I have two entities: Producers and Protocols. Producers has many Protocols.
Problem
Select all producers ordered by their last protocol creation date.
Solution
In order to get this, I tried:
$producers = Producer::join('protocol', 'protocol.producer_id', '=', 'producer.id')
->orderBy('protocol.created_at', 'desc');
But just didn't work as expected mainly because I even limited the last protocol. How can I achieve this using Laravel Eloquent?
Use groupBy():
$producers = Producer::select('producer.*')
->join('protocol', 'protocol.producer_id', '=', 'producer.id')
->groupBy('producer.id')
->orderByRaw('max(protocol.created_at) desc');
first at all describe a relationship for last protocol on Producer model
public function last_protocol(){
return $this->hasOne(Protocol::class)->latest();
}
then, in your controller
$producers = Producer::with('last_protocol')->get()->sortByDesc('last_protocol.created_at');
Related
hello I am new to laravel and maybe I am a bit confused between eloquent and query builder way for writing a query but anyway can you please tell me what could be the best eloquent way to retrieve info like this in laravel 6 or 7
User > hasMany > Recipes
Recipe > belongsTo > User
I want to check if user id 2 present in users table then get only one post which id is 3
Query builder is for explicitly building SQL queries, and does not return instances of your models. Eloquent query builder, is similar but the result will contain the model(s) loaded with all their attributes, and has some handy functions for querying the relations you define in your models.
Given the limited information in your post, I am assuming when you say a post, you mean a recipe:
Query Builder:
DB::table('users')
->join('recipes', 'recipes.user_id', '=', 'users.id')
->select(['users.some_col', ... 'recipes.some_col'])
->where('users.id', 2)
->get();
If you have your models setup with the relations. You can use Eloquent like so:
User::where('id', 2)->with('recipes')->get();
If I understand you correctly it would be like this:
User::whereId($userId) //asuming it is 2
->with(['recipes' => function($q) use($recipeId) {
$q->where('id', $recipeId); //assuming it is 3
}])->first();
you can do this if i understand correctly:
$user = User::findOrFail(2); //auto 404 if user not found
$recipe = $user->Recipes()->where('id',3)->first();
You may use conditional eager loading for better performance.
$userId = 2;
$receiptId = 3;
$user = User::with(['receipts'=> function ($query) use($receiptId){
$query->where('id', $receiptId);
}
])->find($userId)
I have a query which selects some relationships and groups by the accommodations to remove the duplicate accommodations. But now when I want to load the discount realtion on rooms, it doesn't work because I select only the accommodation_id.
Here is my code:
$data = AccommodationRoom::with('accommodation.city', 'accommodation.accommodationFacilities', 'accommodation.gallery','discount')
->select('accommodation_id')
->whereHas('roomCapacityHistory', function ($query) use ($from_date, $to_date) {
$query->whereDate('from_date', '<=', $from_date);
$query->whereDate('to_date', '>=', $to_date);
})
->whereHas('accommodation', function ($query) use ($searchCity) {
$query->where('city_id', $searchCity);
})
->groupBy('accommodation_id')
->get();
Now if I add the id to the select it would be fine, but my groupBy doesn't work and gives me an error in this case. So I need a solution to get all my accomodations with the listed relations.
As you are looking for all your accommodations with some related models, you should actually select from your Accommodation model. This will work out of the box for the first three relations but will require some tweaking for the discount relation. The simpliest solution is to create a HasManyThrough relation on the Accommodation model:
class Accommodation extends Model
{
public function discounts()
{
return $this->hasManyThrough(Discount::class, AccommodationRoom::class);
}
}
Note: this expects your models to use foreign key columns named by convention; for different names you will need to pass the custom foreign key names as additional parameters according to the documentation.
With this relation set up, you can then use a query like the following:
$data = Accommodation::with('city', 'accommodationFacilities', 'gallery', 'discounts')
->whereHas('accommodationRooms.roomCapacityHistory', function ($query) use ($from_date, $to_date) {
$query->whereDate('from_date', '<=', $from_date);
$query->whereDate('to_date', '>=', $to_date);
})
->where('city_id', $searchCity)
->get();
Further explanation as asked for in the comments:
HasManyThrough builds a virtual relation between two models using a third model in between. Imagine you have Post, Comment and Like as models. One Post can have many Comments and one Comment can have many Likes:
has many has many
Post -----------------> Comment -----------------> Like
1 n 1 n
In this case we also know that one Post can have many Likes. And this is exactly the knowledge we utilize when using HasManyThrough. We simply tell Eloquent that a Post has many Likes, connected by the Posts in between:
has many has many
Post -----------------> Comment -----------------> Like
1 n 1 n
has many through Comment
Post ------------------------------------------------> Like
1 n
This may be a dupe but I've been trawling for some time looking for a proper answer to this and haven't found one yet.
