on this post the author made an example helper like this
and the package being used is the Lighthouse
public function graphql(string $query)
{
return $this->post('/graphql', [
'query' => $query
]);
}
and so can be used it like this:
$response = $this->graphql("{articles(first: 10) { edges { node { title } } } }");
but can I implement that on mutation? if example I have a mutation:
type Mutation {
sampleMutation(
id: ID!
)
}
I'm not sure how to do that on a mutation.
It is possible to test mutations.
In the newest version of Lighthouse you now have a trait you can add to your test class called MakesGraphQLRequests.
Taken from their docs is the following example on how to create a test for a mutation.
public function testCreatePost(): void
{
/** #var \Illuminate\Foundation\Testing\TestResponse $response */
$response = $this->postGraphQL([
'query' => '
mutation CreatePost($title: String!) {
createPost(title: $title) {
id
}
}
',
'variables' => [
'title' => 'Automatic testing proven to reduce stress levels in developers'
],
]);
}
The response object you get will contain the json result like from your mutation. So here you can just make any json assertions as you normally would with Laravel.
In their documentation there is some examples on how validation on the json could look like.
Disclaimer: I am the writer of the refered article.
Related
Edit: I was able to see where the relations are being included in my response, but I still don't know why.
On my Customer model, I have:
protected $appends = [
'nps',
'left_feedback',
'full_name',
'url'
];
The accessors are as follows:
/**
* Accessor
*/
public function getNpsAttribute() {
if ($this->reviews->count() > 0) {
return $this->reviews->first()->nps;
} else {
return "n/a";
}
}
/**
* Accessor
*/
public function getLeftFeedbackAttribute() {
if ($this->reviews && $this->reviews->count() > 0 && $this->reviews->first()->feedback != null) {
return "Yes";
} else {
return "No";
}
}
/**
* Accessor
*/
public function getFullNameAttribute() {
return ucwords($this->first_name . ' ' . $this->last_name);
}
/**
* Accessor
*/
public function getUrlAttribute() {
$location = $this->location;
$company = $location->company;
$account_id = $company->account->id;
return route('customers.show', ['account_id' => $account_id, 'company' => $company, 'location' => $location, 'customer' => $this]);
}
So if I comment out the $appends property, I get the response I originally wanted with customer not returning all the relations in my response.
But I do want those appended fields on my Customer object. I don't understand why it would include all relations it's using in the response. I'm returning specific strings.
So is there a way to keep my $appends and not have all the relations it's using in the accessors from being included?
Original Question:
I am querying reviews which belongsTo a customer. I want to include the customer relation as part of the review, but I do not want to include the customer relations.
$reviews = $reviews->with(['customer' => function($query) {
$query->setEagerLoads([]);
$query->select('id', 'location_id', 'first_name', 'last_name');
}]);
$query->setEagerLoads([]); doesn't work in this case.
I've tried $query->without('location'); too, but it still gets included
And I should note I don't have the $with property on the model populated with anything.
Here is the Review model relation:
public function customer() {
return $this->belongsTo('App\Customer');
}
Here is the Customer model relation:
public function reviews() {
return $this->hasMany('App\Review');
}
// I dont want these to be included
public function location() {
return $this->belongsTo('App\Location');
}
public function reviewRequests() {
return $this->hasMany('App\ReviewRequest');
}
In the response, it will look something like:
'review' => [
'id'=> '1'
'customer => [
'somecol' => 'test',
'somecolagain' => 'test',
'relation' => [
'relation' => [
]
],
'relation' => [
'somecol' => 'sdffdssdf'
]
]
]
So a chain of relations ends up being loaded and I don't want them.
As you said in one comment on the main question, you are getting the relations due to the appended accessors.
Let me show you how it should be done (I am going to copy paste your code and simply edit some parts, but you can still copy paste my code and place it in yours and will work the same way but prevent adding the relations) and then let me explain why is this happening:
/**
* Accessor
*/
public function getNpsAttribute() {
if ($this->reviews()->count() > 0) {
return $this->reviews()->first()->nps;
} else {
return "n/a";
}
}
/**
* Accessor
*/
public function getLeftFeedbackAttribute() {
return $this->reviews()->count() > 0 &&
$this->reviews()->first()->feedback != null
? "Yes"
: "No";
}
/**
* Accessor
*/
public function getFullNameAttribute() {
return ucwords($this->first_name . ' ' . $this->last_name);
}
/**
* Accessor
*/
public function getUrlAttribute() {
$location = $this->location()->first();
$company = $location->company;
$account_id = $company->account->id;
return route('customers.show', ['account_id' => $account_id, 'company' => $company, 'location' => $location, 'customer' => $this]);
}
As you can see, I have changed any $this->relation to $this->relation()->first() or $this->relation->get().
