Laravel 5.2 Getting Exception Undefined Variable

Question

I'm working on Laravel and got stuck in this mess of undefined variable $category, I don't know why and where is the exact problem.

I have done this much.

AdminAjaxController

public function category()
{
    $category=DB::select('select category_name,category_id from categories');
    return view('admin.category_table',compact('category'));
}

category_table View

<table id="category" class="table">
    <thead>
        <tr>
            <th>ID</th>
            <th>Category Name</th>
            <th>Delete</th>
            <th>Update</th>
      </tr>
    </thead>
    <tbody>
  @foreach($category as $value)
    <tr>
        <td>{{ $value->category_id}}</td>
        <td>{{ $value->category_name}}</td>
        <th>Delete</td>
        <td>Update</td>
    </tr>
    @endforeach
</tbody>
</table>

Martin · Accepted Answer · 2016-11-14 14:18:23Z

1

You should try this solution:

public function category()
{
    $category=DB::select('select category_name,category_id from categories');
    return view('admin.category_table')->with(['category' => $category]);
}

In addition to this you are getting this error, because you do not send your category variable.

edited Nov 14, 2016 at 14:18

answered Nov 14, 2016 at 14:11

Martin

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Comments

Atiqur · Accepted Answer · 2016-11-14 04:05:49Z

0

Try this :

public function category()
{
    $category=DB::table('categories')->select('category_name', 'category_id')->get();

    return view('admin.category_table',compact('category'));
}

answered Nov 14, 2016 at 4:05

Atiqur

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2 Comments

Mr. Pyramid Over a year ago

still the same thing.. :(

Atiqur Over a year ago

This have to work! please follow the steps

1. make sure the page is .blade.php 2. make sure you are in admin folder 3. run php artisan cache:clear 4. if still the problem give a screenshoot here

Veera Prasad · Accepted Answer · 2016-11-15 08:44:33Z

0

Try this, it will work.

public function category()
{
    $category=
      DB::select('select category_name,category_id from categories');
    return view('admin.category_table',[ 'category'=>$category ]);
}

answered Nov 15, 2016 at 8:44

Veera Prasad

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