Using url parameter in route to use in conditional - php

Question

I'm trying to find the correct way to lay this out in laravel so that I can hit one GET route which calls a single function, but within that function I want to use data from a mysql table to determine which blade to show.
Say this URL is visited with a query string parameter:
www.testsite.com?email=testEmail.com
I hit this route (But not sure how to accept the parameter)
Route::get('register', 'Data\DataController#DataForm')
->name('Data.register');
I have a mysql table called dataTable set up like so
email | type
-------------------------
test1#mail.com A
test2#mail.com B
test3#mail.com C
What's the best way to incorporate the email parameter so that I can hit the single route and single function, then use the email/type columns from mysql to determine the appropriate blade to show?
public function DataForm(Request $request)
{
//query table based on query string parameter 'email'
$email = dataTable::where('email', /*email parameter?*/)->first();
if($email['type']== A){
return view('data.typeA');
}elseif($email['type']== B){
return view('data.typeB');
}elseif($email['type']== C){
return view('data.typeC');
}
}

Answer 1

You can add it as a route parameter :
Route::get('register/{email}', 'Data\DataController#DataForm')->name('Data.register');
And then inside controller :
public function DataForm($email)
{
// Abort to 404 if $email is not a valid email address
if(!filter_var($email, FILTER_VALIDATE_EMAIL)) {
abort(404);
}
$typeData = Model::where('email', $email)->first();
// Email not found in the database
if(!$typeData){
abort(404);
}
switch ($typeData->type) {
case 'A':
return view('data.typeA');
case 'B':
return view('data.typeB');
case 'C':
return view('data.typeAC');
default:
abort(404);
}
}

Answer 2

Using a slash in the route you should be able to pass it like this:
Route::get('register/{email}', 'Data\DataController#DataForm')
->name('Data.register');
Then in your function you can just use the variable $email and it contains the email in the URL. But not sure if this URL is OK, using a slash instead of '?email=' for the passing of the parameter.
More on there here:
https://laravel.com/docs/5.7/routing#required-parameters

Answer 3

You'd need more logic than this, obviously:
public function DataForm(Request $request)
{
$type = (dataTable::where('email', $request->query->get('email'))->first())['type'];
return view('data.type'.$type);
}

Answer 4

As others said you could auto type in Request $request; however you don't have to. You can keep the function definition without the passed parameter, and use the helper functions instead:
if(request()->has('email')) {
$email = request()->input('email');
// Do stuff
} else {
// No email supplied
}