Laravel View Composer Paginating with ajax - php

Question

I managed to find some resources to paginate a normal page with ajax. It's pretty easy. you can simply have this:
if($request->ajax()) {
return [
'posts' => view('ajax.index')->with(compact('posts'))->render(),
'next_page' => $posts->nextPageUrl()
];
}
But for Composer View this won't work, since I can't have $request in View Composer. Can anybody enlighten me. Maybe there is someway to get variables inside composer, since the link for it's pagination is basicaly the current page plus the extension for next contnet.

Answer 1

You can! just type hint the Request class and the service container will inject it at run time.Then you can access all the properties of request object
use Illuminate\Http\Request;
class MyComposer
{
public function __construct(Request $request)
{
$this->request= $request;
}
function compose($view)
{
if($this->request->ajax()) {
return [
'posts' => view('ajax.index')->with(compact('posts'))->render(),
'next_page' => $posts->nextPageUrl()
];
}
}
}
Update:
If you don't pass it from controller to your view but $category object still used in composer you get an error.
To avoid this you can either remove the * and explicitly pass view names those view that use category and create another declaration passing names of those views that do not, so you will need two view composers
//these views use category
View::composer([ 'view1','view2'], 'App\Http\ViewComposers\MyComposerWithCategory' );
//these views don't
View::composer([ 'view3','view4'], 'App\Http\ViewComposers\MyComposerWithoutCategory' );
Or you can just check if $category variable is set in an if statement

Answer 2

So here is an answer to this question. In case of View Composer Pagination. You can not paginate in composer, you technically have the ability to do that, but even if you get passed of doing requests to the provider(Which is obviously a bad idea) the pagination is still not going to work very well.
One of the basic issues you will face immideatly is pagination the page in a different page like posts/postslug, this one is going to throw errors easly. So what you should do instead is to create a very basic function in controller and do the requests to that page.
Here is an example.
Lets say you provide the data with the following query:
$posts = Post::orderBy('created_at', 'desc')
->paginate(15, ["*"], 'sidebar')->withPath('postPaginator');
The function that will paginate this needs to be something like this:
public function postPaginator(){
$posts = Post::orderBy('created_at', 'desc')->paginate(15 , ["*"], 'sidebar');
}
if you would like this pagiantor to be flexible for both ajax request and basic get requests, than you can return the result like:
if($this->request->ajax()) {
return [
'posts' => view('ajax.sidebarindex', ['posts' => $posts])->render(),
'next_page' => $posts->nextPageUrl(),
];
}
$posts->nextPageUrl() is there to provide the next page link so you can assigne it to the button for the next pagination.