Get the value which comes after the domain name in laravel - php

Question

how to create route in laravel for the below 2nd option....
http://localhost:8048/
http://localhost:8048/content/645668/nice-up-civic-poll.html
1st it redirects to home page which is correct for me.
2nd I need to get what ever comes after 8048/
so basically content/645668/nice-up-civic-poll.html is a parameter which I need to deal with it separately and its dynamic link.
Route api in laravel :
Route::get('/', 'HomeController#index');
www.example.com will load home page with all stories.
The below links as an example should get the value after www.example.com/ basically its a story/article link so when that comes specific story will be displayed.
www.example.com/content/645668/nice-up-civic-poll.html
www.example.com/content/283206/something-here.html
www.example.com/content/234323/good-nice.html
www.example.com/content/451425/breakup-actor.html
www.example.com/content/365412/accident-occured.html
So basically get everything after domain name which is using apache server.
.htaccess file
<IfModule mod_rewrite.c>
<IfModule mod_negotiation.c>
Options -MultiViews -Indexes
</IfModule>
RewriteEngine On
# Handle Authorization Header
RewriteCond %{HTTP:Authorization} .
RewriteRule .* - [E=HTTP_AUTHORIZATION:%{HTTP:Authorization}]
# Redirect Trailing Slashes If Not A Folder...
RewriteCond %{REQUEST_FILENAME} !-d
RewriteCond %{REQUEST_URI} (.+)/$
RewriteRule ^ %1 [L,R=301]
# Handle Front Controller...
RewriteCond %{REQUEST_FILENAME} !-d
RewriteCond %{REQUEST_FILENAME} !-f
RewriteRule ^ index.php [L]
</IfModule>

Answer 1

If you want a home route and then every other URI to go to a single Controller method you can make a catch-all route:
Route::get('{catch}', 'SomeController#action')->where('catch', '.*');
This would catch any URI that doesn't match any previously defined route.
If you want everything to go to one place:
Route::get('{catch?}', ....)->where(...); // optional param
Post about creating a catch all route, answer using regex conditions on parameter:
SO - How do I make a Catch-All Route in Laravel 5.2
Update:
If these URIs you need to catch all have the same format,
www.example.com/content/645668/nice-up-civic-poll.html
you can register a route to match that format instead of catching everything possible:
Route::get('content/{id}/{slug}', function ($id, $slug) {
...
});

Answer 2

try using $request helper..
$leftbehind = str_replace($request->getHost(),'', $request->fullUrl());
or try this..
$request->getRequestUri();

Answer 3

Use $request->path()
The path method returns the request's path information. So, if the incoming request is targeted at http://example.com/foo/bar, the path method will return foo/bar
https://laravel.com/docs/5.5/requests#request-path-and-method

Answer 4

You can use The Request::path() to get the current url.
https://laravel.com/api/5.5/Illuminate/Http/Request.html - Check this for all the options available for Request.
eg: If you want to just check whether the users is in some url or not use this - Request::is('/url') // This will return true or false

Answer 5

You can use php's built in function, parse_url to retrieve content/645668/nice-up-civic-poll.html.
parse_url($url, PHP_URL_PATH)