I created a model Game using a condition / constraint for a relation as follows:
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->hasMany('Video')->where('available','=', 1);
}
}
When using it somehow like this:
$game = Game::with('available_videos')->find(1);
$game->available_videos->count();
everything works fine, as roles is the resulting collection.
MY PROBLEM:
when I try to access it without eager loading
$game = Game::find(1);
$game->available_videos->count();
an Exception is thrown as it says "Call to a member function count() on a non-object".
Using
$game = Game::find(1);
$game->load('available_videos');
$game->available_videos->count();
works fine, but it seems quite complicated to me, as I do not need to load related models, if I do not use conditions within my relation.
Have I missed something? How can I ensure, that available_videos are accessible without using eager loading?
For anyone interested, I have also posted this issue on http://forums.laravel.io/viewtopic.php?id=10470
//lower for v4 some version
public function videos() {
$instance =$this->hasMany('Video');
$instance->getQuery()->where('available','=', 1);
return $instance
}
//v5
public function videos() {
return $this->hasMany('Video')->where('available','=', 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.
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 $result;
}
Model (App\Post.php):
/**
* Get all comments for this post.
*/
public function comments($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
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();
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() );
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);
}
Source: Stackoverflow.com