Multiple parameters to crud

Probably you just need to specify the model like this

public function create(Category $category)
{
    return view('articles.create')
    ->with('category', $category);
}

I did specify the Model, but then "Sorry, the page you are looking for could not be found.". And I am still stuck.

the first parameter of Route::resource is the name of the resource, not the url. Try this:

Route::resource('categories', 'CategoriesController');
Route::resource('articles', 'ArticlesController');

then you'll be able to access the routes using the names:

route('articles.show', $article);
...
route('categories.create');

Yes, but to get the url like I listed above, do I need to use a separate Route:get?

The answer @arukomp gave you is perfect, I didn't pick up on that error. I have a feeling you should not be using resource at all, at least until you get used to how they work. You will have more control if you just write one get() method for each route you want.

You cannot have the two routes as you have shown

// Categories
Route::resource('docs', 'CategoryController');

// Articles
Route::resource('docs/{category}/', 'ArticleController');

because the router cannot differentiate between docs/23 and docs/puppies

All your requests will go to the Category controller and not the to Article.

If you want to use Route::resource then you have to recognise that Article and Category are different resources and should have their own URLs

Okay, I understand. So in short, Route::resource gives the curds it's own URLs. So If I want it to be like a did it above I'll be needing to use Route::get except for put and delete?

@dsml you can type

php artisan route:list 

in the console and see all of the routes in your app. This might help you understand what the resource routes are creating for you.

You would see the problem more clearly if you listed the routes out, but you would possibly still need to change the url schema

• docs/23 to show document 23
• docs/puppies to show category 'puppies'

cannot co-exist

• docs/category/puppies
• docs/23

can co-exist if they are in that order since category is matched before trying the variable route with the document ID

Yes, I see. So how can I sort my articles by category and link them like docs/main/some_article_name?

I already applied some hasToMany and BelongsTo in both Models.

What data you return has little to do with the URL

You can pretend to use the category

Route::get('/docs/{category/{slug}','[email protected]');

and then in the controller

public function show($category, $slug)
{

    $article = Article::where('slug',$slug)->first();

and then never use the category if you have unique slugs for the articles.

So this two can't co-exist. That's now pretty clear. But if I create a PagesController and make a page called docs, can I then somehow get the collection of all categories and articles there or do I need I helper function for that? Thank you!

1. Your route calls a controller method.

2. The method gets the data it needs and passes it to the view

3. You return the view.

Your controller can get data from any of your models and pass as many variables to the view as required.

I did everything you suggested. I have now to separate controllers for the articles and categories. I'll make them only accessible for admin. Then I have the PageController for the other stuff, but because I use slug for both of the categories and the articles if I want to say docs/{slug}/{slug} I then get the error "Route pattern "/docs/{slug}/{slug}" cannot reference variable name "slug" more than once." Is there a way around? I'm using laravel-sluggable by spatie. Thank you!

use something other than slug