Ran*_*rki 2 php mysql laravel laravel-5
在laravel5中,我得到以下消息(错误)
No query results for model [App\Products].
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我有一个表格“ products”,其中的每个类别项目都有一列“ category_id” category Id。我想根据显示项目category Id
我的控制器就是这样
public function category()
{
$recordsByCategories=\DB::table('products')
->select('categories','category_id', \DB::raw('count(*) as totalProducts'))
->groupBy('categories')
->get();
return view('dashboard.show',compact('recordsByCategories'));
}
**//I have received error in below function**
public function searchCategoryByTag($id)
{
$category_id = Products::findOrFail($id);
$records=\DB::table('products')->Where('category_id','$category_id');
dd($records);
}
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我的路线是这样的
Route::get('categorys/{id}' ,array(
'as'=>'category',
'uses'=>'GoodsController@searchCategoryByTag'));
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我的看法是这样的
@foreach($recordsByCategories as $recordsByCategory)
<a href="{{URL::route('category',$recordsByCategory->category_id)}}
" id="search">{{$recordsByCategory->categories}}</a>:{{$recordsByCategory->totalProducts}}
@endforeach
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我知道为时已晚,但是对于您(可能是)和其他人的帮助:
在这里您误解了'$category_id',删除了单引号。
public function searchCategoryByTag($id)
{
$category_id = Products::findOrFail($id);
$records=\DB::table('products')->Where('category_id', $category_id);
}
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