How to get the same model id i tried this not working.
$factory->define(App\Item::class, function (Faker\Generator $faker) {
return [
'name' => $faker->name,
'color' => $faker->color_name,
'position' => App\Item::all()->get('id'),
];
});
I want to get the same id for the item so that the item position is the same when the application runs and can be edited to something (integer only).
Related
I was using these codes in my controller to get all the data from my 2 tables and it works fine
$All = Customers::with('order')->paginate(10);
return response()->json([
'code' => 0,
'success' => true,
'data' => $All
], 200);
Here is how I define the relationship between these 2 tables
class Customers extends Model
{
public function order()
{
return $this->hasMany(Orders::class, 'customer_id', 'id');
}
}
class Orders extends Model
{
public function customers()
{
return $this->belongsTo(Customers::class, 'customer_id', 'id');
}
}
Now my desire output is to hide the order id, order timestamps and change the customer_id to customer's name (the customer's name is not in my orders db table).
I'm using 'data' => DataResource::collection($All) in my controller and this is my DataResource
public function toArray($request)
{
return [
'id' => $this->id,
'name' => $this->name,
'created_at' => $this->created_at,
'updated_at' => $this->updated_at,
'order' => $this->order
];
}
and of course the output is same with the image above.
My database structure:
orders table:
customer table:
Can anyone help me with that?
The answer is simple and basically a copy of the official documentation. You simply need to wrap your orders in an OrderResource as well.
// DataResource
public function toArray($request)
{
return [
'id' => $this->id,
'name' => $this->name,
'created_at' => $this->created_at,
'updated_at' => $this->updated_at,
'order' => OrderResource::collection($this->order)
];
}
// OrderResource
public function toArray($request)
{
return [
'items' => $this->items,
'quantity' => $this->quantity
];
}
I don't really understand why you would want to include the customer_name in your orders when it is already present on the customers object one hierarchy above. But if you really want to add it, you should be able to do so with: 'customer_name' => $this->customers->name.
As a side note: you really should be more consistent with your naming. Why is the resource called DataResource when it is about Customers? Why is your model called Customers in plural form rather than Customer in singular, which is the convention (and more logical if you consider that one model represents one customer). Why is your belongsTo relation called customers() in plural when it returns one customer, while your hasMany relation is called order whereas it returns one or more orders?
I am trying to generate a test on Laravel.
What I was trying is to create a fictitious position name, then add 10 people for this position.
PositionsFactory.php
$factory->define(App\Position::class, function (Faker $faker) {
return [
'p_id' => $faker->unique()->randomNumber($nbDigits = 8),
'name' => $faker->word,
'org' => $faker->word,
'user_id' => 1
];
});
Here is my EmployeeFactory.php
$factory->define(App\Employee::class, function (Faker $faker) {
return [
'FirstName' => $faker->name,
'LastName' => $faker->lastName,
'pid' => $position->p_id,
'org'=> $position->org,
'user_id' => 1,
];
});
Well here is one my my trials but it did not work
for ($i=0; $i < 5; $i++ ){
$position = factory('App\Position')->create();
factory('App\Employee',10)->create(
'pid' => $position->pid,
'org' => $position->org
);
}
I am trying to loop for 5 times and for each loop I want to create 10 employees with the same position Id. But obviously I am missing something.
I tried adding $position in the Employee factory, which works great.
$factory->define(App\Employee::class, function (Faker $faker) {
$position = factory('App\Position')->create();
return [
'FirstName' => $faker->name,
'LastName' => $faker->lastName,
'pid' => $position->p_id,
'org'=> $position->org,
'user_id' => 1,
];
});
Is there a way to make something like,
$factory('App\Position',5)->create($factory('App\Employee',10));
Maybe I am missing something with call back but kinda lost. Any help would be appreciated. Thank you!
I think you might be looking for the each method which can be called after create:
// Generate 5 positions and execute a callback
// function for each position created
factory(App\Position::class, 5)->create()->each(function ($position) {
// In the callback, generate 10 employees
// and manually override the foreign key
factory(App\Employee::class, 10)->create([
'pid' => $position->id
]);
});
Further information on each and handling relationships: https://laravel.com/docs/5.6/database-testing#relationships.
Hope it helps!
You can create them separatly and loop through collections.
