I want to populate a table with listings of car offers with dummy data.
The specific is that I want the make and model to be from a table already created in the database. Also the listing's title to be Make and Model. My code is this, but I assume it is totally wrong:
public function definition()
{
$id = DB::table('cars')->select('id')->inRandomOrder()->first();
$make = DB::table('cars')->where('id', $id)->select('make')->first();
$model = DB::table('cars')->where('id', $id)->select('model')->first();
return [
'title' => $make . " " . $model,
'make' => $make,
'model' => $model,
'year' => $this->faker->numberBetween(1950, 2021),
'slug' => $this->faker->word(),
];
}
The query builder in Laravel returns an object representing the database model when you fetch with first()
You should access the attributes when you're setting them as other model attributes
return [
'title' => $make->make . " " . $model->model,
'make' => $make->make,
'model' => $model->model,
'year' => $this->faker->numberBetween(1950, 2021),
'slug' => $this->faker->word(),
];
Note
The answer above only fixes the issue in the current code
The correct implementation would be to create models (if don't exist already) and link them with relationships through foreign keys and then attach them together in a seeder
The factory should only contain a clear and clean definition of fake data
Related
I have difficult to build and return a nested json. I want obtain the information from two differents table joined with an id.
This is my situation:
With this method on my controller:
public function eventOccList(EventOccurrence $eventOccurrence){
return new EventOccurrenceResourceCollection(EventOccurrence::all());
}
and with the mapping in the class EventOccurrenceResource
return [
'type' => 'event',
'id' => (string) $this->id,
'name' => $this->name,
'description' => $this->description,
'location_id' => $this->location_id
];
I obtain this JSON:
{"data":[{"type":"event","id":"1","name":"event_1","description":"event blabla","location_id":11}
If I want to obtain all the informations about the table "location" with the id "location_id" and show in the same json, what is the best way to retrieved this data?
Thanks !
I assume your Event Model have a location relationship :
public function location{
return $this->belongsTo(Event::class);
}
you can do this once you have your event in a Controller :
$event->load('location');
return $event->toJson();
You can then hide or append any attribute you want :)
In a project I have two models, Products and Packages. Packages can be seen as containers of Products and to define the items in a package I've created a model PackageItem (which is basically a Product so its using the same table). Now Products (and so PackageItems) have translatable fields such as as a title and description.
ProductsTable.php contains:
$this->addBehavior('Translate', [
'fields' => ['title', 'description'],
'translationTable' => 'products_translations'
]);
$this->belongsToMany('PackageItems', [
'foreignKey' => 'package_id',
'joinType' => 'LEFT',
'joinTable'=>'products_package_items'
]);
PackageItemsTable contains:
$this->table('products');
$this->addBehavior('Translate', [
'fields' => ['title', 'description'],
'translationTable' => 'products_translations'
]);
$this->belongsTo('Products', [
'foreignKey' => 'package_item_id',
'joinType' => 'LEFT'
]);
Using TranslateBehavior I'm able return the translations on the Product but I can't figure out how to write the query I need to also return the translation on the PackageItems. This is my current query:
$package = $this->Products->find('translations')
->where(['business_id'=>$q['business_id'], 'id'=>$id, 'type'=>'Package'])
->contain([
'PackageItems'=>[
'Prices'=>function($q) {
return $q->where(['product_id'=>$this->product_id]);
}
]
])
->first();
You need two things
1) Set the proper reference name
The translate behavior on the PackageItemsTable class needs to be configured to use the same reference name (the value that is stored in the model column) as the behavior on the ProductsTable class, otherwise you'd never receive any translations, as it would by default look for PackageItems.
This is what the referenceName option can be used for. The reference name is being derived from the class name (not the alias), or for auto-tables, from the database table name or the alias. So for your ProductsTable class it would be Products.
Either set the name manually
$this->addBehavior('Translate', [
'fields' => ['title', 'description'],
'translationTable' => 'products_translations',
'referenceName' => 'Products' // there it goes
]);
or retrieve it dynamically from the behavior on the ProductsTable, like
$referenceName = $this->Products
->target()
->behaviors()
->get('Translate')
->config('referenceName');
This however would need to be done after adding the corresponding belongsTo association for the Products table!
2) Use the translations finder for the containment
You need to configure the PackageItems containment to use the translations finder, which is as simple as
contain([
'PackageItems' => [
'finder' => 'translations', // there you go
'Prices' => function ($q) {
return $q->where(['product_id' => $this->product_id]);
}
]
])
See also
API > \Cake\ORM\Behavior\TranslateBehavior::_referenceName()
API > \Cake\ORM\Behavior\TranslateBehavior::$_defaultConfig
API > \Cake\ORM\Query::contain()
I have mysql table 'test' with three columns,
1.sno 2.name 4.country
this code is easily understandable
$person = \App\Test::find(1);
$person->country; //Defined in Test eloquent model
now i want to do something like this:
$p = ['sno' => 1, 'name' => 'Joe', 'country' => '1' ];
$p->country; //Return relevent column form countries table as defined in Model
The thing to remember is that the user i am trying to map is already present in the database table. How to i convert an array to eloquent model?
