How to show/hide button action depending on value of table field ?

[sergeynilov]

Hallo, In laravel 5.8 app I use yajra/laravel-datatables-oracle": "~8.0 and I need to show in 1 action 2 or 1 buttons, depending of status field value. 2) I define 2 buttons in my control :

            ->editColumn('action', '<a href="/admin/box-rooms/{{$id}}/edit" class="btn btn-default btn-sm btn-icon icon-left"><i class="fa fa-edit icon_dark"></i>Edit</a>' .
                                   '<a href="#" 
onclick="javascript:backendStorageSpace.storageSpaceShowFees({{$id}},\'{{$number}}\',\'{{$status}}\',\''.$current_currency_short.'\',\'{{$selling_range}}\',\'{{$insurance_vat}}\',\'{{$om_vat}}\',\'{{$electricity_vat}}\',\'{{$internet_vat}}\',\'{{$racks_vat}}\',\'{{$hot_desk_vat}}\')" class="btn btn-default btn-sm btn-icon icon-left mt-2"><i
class="fa fa-money-bill icon_dark"></i>Fees</a>')

and I need to show/hide "Fees" depending on $status value. Is there is such a way?

2) I can create 2 actions with different names, but again I did not find a way depending on $status value. Is it possible and if yes how?

Thanks!

• Operating System
• PHP Version
• Laravel Version
• Laravel-Datatables Version

[yajra]

You can use a closure that then builds your desired html or use a view:

->editColumn('action', function($user) {
  if ($user->active) {
   $action = 'active_html';
  }

  return $action;
})

[yajra]

Or if you prefer a polymorphic view:

->editColumn('action', function($user) {
  return view('users.actions.' . $user->status, compact('user'));
})

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[github-actions[bot]]

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