Is it possible to make Laravel relation through belongsToMany relations?
I have 4 tables:
1)Restaurants (id , name) - uses hasManyRelation with Workers table
2)Directors (id , name)
3)Directors_Restaurants (id, director_id, restaurant_id) - pivot table for connecting belongsToMany Restaurants with Directors
3)Workers (id, name, restaurant_id)
With this function in Directors model i can get all connected restaurants
public function restaurants()
{
return $this->belongsToMany('App\Restaurant','director_restaurant');
}
With this function in my code i can get all workers of all restaurants of one director
$director = Director::find(1);
$director->load('restaurants.workers');
$workers = $director->restaurants->pluck('workers')->collapse();
So my question is : can i declare similar relation in my Director model to get all its workers of all its restaurants?
Of course you can have hasMany relationship method on Director model with Eager Loading
just like below
public function restaurants()
{
return $this->hasMany(Restaurant::class)->with('restaurants.workers');
}
i can suggest a solution like this:
Director Model OPTION 1
public function getAllRestaurants(){
return $this->hasMany(Restaurant::class)->with('restaurants.workers');
}
Director Model OPTION 2
public function getAllRestaurants(){
$this->load('restaurants.workers');
return $this->restaurants->pluck('workers')->collapse();
}
You can get all restaurants anywhere
$all_restaurants = Director::find(1)->getAllRestaurants();
You can define a direct relationship by "skipping" the restaurants table:
class Director extends Model
{
public function workers()
{
return $this->belongsToMany(
Worker::class,
'director_restaurant',
'director_id', 'restaurant_id', null, 'restaurant_id'
);
}
}
You can define an accessor method in your model to hide some of the logic
# App/Director.php
// You'll need this line if you want this attribute to appear when you call toArray() or toJson()
// If not, you can comment it
protected $appends = ['workers'];
public function getWorkersAttribute()
{
return $this->restaurants->pluck('workers')->collapse();
}
# Somewhere else
$director = Director::with('restaurants.workers')->find(1);
$workers = $director->workers;
But ultimately, you still have to load the nested relationship 'restaurants.workers' for it to work.
Given your table attributes you could also define a custom HasMany relationship that looks like this
# App/DirectorRestaurant.php
public function workers()
{
return $this->hasMany(Worker::class, 'restaurant_id', 'restaurant_id');
}
# Somewhere else
$director = Director::find(1);
$workers = DirectorRestaurant::where('director_id', $director->id)->get()->each(function($q) { $q->load('workers'); });
But I don't recommend it because it's not very readable.
Lastly, there's the staudenmeir/eloquent-has-many-deep package where you can define that sort of nested relationship.
https://github.com/staudenmeir/eloquent-has-many-deep
Related
I have Games table and which has the following schema
id | status | name
status column has 2 values (Active, Pending)
And GamePlayer table which has the following schema
id | game_id | player_id | request_status
request_status column has 3 values (Pending, Confirm, Rejected)
Now I have to select all game in which the player is involved but with the following constraints:
If the game is Pending state then it will be shown to all game_players
If the game is in Active state then it will be only shown to the game_player whose request_status is Confirm.
Game(Model)
public function GamePlayer()
{
return $this->hasMany('App\Models\GamePlayer', 'game_id', 'id');
}
public function getGames($playerId)
{
$gameList = Game::with(['GamePlayer','Category:id,name'])
->whereHas('GamePlayer', function ($q) use ($playerId) {
$q->where('player_id', $playerId);
})->get();
return $gameList;
}
Controller
$this->gameObj = new Game();
$gameList = $this->gameObj->getGames($player_id);
Please help me out how can I populate data from another table based on condition(parent table as well as the child).
You can use condition in your relationship to get only confirmed game players
public function activeGamePlayers()
{
return $this->hasMany('App\Models\GamePlayer', 'game_id', 'id')
->where('request_status', 'confirm');
}
You can as well use scopes (https://laravel.com/docs/7.x/eloquent#local-scopes) to select only active games
public function scopeActive($query)
{
return $query->where('status', 'active');
// You can then use $game->active()->...
}
So your getGame would look like:
public function getGames($playerId)
{
$player = GamePlayer::find($playerId);
if ($player->status == 'active') {
$gameList = Game::with(['GamePlayer', 'Category:id,name'])
->get();
} else {
$gameList = Game::with(['GamePlayer', 'Category:id,name'])
->active()->get();
}
return $gameList;
}
NOTE:
in your specific case, I would instead get started from the GamePlayer Model to get the games, instead of coming from the Game Model - you can use scopes and conditions in relationships as well to make your code more readable.
You can use Laravel method whereHas. You should read this part of Laravel the documentation.
