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Php laravel 5 where()不工作_Php_Laravel 5 - Fatal编程技术网

Php laravel 5 where()不工作

Php laravel 5 where()不工作,php,laravel-5,Php,Laravel 5,我有两个表,类别和子类别,子类别中的每个项目通过外键“category_id”属于类别 我有两个模型,一个用于类别,一个用于子类别,还有一个控制器 我的类别模型: <?php namespace App; use Illuminate\Database\Eloquent\Model; class Category extends Model { // public function subcategories() { return $this-

我有两个表,类别和子类别,子类别中的每个项目通过外键“category_id”属于类别

我有两个模型,一个用于类别,一个用于子类别,还有一个控制器

我的类别模型:

<?php

namespace App;

use Illuminate\Database\Eloquent\Model;

class Category extends Model
{
    //
    public function subcategories()
    {
        return $this->hasMany('\App\Subcategory');
    }

}

根据请求,您需要返回带有categories变量的视图

 return view('view.name',compact('categories'); 
然后


假设您正在使用blade,并且在类别集合中有名称值。

$categories=Category::with('subcategories')->其中('Category_name'、'='、'Construction Machines')->get();返回$categories还有,在没有返回make()的情况下,如何在视图中访问此变量?因为我需要这些数据的页面已经加载,所以我需要用它填充下拉列表?@VishalDesai它是用futureweb评论解决的还是仍然解决的?是的,但我正在尝试访问我视图中返回的$categories。@futureweb如果您留下答案,我将投票支持它。
<?php

namespace App\Http\Controllers;

use App\Category;
use App\Subcategory;
use Illuminate\Http\Request;
use App\Http\Requests;
use App\Http\Controllers\Controller;

class CategoriesController extends Controller
{
    //return array of categories
    public function categories ()
    {

        $categories =  Category::where('category_name', '=', 'Construction Machines')->subcategories;
        return $categories;
    }
}
 return view('view.name',compact('categories'); 
  {{dd($categories)}} 
 <? print_r($categories); ?>
 @foreach($categories as $category) 
    {{$category->name}} 
 @endforeach