I can't find how to get all of the elements with relations to another element using Eloquent in Laravel.
I have Substances and Contents and want all the contents related to the substances with name 'subst'.
My relation Substance - Content:
public function relation ()
{
return $this->belongsToMany('Substance', 'contents_substances', 'id_contents', 'id_substances');
}
public function relation ()
{
return $this->belongsToMany('Content', 'contents_substances', 'id_contents', 'id_substances');
}
I know I can get all the contents related to one substance:
$content = Substance::find(1)->relation()->get();
but is possible to get all the contents related to a group of substances?
something like:
$sub = Substance::where('name', '=', 'subst'); // get all the substances with that name
$contents =$sub->relation()->get(); // ???
Any help is appreciated!
The easiest is to start the query from the other angle. In this case from Content. whereHas will then add a constraint based on the substances relation:
$contents = Content::whereHas('substances', function($q){
$q->where('name', 'subst');
})->get();
Related
I have a Controller method like this:
public function awaiting()
{
$producers = Producer::where('producer_process',4)->get();
$producers_list = [];
foreach($producers as $producer){
if($producer->brand->brand_rejected == 0){
array_push($producers_list, $producer);
}
}
return view('admin.brands.awaiting', compact('producers_list'));
}
So basically there's One To One relationship between Producer model & Brand model.
In order to get the collection of brands table records that has producer_process of 4 and ALSO the brand_rejected field of related brands table record must be set to 0, I added an array_push and check the condition.
Now this works fine and properly shows the correct data but I wanted to know, what is the shorthand method of doing this with Eloquent relationships?
I mean is there any concise and useful method written in Eloquent relationships that can do this without using array_push or another foreach loop?
You can use whereHas to constrain the result set based on the existence of a relationship. Here we are saying we only want producers that have the field 'produce_process' set to 4 and have a brand with a field of 'brand_rejected' set to 0:
$producers = Producer::where('producer_process', 4)
->whereHas('brand', function ($q) { $q->where('brand_rejected', 0); })
->get();
If you want these producers to have their brand relationship loaded to use you should eager load that. Before the get call you can tell it to load the relationship:
$producers = Producer::where(...)->whereHas(...)->with('brand')->get();
Laravel 5.8 Docs - Eloquent - Relationships - Querying Relationship Existence whereHas
Laravel 5.8 Docs - Eloquent - Relationships - Eager Loading with
You can try this:
public function awaiting()
{
$producers = Producer::where('producer_process',4)
->with('brand', function($q) {
$q->where('brand_rejected', 0);
})->get();
// dd($producers);
dd($producers->pluck(brand));
Sure you can use the method with() also with the where() clause to can apply some conditions to the relationship
Example
$yourQuery->with('brand', function($query){
$query->where('brand_rejected', 0);
});
check this for more info
https://laravel.com/docs/9.x/eloquent-relationships#constraining-eager-loads
I hope it's helpful
I have comments table where has parent_id
This is Comment table sub_comments relation.
public function sub_comments()
{
return $this->hasMany(self::class, 'parent_id');
}
This code return all comments with related all sub-comments
Comment::with('sub_comments')->get();
But I want to get all comments also sub-comments when sub-comments is single. That mean if comment have 2 or more comments for that comment I did not want get that sub-comments.
Now I use this code
$oneSubcommentCommentIds = Comment::has('sub_comments', '=', 1)->pluck('id');
Comment::with([
'sub_comments' => function ($q) use ($oneSubcommentCommentIds) {
$q->whereIn('parent_id', $oneSubcommentCommentIds);
}
])->get();
but this make one additional query.
Try this:
Comment::with('sub_comments')->has('sub_comments', '=', 1)->get();
Update
Your question wasn't clear, I can't imagine another way to doing this without previosly loaded the relationship or the count of the relationship.. so I'd do this:
// First get all your comments with an aditional count field
$comments = Comments::withCount('sub_comments')->get();
// separate the ones with just one sub_comment from the rest
list($oneSubComment, $theRest) = $collection->partition(function ($comment) {
return $comment->sub_comments_count == 1;
});
// Then load the relationship on just the selected elements
$oneSubComment->load('sub_comments');
// re-join the collection
$comments = $oneSubComment->union($theRest);
What am I doing here?
Adding an additional field to each $comment with the relationship count (it should be something like sub_comments_count)
Partition the resulting collection in two parts: the ones with one comment and the rest. Using the partition() method.
Lazy eager loading the collection.
Re-joining the two collections using the union() method.
sorry for the title of this question but I am not sure how to ask it...
I am working on a project where I have two Models Trains and Cars, to this model I have a belonging Route.
I want to make a query and check if the routeable_type is App\Car than with the selected routeable_id to get the data from the Car. And if the routeable_type is Train then with the ID to get the data from the Tran.