Note that you should have relationships declared on every model.
I'm pretty new to Laravel and i got stuck with building a more complex query:
I've 3 tables with their models:
Codes:
id
description
factor
public function purposes() {
return $this->hasMany('App\Purpose', 'code_purposes', 'code_id', 'purpose_id');
//I could be wrong here
}
Purpose:
id
name
description
Code_purposes:
code_id
purpose_id
public function codes() {
$this->belongsToMany('App\Code'); //I could be wrong here
}
public function purposes() {
$this->belongsToMany('App\Purpose'); //I could be wrong here
}
What I want is to fetch all the codes with the condition where the purposes name = 'some_name'
I thought this would be easy with the relationships, but I can't figure it out.
So how do I do this in Laravel?
In Code model:
public function purposes() {
return $this->belongsToMany('App\Purpose');
}
In Purpose model:
public function codes() {
return $this->belongsToMany('App\Code');
}
Now you can get data like:
$codes = Purpose::where('name', 'some_name')->first()->codes;
Relation table name must be code_purpose. And no need any model for this table.
Source: Laravel Many To Many Relationships
I got the following:
User, Role, with a role_user pivot table and a belongsToMany
relationship
User, Location, with a location_user pivot table and a belongsToMany relationship
There's 2 roles for the user: owner & gardener
Location has a 'gardeners_max' field
In model Location:
protected $appends = ['is_full'];
public function getIsFullAttribute()
{
return $this->attributes['name'] = $this->remainingGardeners() <= 0;
}
public function countGardeners()
{
return $this->gardeners()->count();
}
public function remainingGardeners()
{
return $this->gardeners_max - $this->countGardeners();
}
Now, doing that :
Location::all();
I get that :
[
{
name: 'LocationA',
gardeners_max: 3,
owners: [...],
garderners: [...]
...
is_full: false
}
]
which is cool. BUT... it's not possible to do a WHERE clause on the appended attribute.
Location::where('is_full',true)->get() // Unknown column 'is_full' in 'where clause'
So i'd like to write a join query so I can do a where clause on is_full
And I just can't find the way. Any help will be greatly appreciated!
IMPORTANT:
I know the filter() method to get the results but I need to do a single scopeQuery here
You could try to manipulate the Collection after loading the object from database:
Location::get()->where('is_full', true)->all();
(You have to use get first then all, not sure it works otherwise)
Not sure it's optimized thought.
You can make scope in your location model like this
public function scopeFull(Builder $query)
{
return $query->where('is_full', true);
}
Now you just get all location like this
Location::full()->get();
Is it possible to return an value for an hasOne relation directly with an Model?
For example:
$List = Element::orderBy('title')->get();
The Element has a "hasOne" Relation to an column:
public function type()
{
return $this->hasOne('App\Type', 'id', 'type_id');
}
How can i now return automatically the "type" for the Model?
At the Moment i am looping through all Elements, and build my own "Array" of Objects, including the Key of "type" in this example. But ill prefer to do this only in my Model.
Ill know how to add a "normal" property, but can it be someone from an relation?
public function getTypeAttribute($value)
{
return // how ?
}
protected $appends = array('type');
Is this possible?
Edit:
A workaround could be to use DB:: to return the correct value - but ill dont thing thats a good workaround: like:
public function getTypeAttribute($value)
{
// make a query with $this->type_id and return the value of the type_name
}
protected $appends = array('type');
you need to eager load your relations when getting the Element:
$list = Element::with('type')->orderBy('title')->get();
then access the type using
foreach ($list as $item) {
echo $item->type->type_name;
}
where type_name would be the name of a column in the types table
Make a query scope and in that scope, join your attributes from other tables.
http://laravel.com/docs/5.1/eloquent#query-scopes
http://laravel.com/docs/5.1/queries#joins
how can i do that with function in model?
Select * from friends where (user_id = 22 or content_id = 22) and type=2 and situation = 1;
i try that, but it gives error;
public function friends() {
return $this->hasMany('Friends',function ($query) {
$query->where('content_id', $this->attributes['id']);
$query->orWhere('user_id',$this->attributes['id']);
})->where('type',2)->where('situation',1)->orderBy('id','DESC');
}
First, is your model name 'Friends' or just 'Friend'? Assuming you're following convention and using the singular tense for your model names, this should work:
return $this->hasMany('Friend')->where(function($query) {
$query->where('content_id', $this->attributes['id']);
$query->orWhere('user_id',$this->attributes['id']);
})->where('type',2)->where('situation',1)->orderBy('id','DESC');
in other words, add your additional query parameters following the hasMany() relationship, not as a closure for the second argument.