No query results for model [App\WhatTodoModel] - php

Question

I am newbie in laravel and I try to insert a data form a form having foreign key by using hide such as the code is mention below:-
<form class="form-horizontal" role="form" action="/WhatTodo/store" method="POST">
<input type="hidden" name="_token" value="{{ csrf_token() }}">
<input type="hidden" name="task_id" value=" {{$what->task_id}}">
<input type="hidden" name="work_id" value="{{$what->work_id}}">
<div class="form-group">
<label class="control-label col-sm-2" for="name"> Name</label>
<div class="col-sm-5">
{!!Form::select('name',$name)!!}
</div>
</div>
<div class="form-group">
<label class="control-label col-sm-2" for="work">work:</label>
<div class="col-sm-5">
<input type="text" class="form-control" name="work" value="">
</div>
</div>
<div class="form-group">
<div class="col-sm-offset-2 col-sm-10">
<input type="submit" class="btn btn-default" value="Submit">
</div>
</div>
</form>
I have the controller with function:-
public function create($id)
{
$what=WhatTodoModel::findorFail($id);
$name=WOrk::lists('name','name');
return view('what/create',compact('what','name'));
}

Answer 1

You haven't really told us what your issue is or what error you're getting, but my guess given the current question is:
Assuming you're trying to implement a resource route and resourceful controller, the create method is used to show a form to create a new object, not edit an existing one. The create method does not take any parameters, therefore $id will be blank and WhatTodoModel::findorFail($id); will throw an exception.
If you want to edit an existing record, you do that using the edit action.
For creating a new record, create shows the form, store saves the record.
For editing an existing record, edit shows the form, update saves the record.