laravel getting dynamic view based on url - php

Question

I have made dynamic category routes by adding custom class to my app (how i did it is here) now i need to make my blade works with this dynamic path.
Logic
based on categories deeps my url will create such as:
site.com/category/parent
site.com/category/parent/child
site.com/category/parent/child/child
etc.
so far my view is just loading for site.com/category/parent for other urls it return 404 error.
code
CategoryRouteService
class CategoryRouteService
{
private $routes = [];
public function __construct()
{
$this->determineCategoriesRoutes();
}
public function getRoute(Category $category)
{
return $this->routes[$category->id];
}
private function determineCategoriesRoutes()
{
$categories = Category::all()->keyBy('id');
foreach ($categories as $id => $category) {
$slugs = $this->determineCategorySlugs($category, $categories);
if (count($slugs) === 1) {
$this->routes[$id] = url('category/' . $slugs[0]);
}
else {
$this->routes[$id] = url('category/' . implode('/', $slugs));
}
}
}
private function determineCategorySlugs(Category $category, Collection $categories, array $slugs = [])
{
array_unshift($slugs, $category->slug);
if (!is_null($category->parent_id)) {
$slugs = $this->determineCategorySlugs($categories[$category->parent_id], $categories, $slugs);
}
return $slugs;
}
}
CategoryServiceProvider
class CategoryServiceProvider
{
public function register()
{
$this->app->singleton(CategoryRouteService::class, function ($app) {
// At this point the categories routes will be determined.
// It happens only one time even if you call the service multiple times through the container.
return new CategoryRouteService();
});
}
}
model
//get dynamic slug routes
public function getRouteAttribute()
{
$categoryRouteService = app(CategoryRouteService::class);
return $categoryRouteService->getRoute($this);
}
blade
//{{$categoryt->route}} returning routes
<a class="post-cat" href="{{$category->route}}">{{$category->title}}</a>
route
//show parent categories with posts
Route::get('/category/{slug}', 'Front\CategoryController#parent')->name('categoryparent');
controller
public function parent($slug){
$category = Category::where('slug', $slug)->with('children')->first();
$category->addView();
$posts = $category->posts()->where('publish', '=', 1)->paginate(8);
return view('front.categories.single', compact('category','posts'));
}
Note: I'm not sure about this but i think i my route is kinda static! I mean it just getting 1 slug with it while my category can goes 2, 3 or 4 slug deep and it doesn't make sense to me to make several route and keep repeating Route::get('/category/{slug}/{slug}/{slug} like that.
As I said I'm not sure about this, please share your idea and solutions if you may.
UPDATE
based on Leena Patel answer I changed my route but when I get more than 1 slug in my url it returns error:
Example
route: site.com/category/resources (works)
route: site.com/category/resources/books/ (ERROR)
route: site.com/category/resources/books/mahayana/sutra (ERROR)
error
Call to a member function addView() on null
on
$category->addView();
when I comment that it returns error for $posts part. then error for my blade where i returned category title {{$category->title}}
So basically it seem doesn't recognize this function for returning view of category routes.
here is my function
public function parent($slug){
$category = Category::where('slug', $slug)->with('children')->first();
$category->addView();
$posts = $category->posts()->where('publish', '=', 1)->paginate(8);
return view('front.categories.single', compact('category','posts'));
}
any idea?

Answer 1

You can try using Route Pattern like below
Route::get('/category/{slug}', 'Front\CategoryController#parent')->where('slug','.+')->name('categoryparent')
So if you have more than one slugs in your url like /category/slug1/slug2
Your addView() method will work for one record and not for Collection So add foreach loop to achieve this.
public function parent($slug){
// $slug will be `slug1/slug2`
$searchString = '/';
$posts = array();
if( strpos($slug, $searchString) !== false ) {
$slug_array = explode('/',$slug);
}
if(isset($slug_array))
{
foreach($slug_array as $slug)
{
$category = Category::where('slug', $slug)->with('children')->first();
$category->addView();
$posts_array = $category->posts()->where('publish', '=', 1)->paginate(8);
array_push($posts,$posts_array);
}
}
else
{
$category = Category::where('slug', $slug)->with('children')->first();
$category->addView();
$posts = $category->posts()->where('publish', '=', 1)->paginate(8);
}
return view('front.categories.single', compact('category','posts'));
}
Hope it helps!
Documentation : https://laravel.com/docs/4.2/routing#route-parameters

Answer 2

Create route
Route::get('category/{cat}', 'YourController#mymethod');
Add this to your Providers/RouteServiceProvider.php 's boot method
public function boot()
{
Route::pattern('cat', '.+'); //add this
parent::boot();
}
In your method:
public function mymethod($cat){
echo $cat; //access your route
}

Answer 3

You can use optional URL sections in the route and use conditionals in controllers. Try this:
In your route:
Route::get('/category/{parent?}/{child1?}/{child2?}', 'Front\CategoryController#parent')->name('categoryparent');
In your controller:
public function mymethod($category, $parent, $child1, $child2){
if(isset($child2)){
//use $category, $parent, $child1, $child2 and return view
} else if(isset($child1)){
//use $category, $parent, $child1 and return view
} else if(isset($parent)){
//use $category, $parent and return view
} else {
//return view for $category
}
}