Laravel 资源条件返回

Posted 2023-02-16

【中文标题】Laravel 资源条件返回

【英文标题】：Laravel resource conditional return

【发布时间】：2018-12-10 00:50:40

【问题描述】：

我有简单的 laravel 资源：

<?php

namespace App\Http\Resources;

use Illuminate\Http\Resources\Json\JsonResource;

class UserResource extends JsonResource

    /**
     * Transform the resource into an array.
     *
     * @param  \Illuminate\Http\Request  $request
     * @return array
     */
    public function toArray($request)
    
        return [
            'id' => $this->id,
            'unread' => $this->unread,
            'details' => new EmployeeAddressResource($this->employeeAddress),
        ];
    

这工作正常，现在我想详细说明：

       '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，如果角色是user，我想获取资源：CompanyAddressResource

我试过了：

       '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 登录时，它没有给出details

我怎样才能做到这一点？

【参考方案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

【讨论】：

感谢这个例子，非常有帮助，而且不是很明显