Warning: file_get_contents(/data/phpspider/zhask/data//catemap/8/mysql/66.json): failed to open stream: No such file or directory in /data/phpspider/zhask/libs/function.php on line 167

Warning: Invalid argument supplied for foreach() in /data/phpspider/zhask/libs/tag.function.php on line 1116

Notice: Undefined index: in /data/phpspider/zhask/libs/function.php on line 180

Warning: array_chunk() expects parameter 1 to be array, null given in /data/phpspider/zhask/libs/function.php on line 181
Php 获取laravel中未定义的变量错误_Php_Mysql_Laravel_Model_Migration - Fatal编程技术网

Php 获取laravel中未定义的变量错误

Php 获取laravel中未定义的变量错误,php,mysql,laravel,model,migration,Php,Mysql,Laravel,Model,Migration,我有两个模型:酒店和评论,随着它们各自在酒店内部的迁移,有一个功能: 酒店模式: <?php namespace App; use Illuminate\Database\Eloquent\Model; class Hotel extends Model { public function Comments(){ return $this->hasMany(Comment::class); } } <?php namespace App; us

我有两个模型:酒店评论,随着它们各自在酒店内部的迁移,有一个功能:

酒店模式:

<?php
namespace App;
use Illuminate\Database\Eloquent\Model;
class Hotel extends Model
{
    public function Comments(){
        return $this->hasMany(Comment::class);
    }
}
<?php
namespace App;
use Illuminate\Database\Eloquent\Model;
class Comment extends Model
{
    public function hotel(){
        return $this->belongsTo(hotel::class);
    }
}

通过执行以下操作,确保
$hotel
是已初始化的对象:

$hotel = new hotel();
$hotel->name;

它是如何从控制器传递到视图的?是否将$hotel变量从控制器传递到视图?
$hotel = new hotel();
$hotel->name;