95

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

Remluben
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  • Best way to implement role based user management in laravel is to use the sentry package. So give it a try. – harishannam Aug 30 '13 at 01:06
  • As I said in my description above, the Model names are just an example, my problem has got nothing to do with a user management. I am going to edit my question and post my real wordl example. – Remluben Sep 03 '13 at 19:13
  • Awesome question thank you! And great answers to the guys below. Saved me time on my project. – Qazi Ammar Sep 27 '22 at 11:14

8 Answers8

116

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() );
Antonio Carlos Ribeiro
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  • Thanks for your reply, your example perfectly works, but so does mine when calling $game->available_videos()->get(). The point is, that I would like to use $game->available_videos as I can also use $game->videos without running into trouble. As far as I know Eloquent should load the related models if not eager loaded before, or am I wrong? – Remluben Sep 03 '13 at 21:19
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    update function name to `availableVideos`. If you only need the videos to return available videos you could do: `return $this->hasMany('Video')->where('available','=', 1);` – Connor Leech Feb 28 '18 at 19:25
45

//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);
}
xsilen T
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44

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();
Sabrina Leggett
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17

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;
    }
Nikunj K.
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16

Just in case anyone else encounters the same problems.

Note, that relations are required to be camelcase. So in my case available_videos() should have been availableVideos().

You can easily find out investigating the Laravel source:

// Illuminate\Database\Eloquent\Model.php
...
/**
 * Get an attribute from the model.
 *
 * @param  string  $key
 * @return mixed
 */
public function getAttribute($key)
{
    $inAttributes = array_key_exists($key, $this->attributes);

    // If the key references an attribute, we can just go ahead and return the
    // plain attribute value from the model. This allows every attribute to
    // be dynamically accessed through the _get method without accessors.
    if ($inAttributes || $this->hasGetMutator($key))
    {
        return $this->getAttributeValue($key);
    }

    // If the key already exists in the relationships array, it just means the
    // relationship has already been loaded, so we'll just return it out of
    // here because there is no need to query within the relations twice.
    if (array_key_exists($key, $this->relations))
    {
        return $this->relations[$key];
    }

    // If the "attribute" exists as a method on the model, we will just assume
    // it is a relationship and will load and return results from the query
    // and hydrate the relationship's value on the "relationships" array.
    $camelKey = camel_case($key);

    if (method_exists($this, $camelKey))
    {
        return $this->getRelationshipFromMethod($key, $camelKey);
    }
}

This also explains why my code worked, whenever I loaded the data using the load() method before.

Anyway, my example works perfectly okay now, and $model->availableVideos always returns a Collection.

Remluben
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  • No that was not the reason definitely. You should always use function name with underscore `_` because, in MVC there should always become a URL structure like `/controller/function_name`. If you did not set proper routing for the function you will get the camel cased function name in the URL, which is not SEO friendly. – Vineet Apr 10 '16 at 16:46
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    @Vineet sorry, no. MVC has nothing to do with the URL structure. The routes.php file defines the mapping from url to controller. – Greg Oct 27 '16 at 21:56
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    Not to mention that this is in the model, not the controller. It has absolutely zero correspondence to the URL. – Charles Wood May 28 '19 at 13:25
9
 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);
}
kush
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7

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.

Vitalii
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2

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
Santo Boldizar
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