I am working a project and I would want to use a where clause, paginate and then sort in the collection in specific order. I have tried the result below but keeps throwing the errors below Method:
Illuminate\Database\Eloquent\Collection::links does not exist. (View:
/Applications/XAMPP/xamppfiles/htdocs/vermex/resources/views/equipments.blade.php)
The Product model is where I am getting the data and store in a variable called $equipment. If there is a better way of doing this, please help.
public function equipments()
{
$equipments = Product::where('product_category_id', 3)->paginate(2)-
>sortByDesc('id');
return view('equipments', compact('equipments'));
}
Try putting the orderBy before the paginate
$equipments = Product::where('product_category_id', 3)->orderBy('id', 'desc')->paginate(2);
sortByDesc is a collection method.
paginate will need to be last for links to be available in the blade view.
This is my model
User
Role
and relationship between of those models are many to many.
I want to create query like this:
return User::with('roles')->orderBy('roles.id')->paginate();
I don't want join because I created a base class for every model. I also don't want use orderBy after get because it must load all of my data and after that I should be able to sort and paginate it. So it is not a very good idea.
You can try something like this:
return User::with(['roles' => function ($query) {
$query->orderBy('id', 'desc');
}])->paginate();
But this will only order the eager loading attributes, but if you are interested to use join you can have something like this:
return User::with('roles')
->join('roles', 'user.id', '=', 'roles.user_id')
->orderBy('roles.id', 'desc')
->paginate();
In this you can easily use paginate which is your main concern.
Hope this helps.
User::where('role_id','!=','0')->orderBy('role_id','DESC')->paginate(10);
If you have a query that uses eager loading like this:
Brand::with('tags')
->where('id', $id)
->get();
A brand can have many tags.
I then also have an array of tag ids like this [2,4]. How do I add a condition to this query where it returns only those brands whose tags are in the array?
I tried the eager load constraints but that condition is then placed on the tags model, not the Brand.
I tried this also but it returns an unknown method error:
public function tagsIn($allTags){
return $this->belongsToMany('App\Tag', 'brand_tags')
->whereIn('tags.id', $allTags);
}
Brand::with('tags')
->tagsIn('[2,4]')
->get();
I suspect a possible limitation to getting it to work is the fact that Eloquent makes two separate database calls. But is there a way nevertheless?
DB::table('Brands')
->join('brand_tag','brands.id','=','brand_tag.brand_id')
->join('tags','brand_tag.tag_id','=','tags.id')
->whereIn('tags.id',$allTags)
->get();
Try this DB::table('name')->whereIn('column', array(1, 2, 3))->get();
I think you should use this package to handle tag. I used it in my projects. laravel-tagging
I'm trying to get a single column value from the first result of a Model's belongsToMany relationship query, as i'm returning the ->first() result of the relationship I was hoping $code->reward->title would work but it doesn't.
I get an Relationship method must return an object of type Illuminate\Database\Eloquent\Relations\Relation error
What I'm trying to do is the get the title of the current reward that is linked to a specific code - the code_reward pivot table has a valid_from and expires_at date as the reward linked to a code will change as time goes by, hence the need to get the currently active reward for that code.
Here's my code:
Model: Code
public function rewards()
{
return $this->belongsToMany('App\Reward')->withPivot('valid_from', 'expires_at')->withTimestamps();
}
public function reward()
{
$now = Carbon::now();
return $this->rewards()
->wherePivot('valid_from', '<', $now)
->wherePivot('expires_at', '>', $now)
->first();
}
Controller: CodeController
public function index()
{
$codes = Code::all();
return view('codes/index')->with('codes', $codes);
}
View: Codes/index
#foreach ($codes as $code)
{{$code->id}}
{{$code->reward->title}}
#endforeach
Any help is really appreciated!
Update
Unfortunately both suggestions below ($code->reward()->title and getRewardAttribute() return an Trying to get property of non-object error.
If I remove ->first() from the Code->reward() method and replace $code->reward->title with $code->reward->first() in the view it echoes out the whole reward model as json, however $code->reward->first()->title still returns the Trying to get property of non-object error
Update 2
If I do {{dd($code->reward->title)}} in the view I get the reward title but if I just do {{$code->reward->title}}, I don't!
AND the $code->reward->title works as expected in a #Show view, so could it be that the collection of codes supplied by the controller's #index method isn't passing the necessary data or not passing it in a necessary format??
SOLVED
The issue was caused by one of the $code->rewards in the foreach loop in the index view returning null! The first one didn't, hence the dd() working but as soon as the loop hit a null it crashed.
Once I wiped and refreshed the db (and made sure my seeds where adding only valid data!) it worked. Doing {{$code->reward ? $code->reward->title : ''}} fixed the issue. Grrr.
Your statement is failing because $code->reward->title tells Laravel that you have defined a relationship on your Code model in a method called reward(). However, your relationship is actually defined in the method rewards(). Instead, reward() is a custom method on the model that you have made up. Calling it as a method and not a relation is the quickest way to get what you want.