So essentially all I want to do is join two tables and attach a where condition to the entire collection based on a field from the joined table.
So lets say I have two tables:
users:
-id
-name
-email
-password
-etc
user_addresses:
-address_line1
-address_line2
-town
-city
-etc
For the sake of argument (realising this may not be the best example) - lets assume a user can have multiple address entries. Now, laravel/eloquent gives us a nice way of wrapping up conditions on a collection in the form of scopes, so we'll use one of them to define the filter.
So, if I want to get all the users with an address in smallville, I may create a scope and relationships as follows:
Users.php (model)
class users extends Eloquent{
public function addresses(){
return $this->belongsToMany('Address');
}
public function scopeSmallvilleResidents($query){
return $query->join('user_addresses', function($join) {
$join->on('user.id', '=', 'user_addresses.user_id');
})->where('user_addresses.town', '=', 'Smallville');
}
}
This works but its a bit ugly and it messes up my eloquent objects, since I no longer have a nice dynamic attribute containing users addresses, everything is just crammed into the user object.
I have tried various other things to get this to work, for example using a closure on the relationship looked promising:
//this just filters at the point of attaching the relationship so will display all users but only pull in the address where it matches
User::with(array('Addresses' => function($query){
$query->where('town', '=', 'Smallville');
}));
//This doesnt work at all
User::with('Addresses')->where('user_addresses.town', '=', 'Smallville');
So is there an 'Eloquent' way of applying where clauses to relationships in a way that filters the main collection and keeps my eloquent objects in tact? Or have I like so many others been spoiled by the elegant syntax of Eloquent to the point where I'm asking too much?
Note: I am aware that you can usually get round this by defining relationships in the other direction (e.g. accessing the address table first) but this is not always ideal and not what i am asking.
Thanks in advance for any help.
At this point, there is no means by which you can filter primary model based on a constraint in the related models.
That means, you can't get only Users who have user_address.town = 'Smallwille' in one swipe.
Personally I hope that this will get implemented soon because I can see a lot of people asking for it (including myself here).
The current workaround is messy, but it works:
$products = array();
$categories = Category::where('type', 'fruit')->get();
foreach($categories as $category)
{
$products = array_merge($products, $category->products);
}
return $products;
As stated in the question there is a way to filter the adresses first and then use eager loading to load the related users object. As so:
$addressFilter = Addresses::with('Users')->where('town', $keyword)->first();
$users= $addressFilter->users;
of course bind with belongsTo in the model.
///* And in case anyone reading wants to also use pre-filtered Users data you can pass a closure to the 'with'
$usersFilter = Addresses::with(array('Users' => function($query) use ($keyword){
$query->where('somefield', $keyword);
}))->where('town', $keyword)->first();
$myUsers = $usersFilter->users;
Using Laravel 4 I have the following models and relations: Event which hasMany Record which hasMany Item. What I would like to do is something like this
Item::where('events.event_type_id', 2)->paginate(50);
This of cause doesn't work as Eloquent doesn't JOIN the models together when retrieving the records. So how do I go about this without just writing the SQL myself (which I would like to avoid as I want to use pagination).
What you want is eager loading.
It works like this if you want to specify additional constraints:
Item::with(array('events' => function($query) {
return $query->where('event_type_id', 2);
}))->paginate(50);
There is a pull request pending here https://github.com/laravel/framework/pull/1951.
This will allow you to use a constraint on the has() method, something like this:
$results = Foo::has(array('bars' => function($query)
{
$query->where('title', 'LIKE', '%baz%');
}))
->with('bars')
->get();
The idea being you only return Foos that have related Bars that contain the string 'baz' in its title column.
It's also discussed here: https://github.com/laravel/framework/issues/1166. Hopefully it will be merged in soon. Works fine for me when I update my local copy of the Builder class with the updated code in the pull request.
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.