You can modify the condition inside the where callback function to suit your specific needs.