本文介绍了试图在laravel 5.4中获取非对象的属性的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我正试图回显学校里用户的名字.
I'm trying to echo out the name of the user in my school.
在学校餐桌上
Schema::create('schools', function (Blueprint $table) {
$table->increments('id');
$table->integer('user_id')->unsigned();
$table->foreign('user_id')->references('id')->on('users');
$table->string('school_name');
$table->string('status');
$table->string('gender');
$table->string('notes');
$table->string('member_name');
$table->string('type');
$table->string('file_number');
$table->string('phone');
$table->string('address');
});
对于SchoolController
For SchoolController
public function show(School $school)
{
$province_names = Province::all();
$city_names = City::all();
$center_names = City::all();
return view('school.all', compact('school','city_names', 'province_names', 'center_names'));
}
对于示范学校
public function user()
{
return $this->belongsTo(User::class, 'user_id');
}
all.blade.php
all.blade.php
{{ $school->user->firstـname }}
我收到此错误
但是我写了{{ dd($school->user) }}
显示null
.
也许我错了
推荐答案
试试这个:
public function show($id) {
$school = School::with('user')->where('id', $id)->first();
// ...
}
或在模型上自动加载关系:
Or auto-load relation on your model:
class School extends Model {
protected $with = ['user'];
}
这篇关于试图在laravel 5.4中获取非对象的属性的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!