尝试获取非对象属性 'cname' Laravel php 5.8

Question

我正在学习laravel 5.8，我正在建设一个求职门户网站。

我的问题是为什么会出现这个错误。我已经检查了我的 table，显然，我在公司 table 中有一个列 cname（公司名称）。而且我还有一个对象。

以下是我的代码。如果有人帮助我，我真的很高兴。

此代码用于识别用户是求职者还是雇主。我想报名求职

this is my error page.

app.blade.php

<li class="nav-item dropdown">
                                <a id="navbarDropdown" class="nav-link dropdown-toggle" href="#" role="button" data-toggle="dropdown" aria-haspopup="true" aria-expanded="false" v-pre>
                                    @if(Auth::user()->user_type='company')
                                        {{Auth::user()->company->cname}}
                                    @else
                                        {{Auth::user()->name}}
                                    @endif
                                    <span class="caret"></span>
                                </a>

user.php

use App\Profile;
use App\Company;

public function profile(){
        return $this->hasOne(Profile::class);
        }

    public function company(){
        return $this->hasOne(Company::class);
        }

web.php

Route::POST('employer/register', 'EmployerRegisterController@employerRegister')->name('emp.register');

Route::view('employer/register','auth.employer-register')->name('employer.register');

CompanyController.php

public function store(Request $request){

        $user_id = auth()->user()->id;
        Company::where('user_id',$user_id)->update([
            'address'=>request('address'),
            'phone'=>request('phone'),
            'website'=>request('website'),
            'slogan'=>request('slogan'),
            'description'=>request('description')
        ]);
        return redirect()->back()->with('message','Company Successfully Updated !');
    }

感谢任何帮助。

Answer 1

错误抛出，因为 Auth::user()->company 是 Null。尝试获取这样的关系属性时使用三元运算符 Auth::user()->company->cname ?? ''

<li class="nav-item dropdown">
    <a id="navbarDropdown" class="nav-link dropdown-toggle" href="#" role="button" data-toggle="dropdown" aria-haspopup="true" aria-expanded="false" v-pre>
        @if(Auth::user()->user_type == 'company')
            {{Auth::user()->company->cname ?? ''}}
        @else
            {{Auth::user()->name}}
        @endif
        <span class="caret"></span>
    </a>
</li>

Answer 2

结果似乎 return 什么都没有。因为单个 = 而不是 == 尝试这样的事情：

<li class="nav-item dropdown">
                                <a id="navbarDropdown" class="nav-link dropdown-toggle" href="#" role="button" data-toggle="dropdown" aria-haspopup="true" aria-expanded="false" v-pre>
                                    @if(Auth::user()->user_type=='company')
                                        {{Auth::user()->company->cname}}
                                    @else
                                        {{Auth::user()->name}}
                                    @endif
                                    <span class="caret"></span>
                                </a>

尝试获取非对象属性 'cname' Laravel php 5.8

Trying to get property 'cname' of non-object Laravel php 5.8

php

error-handling

object

laravel

laravel-5.8

尝试获取非对象 属性 'cname' Laravel php 5.8

Trying to get property 'cname' of non-object Laravel php 5.8

php

error-handling

object

laravel

laravel-5.8

尝试获取非对象属性 'cname' Laravel php 5.8