{{$code->reward()->title}}
As #andrewtweber points out below, you could also make your custom reward() method into an attribute accessor. To do that, just rename the function to getRewardAttribute() and then you can call it in your view like you originally did.
Alternatively, you could get rid of that reward() method entirely and move all of that logic to the controller, where it probably makes more sense. You'd have to use constrained eager loading to pull that off. So in your controller you'd have something like this:
$codes = App\Code::with(['rewards' => function ($query) {
$query->wherePivot('valid_from', '<', $now)
->wherePivot('expires_at', '>', $now);
])->get();
Of course, this would return all of your filtered codes. This is because you cannot apply a sql limit inside a nested eager relationship as outlined here. So in your view, you would then have to do something like this:
{{$code->rewards->first()->title}}
However, it will be simpler to go with my first solution, so that's entirely up to you.
Try to set this method in Code Model, because query builder treats valid_from and expired_at as string, not date?
public function getDates()
{
return ['valid_from','expired_at'];
}
I have got 2 joined tables in Eloquent namely themes and users.
theme model:
public function user() {
return $this->belongs_to('User');
}
user model:
public function themes() {
return $this->has_many('Theme');
}
My Eloquent api call looks as below:
return Response::eloquent(Theme::with('user')->get());
Which returns all columns from theme (that's fine), and all columns from user (not fine). I only need the 'username' column from the user model, how can I limit the query to that?
Change your model to specify what columns you want selected:
public function user() {
return $this->belongs_to('User')->select(array('id', 'username'));
}
And don't forget to include the column you're joining on.
For Laravel >= 5.2
Use the ->pluck() method
$roles = DB::table('roles')->pluck('title');
If you would like to retrieve an array containing the values of a single column, you may use the pluck method
For Laravel <= 5.1
Use the ->lists() method
$roles = DB::table('roles')->lists('title');
This method will return an array of role titles. You may also specify a custom key column for the returned array:
You can supply an array of fields in the get parameter like so:
return Response::eloquent(Theme::with('user')->get(array('user.username'));
UPDATE (for Laravel 5.2)
From the docs, you can do this:
$response = DB::table('themes')
->select('themes.*', 'users.username')
->join('users', 'users.id', '=', 'themes.user_id')
->get();
I know, you ask for Eloquent but you can do it with Fluent Query Builder
$data = DB::table('themes')
->join('users', 'users.id', '=', 'themes.user_id')
->get(array('themes.*', 'users.username'));
This is how i do it
$posts = Post::with(['category' => function($query){
$query->select('id', 'name');
}])->get();
First answer by user2317976 did not work for me, i am using laravel 5.1
Using with pagination
$data = DB::table('themes')
->join('users', 'users.id', '=', 'themes.user_id')
->select('themes.*', 'users.username')
->paginate(6);
Another option is to make use of the $hidden property on the model to hide the columns you don't want to display. You can define this property on the fly or set defaults on your model.
public static $hidden = array('password');
Now the users password will be hidden when you return the JSON response.
You can also set it on the fly in a similar manner.
User::$hidden = array('password');
user2317976 has introduced a great static way of selecting related tables' columns.
Here is a dynamic trick I've found so you can get whatever you want when using the model:
return Response::eloquent(Theme::with(array('user' => function ($q) {
$q->addSelect(array('id','username'))
}))->get();
I just found this trick also works well with load() too. This is very convenient.
$queriedTheme->load(array('user'=>function($q){$q->addSelect(..)});
Make sure you also include target table's key otherwise it won't be able to find it.
This Way:
Post::with(array('user'=>function($query){
$query->select('id','username');
}))->get();
I know that this is an old question, but if you are building an API, as the author of the question does, use output transformers to perform such tasks.
Transofrmer is a layer between your actual database query result and a controller. It allows to easily control and modify what is going to be output to a user or an API consumer.
I recommend Fractal as a solid foundation of your output transformation layer. You can read the documentation here.
In Laravel 4 you can hide certain fields from being returned by adding the following in your model.
protected $hidden = array('password','secret_field');
http://laravel.com/docs/eloquent#converting-to-arrays-or-json
On Laravel 5.5, the cleanest way to do this is:
Theme::with('user:userid,name,address')->get()
You add a colon and the fields you wish to select separated by a comma and without a space between them.
Using Model:
Model::where('column','value')->get(['column1','column2','column3',...]);
Using Query Builder:
DB::table('table_name')->where('column','value')->get(['column1','column2','column3',...]);
If I good understood this what is returned is fine except you want to see only one column. If so this below should be much simpler:
return Response::eloquent(Theme::with('user')->get(['username']));
#You can get selected columns from two or three different tables
$users= DB::Table('profiles')->select('users.name','users.status','users.avatar','users.phone','profiles.user_id','profiles.full_name','profiles.email','profiles.experience','profiles.gender','profiles.profession','profiles.dob',)->join('users','profiles.user_id','=','users.id')
->paginate(10);
Check out, http://laravel.com/docs/database/eloquent#to-array
You should be able to define which columns you do not want displayed in your api.