试图获取非对象的属性“员工”（查看：/Applications/MAMP/htdocs

Question

一直试图从我的 vew 中获取这些代码的输出，但它给我带来了问题，请我真的很乐意寻求帮助。在我的控制器中<?phpnamespace App\Http\Controllers;use Illuminate\Http\Request;use Illuminate\Support\Facades\Input;use App\leaveType;use App\allLeave;use App\leaveDepartment;class LeavesController extends Controller{    public function getAllLeave()    {        $data = App\allLeave::find(1)->full_name;        return view('leave/allLeave',["data"=>$data]);    }}在我的员工模型中<?phpnamespace App;use Illuminate\Database\Eloquent\Model;/** * Class Personnel * @package App */class Employee extends Audit{    public function leave()    {        return $this->belongsTo('App\allLeave');    }}在所有离开模型<?phpnamespace App;use Illuminate\Database\Eloquent\Model;class allLeave extends Model{    public function empolyee()    {        return $this->hasMany('App\Employee');    }}在刀片{{$data->employee->full_name}}

Answer 1

您已经为控制器中的数据分配了全名。你只需要{{ $data }}在刀片

Answer 2

如果我必须做同样的事情，我会这样做

 <?php

namespace App\Http\Controllers;

use Illuminate\Http\Request;

use Illuminate\Support\Facades\Input;

use App\leaveType;

use App\allLeave;

use App\leaveDepartment;

class LeavesController extends Controller

{

    public function getAllLeave()

    {

           /* No need to use App\allLeave because you already have that used in 

              top of the project */

           $data=allLeave::findorFail(1);

           return view('leave.allLeave')->with('data', $data);

    }

}

在前端只使用

{{$data->first_name}} //same column as in database table

注意：确保使用 laravel eloquent 模型关系

Answer 3

在你的控制器中应该是这样的：

$data = App\allLeave::find(1)->empolyee();

还有你的刀片：

{{$data->full_name}}