how to call links() with after ajax update?

Your pagination is integrated into ajax call, see http://laraget.com/blog/how-to-create-an-ajax-pagination-using-laravel

As @jlrdw said.

You can not regenerate php on the client side. You can manipulate the rendered (HTML) output but there is no guarantee that it'll be the same so that causes bugs and a whole lot of checks in JavaScript.

Much simpler to use the data from the ajax object to set up links.

Another I had bookmarked a while back, I remembered https://gist.github.com/FilipQL/7a569c6ea96d1800974e59e6b7c6d5ed

@jlrdw

great guys thanks, this is what I did but have run into a simple problem, please read:

public function status(Request $request)
{
    if($request->ajax()) {
        $comment = Comment::where('id', $request->comment_id)->first();
        $comments = Comment::latest()->paginate(3);
        $comment->status = $request->status;
        $comment->save();
        return response()->json([
            'status' => 'success',
            'comment' => $comment,
            'pagination' => view('core.comments.pagination', compact('comments'))->render(),
        ], 201);
    }
}

my routes:

        Route::get('/', ['as' => 'backend.comments.index', 'uses' => 'CommentController@index']);
        Route::post('status', ['as' => 'backend.comments.status', 'uses' => 'CommentController@status']);

I have a problem now as you can see function status is called, and the problem is that pagination generated applys status in the url so the link ends up being

http://larapress.dev/backend/comments/status?page=2

instead of

http://larapress.dev/backend/comments?page=2

second one works and is generated corrctly when I first enter the page, but as soon as i update the status on the model and new pagination is sent back it has added status in the url and that fails as you can understand. Is there a way around this?

nevermind got it figured out :)

Can you elaborate? @nikocraft