If you access any Model's relation as $this->relation it will add it to the eager load (loaded) so it will really get the relation data and store it in the Model's data so next time you do $this->relation again it does not have to go to the DB and query again.
So, to prevent that, you have to access the relation as $this->relation(), that will return a query builder, then you can do ->count() or ->exists() or ->get() or ->first() or any other valid query builder method, but accessing the relation as query builder will prevent on getting the data and store it the model (I know doing ->get() or ->first() will get the data, but you are not directly getting it through the model, you are getting it through the query builder relation, that is different).
This way you will prevent on storing the data on the model, hence giving you problems.
You can also use API Resources, it is used to map a Model or Collection to a desired output.
One last thing, if you can use $this->relation()->exists() instead of $this->relation()->count() > 0 it will help on doing it faster, mostly any DB is faster on looking if data exists (count >= 1) than really counting all the entries it has, so it is faster + more performant on using exists.
Try :
$review->with(‘customer:id,location_id,first_name,last_name’)->get();
Or :
$review->withOnly(‘customer:id,location_id,first_name,last_name’)->get();
I am using OctoberCMS and I have created a custom component. I am trying to create a frontend filter to filter Packages by the Tour they are assigned to.
This is what I have so far. The issue is that the code is looking for a tour field within the packages table rather than using the tour relationship. Does anyone have any ideas?
<?php namespace Jakefeeley\Sghsportingevents\Components;
use Cms\Classes\ComponentBase;
use JakeFeeley\SghSportingEvents\Models\Package;
use Illuminate\Support\Facades\Input;
class FilterPackages extends ComponentBase
{
public function componentDetails()
{
return [
'name' => 'Filter Packages',
'description' => 'Displays filters for packages'
];
}
public function onRun() {
$this->packages = $this->filterPackages();
}
protected function filterPackages() {
$tour = Input::get('tour');
$query = Package::all();
if($tour){
$query = Package::where('tour', '=', $tour)->get();
}
return $query;
}
public $packages;
}
I really appreciate any help you can provide.
Try to query the relationship when the filter input is provided.
This is one way to do it;
public $packages;
protected $tourCode;
public function init()
{
$this->tourCode = trim(post('tour', '')); // or input()
$this->packages = $this->loadPackages();
}
private function loadPackages()
{
$query = PackagesModel::query();
// Run your query only when the input 'tour' is present.
// This assumes the 'tours' db table has a column named 'code'
$query->when(!empty($this->tourCode), function ($q){
return $q->whereHas('tour', function ($qq) {
$qq->whereCode($this->tourCode);
});
});
return $query->get();
}
If you need to support pagination, sorting and any additional filters you can just add their properties like above. e.g;
protected $sortOrder;
public function defineProperties(): array
{
return [
'sortOrder' => [
'title' => 'Sort by',
'type' => 'dropdown',
'default' => 'id asc',
'options' => [...], // allowed sorting options
],
];
}
public function init()
{
$filters = (array) post();
$this->tourCode = isset($filters['tour']) ? trim($filters['tour']) : '';
$this->sortOrder = isset($filters['sortOrder']) ? $filters['sortOrder'] : $this->property('sortOrder');
$this->packages = $this->loadPackages();
}
If you have a more complex situation like ajax filter forms or dynamic partials then you can organize it in a way to load the records on demand vs on every request.e.g;
public function onRun()
{
$this->packages = $this->loadPackages();
}
public function onFilter()
{
if (request()->ajax()) {
try {
return [
"#target-container" => $this->renderPartial("#packages",
[
'packages' => $this->loadPackages()
]
),
];
} catch (Exception $ex) {
throw $ex;
}
}
return false;
}
// call component-name::onFilter from your partials..
You are looking for the whereHas method. You can find about here in the docs. I am not sure what your input is getting. This will also return a collection and not singular record. Use ->first() instead of ->get() if you are only expecting one result.