So my models go like this:
Train:
class Train extends Model
{
public function routes()
{
return $this->morphMany('App\Route', 'routeable');
}
}
Car:
class Car extends Model
{
public function routes()
{
return $this->morphMany('App\Route', 'routeable');
}
}
Route:
class Route extends Model
{
public function routeable()
{
return $this->morphTo();
}
}
And the query I have at the moment is:
$data = Route::leftjoin('cars', 'cars.id', '=', 'routes.routeable_id')
->leftjoin('trains', 'trains.id', '=', 'routes.routeable_id')
->select('routes.id', 'cars.model AS carmodel', 'trains.model AS trainmodel', 'routeable_type', 'routes.created_at');
With this query if I have the same ID in cars and trains I get the data from both and all messes up. How do I check if routeable_type is Car ... do this, if routeable_type is Train .. do that?
Will something like this be possible in a 1 single query:
$data = Route::select('routes.id', 'routeable_type', 'routes.created_at');
if(routeable_type == 'Car'){
$data = $data->leftjoin('cars', 'cars.id', '=', 'routes.routeable_id')->select('routes.id', 'cars.model AS carmodel', 'routeable_type', 'routes.created_at');
}else{
$data = $data->leftjoin('trains', 'trains.id', '=', 'routes.routeable_id')->select('routes.id', 'trains.model AS trainmodel', 'routeable_type', 'routes.created_at');
}
Maybe this is what you are looking for?
DB::table('routes')
->leftJoin('cars', function ($join) {
$join->on('cars.id', '=', 'routes.routeable_id')
->where('routes.routeable_type', 'App\Car');
})
->leftJoin('trains', function ($join) {
$join->on('trains.id', '=', 'routes.routeable_id')
->where('routes.routeable_type', 'App\Train');
})
->select('routes.id', 'cars.model AS car_model', 'trains.model AS train_model', 'routes.routeable_type', 'routes.created_at');
->get();
I think you may want to follow the morphedByMany design.
https://laravel.com/docs/5.7/eloquent-relationships#many-to-many-polymorphic-relations
This was also a neat visual for the different relation types.
https://hackernoon.com/eloquent-relationships-cheat-sheet-5155498c209
I was faced with a similar issue though I failed to follow the correct design initially and was forced to query the many possible relations then wrote custom logic after to collect the relation types and ids then do another query and assign them back through iteration. It was ugly but worked... very similar to how Eloquent does things normally.
i don't have enough repo, so i can't comment. that's why i am putting as an answer.
You should use 2 different queries, for each model.
This will be better, code wise as well as performance wise. also if both models have similar fields you should merge them to 1 table and add a 'type' column.
and put non-similar fields in a 'meta' column.
( in my opinion )
I have two tables website_link and website_links_type. website_link is related website_links_type with hasmany relationship.
$this->website_links->where('id',1)->Paginate(10);
and relationship
public function broken()
{
return $this->hasMany('App\Website_links_type')->where('status_code','!=',"200");
}
Now I want to get result from website_link table but Orderby that result on count of broken relationship result.
There are many ways to solve this problem. In my answer I'll use two I know.
You can eagerload your relationship and use the function sortBy(). However I don't think you can use the paginate() functionality with this solution.
Example:
$results = Website_link::with('website_links_type')->get()->sortBy(function ($website_link) {
return $website_link->website_links_type->count();
});
See this answer
You can also use raw queries to solve this problem. With this solution you can still use the pagination functionality (I think).
Example:
$results = Website_link
::select([
'*',
'(
SELECT COUNT(*)
FROM website_links_type
WHERE
website_links_type.website_link_id = website_link.id
AND
status_code <> 200
) as broken_count'
])
->orderBy('broken_count')
->paginate(10);
You may have to change the column names to match your database.
You can not put WHERE condition in model file.
You just give relationship hasMany in model file.
And use where condition in controller side.
Refer this document.
Try this
Model file:
public function broken()
{
return $this->hasMany('App\Website_links_type');
}
Controller file:
$model_name->website_links->where('id',1)
->where('status_code','!=',"200")
->orderBy('name', 'desc')
->Paginate(10);
I am not able to construct simple Laravel query.
I have translation with categories (translation.category_id is foreign key to category.id). Moroever category also has property is_technical.
What I need is:
- get all translations where translation's category.is_technical = 1.
Currently I am haveing this query:
$match = ['lang1_code' => $langfrom, 'lang2_code' => $langto];
$translation = Translation::where($match)->orderByRaw("RAND()")->take(4)->get();
But this query doesn't join category (I have relationship in my db and also in my models). Thus how to join category and set where is_Technical = 1?
I believe this is basic question, but I am new to Laravel and I cannot find answer in documentation.
you need whereHas. See Laravel Document for more info
http://laravel.com/docs/5.1/eloquent-relationships
Here is an example, correct your model name.
//Translation.php
public function category() {
return $this->belongsTo('Category', 'category_id');
}
// query
$translation = Translation::whereHas('category', function($q) {
$q->where('is_technical', '=', 1);
})->where($match)->orderByRaw("RAND()")->take(4)->get();