拉拉维尔资源条件返回

Question

我有一个简单的拉拉维尔资源：

<?php
namespace AppHttpResources;
use IlluminateHttpResourcesJsonJsonResource;
class UserResource extends JsonResource
{
/**
* Transform the resource into an array.
*
* @param  IlluminateHttpRequest  $request
* @return array
*/
public function toArray($request)
{
return [
'id' => $this->id,
'unread' => $this->unread,
'details' => new EmployeeAddressResource($this->employeeAddress),
];
}
}

这工作正常，现在我想使细节有条件：

'details' => $this
->when((auth()->user()->role == 'company'), function () {
return new EmployeeAddressResource($this->employeeAddress);
}),

它也可以正常工作，但是如何添加其他条件以返回其他资源？例如，如果角色user我想获取资源：CompanyAddressResource

我试过这个：

'details' => $this
->when((auth()->user()->role == 'company'), function () {
return new EmployeeAddressResource($this->employeeAddress);
})
->when((auth()->user()->role == 'user'), function () {
return new CompanyAddressResource($this->companyAddress);
}),

但这不起作用，当我以company身份登录时，它不会给出任何details

我怎样才能做到这一点？

Answer 1

你可以这样做

public function toArray($request)
{
$arrayData = [
'id' => $this->id,
'unread' => $this->unread
];
if(auth()->user()->role == 'company'){
$arrayData['details'] = new EmployeeAddressResource($this->employeeAddress);
}else {
$arrayData['details'] = new CompanyAddressResource($this->companyAddress);
}
return $arrayData
}