$package = Package::whereHas('tour', function ($query) {
$query->where('id', $tour);
})->get();
I'm about to ...
extend my App/Orm/MyModel.php with Http/Json/V1/MyModel.php so I can keep the $appends, $hides, toArray() neatly tucked away in a V1
namespace and prefix some routing for V1
probably do some custom resolvers for route model binding
And I'm thinking ... really? They haven't built this in... what am I missing here? There's gotta be a quick, turn-key for this. I'm interested in knowing how other people are doing this, so please chime in.
Try Resources instead of Models
Have a look at resources:
https://laravel.com/docs/5.7/eloquent-resources
And add your logic to resources so that you display different versions of a model depending on the API version. You can still make use of $appends and $hidden.
With this approach we return a Resource of a model rather than the model itself.
Here is an example of a UserResource for different API versions:
class UserResource extends JsonResource
{
private $apiVersion;
public function __construct($resource, int $apiVersion = 2) {
$this->apiVersion = $apiVersion; // OPTION 1: Pass API version in the constructor
parent::__construct($resource);
}
public function toArray($request): array
{
// OPTION 2: Get API version in the request (ideally header)
// $apiVersion = $request->header('x-api-version', 2);
/** #var User $user */
$user = $this->resource;
return [
'type' => 'user',
'id' => $user->id,
$this->mergeWhen($this->apiVersion < 2, [
'name' => "{$user->first_name} {$user->last_name}",
], [
'name' => [
'first' => $user->first_name,
'last' => $user->last_name
],
]),
'score' => $user->score,
];
}
}
The you can call:
$user = User::find(5);
return new UserResource($user);
If you need a different connection you can do:
$user = User::on('second_db_connection')->find(5);
So V1 API gets:
{
id: 5,
name: "John Smith",
score: 5
}
and V2 API gets:
{
id: 5,
name: {
first: "John",
last: "Smith",
},
score: 5
}
Now if later you wanted to rename score to points in your DB, and in V3 of your API you also wanted to change your JSON output, but maintain backwards compatibility you can do:
$this->mergeWhen($this->apiVersion < 3, [
'score' => $user->points,
], [
'points' => $user->points,
])
Prefix routes
You can easily prefix routes as mentioned here: https://laravel.com/docs/5.7/routing#route-group-prefixes
Route::prefix('v1')->group(function () {
Route::get('users', function () {
// ...
});
});
Explicit Route Model Binding
To do custom route model bindings have a look at: https://laravel.com/docs/5.7/routing#route-model-binding
e.g.
Route::bind('user', function ($value) {
return App\User::where('name', $value)->first() ?? abort(404); // your customer logic
});
I'm trying to get one to many relationship objects with transformers. I want to get include metas but i only get just regular transform fields.
my transformer:
class AssistantTransformer extends TransformerAbstract
{
protected $availableIncludes = [
'assistantmetas'
];
public function transform(User $user)
{
return [
'id' => (int) $user->id,
'firstname' => ucfirst($user->first_name),
'lastname' => ucfirst($user->last_name),
];
}
public function includeMetas(User $user)
{
$assistantmetas = $user->userMetas;
return $this->item($assistantmetas, new AssistantsMetaTransformer);
}
}
Just use defaultIncludes not available includes, because it needs to send request via url? include=assistantmetas to get result like this.
Yii2. How can i get data from the related table and add it to Sluggable behavior.
In the example below, I want for every book add slug-title like "Book House, author Greenberg".
class Books extends \yii\db\ActiveRecord
{
public function behaviors()
{
return [
[
'class' => SluggableBehavior::className(),
'attribute' => "Book" . $this->name . ", author " . $this->getAuthor->name,
//'slugAttribute' => 'slug',
],
];
}
public function getAuthor()
{
return $this->hasOne(Author::className(), ['id' => 'author_id']);
}
}
Simply use a callback for the sluggable behavior like so:
public function behaviors()
{
return [
[
'class'=>SluggableBehavior::className(),
'value'=>function ($event) {
$parts = ['Book', $this->name, 'author', $this->author->name];
return Inflector::slug(implode('-', $parts));
},
],
];
}
My example will output a slug like this: book-house-author-greenberg which is more best practice for slugs than your version. Anyway...if you still prefer the way you described above, simply change the return value.
The doc for the feature used in my example is here.