You could instantiate the model class with no attributes:
$dummy = new \App\Test;
Then you can call the newInstance() method:
$attributes = ['sno' => 1, 'name' => 'Joe', 'country' => '1' ];
$desiredResult = $dummy->newInstance($attributes, true);
The true flag in the method is telling eloquent that the instance already exists in database, so you can continue working with it normally. Now you can do:
$desiredResult->country //'1'
i was looking for a previous answer, but the ones i've found are related to older cakephp versions
i have two tables, 'magazines' and 'issues' where there is a relation 'issues' BelongsTo 'magazines', this is what IssuesTable looks like:
public function initialize(array $config){
$this->belongsTo('Magazines', [
'foreignKey' => 'id'
]);
}
table magazines has two fields, magazines.id and magazines.name
table issues has two fields, issues.id, issues.magazine_id where issues.magazine_id is the foreign key
to populate a select input in the issues view with the magazine.name values and save the issues.magazine_id, i've set the controller like this
$this->set('magazines', $this->Issue->Magazine->find('list'));
then i've added the following code to the issue view add.cpt
<?php
echo $this->Form->input('name', [
'type' => 'select',
'multiple' => false,
'options' => $magazines,
'empty' => true]);
?>
but i get the input select with the issues.magazine_id as values instead of magazines.name
thanks for your help and comments
You want to use find('list') as this will return the primary key and display field:-
$this->set(
'magazines',
$this->Issues->Magazines->find('list')
);
Then in your form you need the input name to be magazine_id if you're wanting to set the foreign key for the association:-
echo $this->Form->input(
'magazine_id',
[
'type' => 'select',
'multiple' => false,
'options' => $magazines,
'empty' => true
]
);
See the docs for more info.
Update
If you're experiencing issues with find('list') it is perhaps because your model's displayField is not getting set correctly. Cake normally determines the displayField for the model on initialisation. If this isn't working, or you want a different field you can set this manually in the model's initialize() method. E.g.:-
class MagazinesTable extends Table
{
public function initialize(array $config)
{
$this->displayField('name');
}
}
Changing 'name' to the appropriate field.
Alternatively, you can choose which field Cake will use for the values returned by find('list') (this is particularly useful when you want to override the default displayField). E.g.:-
$this->Issues->Magazines->find('list', [
'keyField' => 'id',
'valueField' => 'name'
]);
Display selected option in month helper
$this->Form->month('students.month', [
'label' => false,
'value'=>date('m'),
'required'=>true
]);
This actually helped me.
$this->Issues->Magazines->find('list', [
'keyField' => 'id',
'valueField' => 'name'
]);
Im trying to save data inside a pivot table with an extra field called data.
when i save i have this array:
[
5 => "files"
4 => "pictures"
3 => "tags"
1 => "thumbs"
]
My table looks like this:
project_id
option_id
name
The ids shown above refer to option_id and the string to name inside the database.
When i try to use sync like this: $project->options()->sync($data);
$data is the array shown above
Im getting a error thats its trying to save the option_id with "files".
Here is how i build up the data that i use for sync:
Im trying to get what you suggested but dont know how to achieve it:
here is how i build up the array:
foreach($request->input('option_id') as $id) {
$option['option_id'][] = $id;
$option['data'][] = $request->input('data')[$id];
}
$data = array_combine($option['option_id'], $option['data']);
This is covered in the manual:
Adding Pivot Data When Syncing
You may also associate other pivot table values with the given IDs:
$user->roles()->sync(array(1 => array('expires' => true)));
In your example, you would have to change your array to look something like below but I believe this would translate to:
$data = [
5 => [ 'name' => "files" ],
4 => [ 'name' => "pictures" ],
3 => [ 'name' => "tags" ],
1 => [ 'name' => "thumbs" ],
];
$project->options()->sync($data);
I believe you may also need to modify how your Project model relates itself to your Options model:
// File: app/model/Project.php
public function options()
{
return $this->belongsToMany('Option')->withPivot('name');
}
This is also noted in the linked-to manual page:
By default, only the keys will be present on the pivot object. If your pivot table contains extra attributes, you must specify them when defining the relationship.
Update
Try creating your $data array like this:
$data = [];
foreach($request->input('option_id') as $id) {
$data[$id] = [ 'name' => $request->input('data')[$id] ];
}