#if statement using URL parameter laravel blade - if-statement

Question

I am new to laravel and blade. I'm a student doing a simple 'job seeker' assignment. I have 2 different types of users - jobseekers (category 1) and employers (category 2). When I create a new user from buttons in layout.blade.php, users will click on a register (category 1) link or an Employer Button (category 2), I want to pass the category on to the create.blade.php so that I can stylise it depending on what category they are, and of course keep that information hidden from the actual User.
I'm not sure what code you want to see, but I'll start with my layout.blade.php - when the link or button is clicked, it redirects to create.blade.php and the url updates to either category 1 or categoy 2 - depending on what is clicked. I want to add an #if statement as to which create for gets displayed, one jobseeker or one for employer (they have slightly different options)
layout.blade.php
<div class="col-sm-9">
#if (!Auth::check())
<div class="login-form">
{{ Form::open(array('action' => 'UserController#login')); }}
{{ Form::text('username', null, array('class' => 'input-small', 'placeholder' => 'Email')); }}
{{ Form::password('password', array('class' => 'input-small', 'placeholder' => 'Password')); }}
{{ Form::submit('Sign in', array('class' => 'btn btn-danger')); }}
{{ Form::close(); }}
{{ Form::open(array('action' => 'UserController#create')); }}
{{link_to_route('user.create', 'or Register here', ['category' => 1] )}}
</div>
{{link_to_route('user.create', 'Employers', ['category' => 2], array('class' => 'btn btn-primary')) }}
#endif
#yield('content1')
create.blade.php
#extends('job.layout')
#section('content1')
#if('category' == 2)
<h1>New Employer page</h1>
{{ Form::open(array('action' => 'UserController#store')); }}
{{ Form::text('username', null, array('class' => 'input-small', 'placeholder' => 'Email')); }}
<p>{{ Form::password('password', array('class' => 'input-small', 'placeholder' => 'Password')); }}
{{ Form::hidden('category', 2) }}
{{ Form::label('name', 'Name:', array('class' => 'col-sm-3')) }}
{{ Form::text('name') }}
{{ Form::label('description', 'Company Description:', array('class' => 'col-sm-3')) }}
{{ Form::text('description') }}
{{ Form::label('industry', 'Industry', array('class' => 'col-sm-3')) }}
{{ Form::text('industry') }}
{{ Form::label('phone', 'Phone Number:', array('class' => 'col-sm-3')) }}
{{ Form::text('phone') }}
<p>{{ Form::submit('Sign in'); }}
{{ Form::close(); }}
#else
<p>just a test for New User Page
#endif
#stop
The create page so far just results in getting the #else condition back. ie: "just a test for New User Page"
Thanks in advance

Answer 1

i will avoid all your codes because you are making it way complicated than it is.
i will explain in step by step.
make two views. one for job seeker, one for employer.
depending upon the category, load the respective view. that's all you want.
let's come to code.
Routes.php
Route::get('create/{category}', array(
'as' => 'create',
'uses' => 'UserController#create'
));
UserController
public function create($category)
{
if($category==1)
return View::make('seeker');
elseif($category==2)
return View::make('employer');
else
App::abort(400);
}
that's it. no need to touch the layout. Avoid putting logics in layout as possible. in longer run, it will be a mess.