$positions = factory('App\Position', 3)->make();
foreach ($positions as $position){
$employes = factory('App\Employee', 3)->make();
foreach ($employes as $employee){
$employee->p_id = $position->id;
//etc.. watever you want to connect
}
}
now you have 1 collection of positions and 1 collection of employes devided of the positions
note that the make method does not save to database you need to manually save them
you could also change your factory as is stated in the documentation https://laravel.com/docs/5.6/database-testing#using-factories
yours would look like:
$factory->define(App\Employee::class, function (Faker $faker) {
return [
'FirstName' => $faker->name,
'LastName' => $faker->lastName,
'org'=> $position->org,
'user_id' => 1,
'pid' => function () {
return factory('App\Position')->create()->id;
}
];
});
This will create a position for each user the factory creates.
You could also use existing eloquent models instead if you have existing positions.
Okay, let's say that I have a Post model with the attributes name, slug and content. I'd like to generate models with my ModelFactory, but want to set a specific name, which I do by overwriting the value:
factory(App\Post::class)->create(["name" => "Something here"]);
But now I want the slug to auto-generate by using the (new) name and without passing it as argument. Like
"slug" => str_slug($name);
Is this possible or do I need to write the slug manually?
When using the factory below with ->create(['name' => 'anything']); the slug is not created.
My current factory
$factory->define(App\Post::class, function (Faker\Generator $faker) {
static $name;
return [
'name' => $faker->name,
'slug' => $name ?: str_slug($name),
'content' => $faker->sentences(),
];
});
This should do the trick. You can pass a name in manually or let Faker handle it.
$factory->define(App\Post::class, function (Faker\Generator $faker) {
return [
'name' => $faker->name,
'slug' => function (array $post) {
return str_slug($post['name']);
},
'content' => $faker->sentences(),
];
});
Have you tried
$name ="Something here";
factory(App\Post::class)->create(["name" => $name, "slug" => str_slug($name)]);
I'm trying to seed the users table on Laravel 5.2. I've a custom column called role. I'm using Faker to seed the users table. However, I need to populate the role column either with Administrator or Customer. So, I've passed an array and that returned Parameter mismatch error.
Here's the relevant code I'm trying with:
$factory->define(App\User::class, function (Faker\Generator $faker) {
return [
'name' => $faker->name,
'email' => $faker->email,
'password' => 123456,
'role' => ['Administrator', 'Customer'],
'remember_token' => str_random(10),
];
})
I know that it only accept string instead of an array. So, how do you do that when you want to seed the database with multiple particular data.
You can use Faker to get a random element:
$factory->define(App\User::class, function (Faker\Generator $faker) {
return [
'name' => $faker->name,
'email' => $faker->email,
'password' => 123456,
'role' => $faker->randomElement(['Administrator', 'Customer']),
// ^^^^^^^ here ^^^^^^^
'remember_token' => str_random(10),
];
})
I suggest using random index generation (in your case 0 or 1) and retrieving value from the array of possible values (['Administrator', 'Customer']) by this index.
$i = rand(0, 1);
$roleOptions = ['Administrator', 'Customer'];
$role = $roleOptions[$i];
Also if you would like to have much more users than administrators, you can look for something like this: changing probability of getting a random number
I am beginner. I explain the topic:
there is this relationship in the Ticket model:
public function getTyp()
{
return $this->hasOne(Typology::className(), [ 'id' =>'typ_id']);
}
and in the ticket table there is the typ_id column (it is in relationship with the id of the Typology table).
In the view views/ticket/index.php there is GridView::widgetwith these columns:
[
'attribute' => 'typ_id',
'value' => 'typ.typology'
],
I want to anchor the value of the relationship.
I have tried this:
[
'attribute' => 'typ_id',
'value' => function ($model) {
return Html::a (
'typ.typology',
'/typology/view?id='.$model->typ_id
);
}
]
but it doesn't work
someone can help me?
Html::a() interprets typ.typology as raw string. Use $model in value closure to get necessary property through relation.
Also instead of manually concatenate the url with its parameters, just pass them in array (see Url::to() to understand how link is constructed).
[
'attribute' => 'typ_id',
'value' => function ($model) {
return Html::a($model->typ->typology, ['/typology/view', 'id' => $model->typ_id]);
},
],