CODEWITHSUNDEEP
phplaravellaravel-5
Bas
Bas

Reputation: 2400

Pass setting data from database to layout

I want to pass data from de setting table from the database to my layout view.

How do I get it done? 

$item = Setting::find(1);
return view($this->controller.'/show')->with( 'item', $item);

Solution:

public function boot() {
    if( !isset( $_SESSION['adminTitle'] ) ){
        $item = Setting::find(1);
        $item = $item->toArray();
        $_SESSION['adminTitle'] = $item['title'];
        $_SESSION['adminEmail'] = $item['email'];
        $_SESSION['adminLogo'] = $item['logo'];
    }
}

Upvotes: 1

Views: 64

Answers (2)

Andrea
Andrea

Reputation: 16580

Why not simply this?: 

// File app/Http/Controllers/ExampleController.php

//

class ExampleController extends Controller
{
  public function show() 
  {
    //

    $setting = Setting::find(1);
    return view('example', ['setting' => $setting]);
  }

}

Within the Blade view:

<!-- resources/views/example.blade.php -->

{{ $setting->title }}
{{ $setting->logo }}
...

But, if you want to share settings between all your views, you can add this middleware:

// File app/Http/Middleware/ViewShareSettingMiddleware

//

class ViewShareSettingMiddleware
{
  public function handle($request, Closure $next)
  {
     $setting = Setting::find(1);
     view()->share('setting', $setting);
     return $next($request);
  }

}

Upvotes: 1

rome 웃
rome 웃

Reputation: 1901

Create your view in:

\resources\views\

Example: \resources\views\index.blade.php

$data['item'] = Setting::find(1);
return view('index')
            ->with( $data);

Or 

$item = Setting::find(1);
return view('index', compact('item');

View: {{$item}}

Upvotes: 0

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