How to limit pagination to X items/pages

Why not add in a column to the videos called ranking? Add in a where clause like

->andWhere('rank', '<=', 100)

That then will fetch all items with a rank under 100 for a specified genre

Edit* If rating is your rank then simply do...

$top100videos = Videos::where('genre', $genre)
                    ->andWhere('rating', '<=', 100)
                    ->orderby('rating', 'desc')
                    ->with('actors')
                    ->paginate(20);

Maybelimit(100) will work?

Please dig in the docs, skip take is all there:

skip / take

To limit the number of results returned from the query, or to skip a given number of results in the query, you may use the skip and take methods:

$users = DB::table('users')->skip(10)->take(5)->get();

Alternatively, you may use the limit and offset methods:

$users = DB::table('users')
                ->offset(10)
                ->limit(5)
                ->get();

@gofish543 I don't want all videos with 100 rank/score than first 100 videos ordered by rank/score.

@topvillas I have tried that too - but it doesn't work :(

@jirdw I know how skip/take and offset/limit works but I don't think that it will work with paginate, and also I don't want to paginate query builder results collection than eloquent objects collection.

Yes, if you use the <= sign in the query it will take rank/score of 1-100 and then paginate it by 20. This gives 5 blocks of 20 scores with each block representing 1-20, 21-40, 41-60, 61-80, and 81-100 respectivly

@goldfish543 This table can have thousands of clips... The rating is used just to order the clips by user rating and nothing else. It can't be used to fetch 100 videos since you don't know their ratings...

https://github.com/laravel/framework/issues/4629

$top100videos = Videos::where('genre', $genre)
                    ->orderby('rating', 'desc')
                    ->with('actors')
                    ->take(100)
                    ->get()

$videos = Paginator::make($top100videos->toArray(), $top100videos->count(), 100);

What appears to be happening is you are forced to make the query, and limit your results. Then take that collection and make a paginator out of it.

I have seen that but that's old (laravel 4) and Paginator::make doesn't work now.

It appears that I can create a paginator object but the code looks really messy and uses array (not eloquent objects so I can't call methods on them) so I'm looking if there's some better and cleaner way to do it...

Edit* Have you tried using a LengthAwarePaginator?

use Illuminate\Pagination\LengthAwarePaginator;
use Illuminate\Support\Collection;

$currentPage = LengthAwarePaginator::resolveCurrentPage();

$top100videos = Videos::where('genre', $genre)
                      ->orderby('rating', 'desc')
                      ->with('actors')
                      ->take(/* How many you want */)
                      ->get()

$perPage = /* Per page Number */;

$currentPageSearchResults = $collection->slice($currentPage * $perPage, $perPage)->all();

$paginatedTop100 = new LengthAwarePaginator($currentPageSearchResults, count($collection), $perPage);

return view(/* Insert View */, ['top100' => $paginatedTop100 ]);

and then the view

@foreach ($top100 as $top)
    {{ $top }}
@endforeach

{{ $top100->render() }}

I'll give it a try...

Still looks far more comples than simple ->take(100)

:)

This post has gotten funny I gave the answer above all of those query-builder methods are available in the orm also.

Yes, but these don't work with Paginate :) https://github.com/laravel/framework/issues/4629

The LengthAwarePaginator woks... I had just to modify this line since offsets were wrong:

$currentPageSearchResults = $collection->slice($start, $perPage)->all();

Also had to add this above it to calculate the start correctly:

if ($currentPage==1) {
    $start = 0;
}
else {
    $start = ($currentPage-1) * $perPage;
}

Since for the first page $currentPage * $perPage equals 20 so on the first page it would start from 20th item (instead from 1st - zero one) and then at last page it would have nothing to show...

Anyway, this solution works... I hope that Paginator would soon be compatible with take() or limit() so this can be done easier...

In the meantime here's the entire code:

$collection = Videos::where('genre', $genre)
                      ->orderby('rating', 'desc')
                      ->with('actors')
                      ->take(100)
                      ->get()


$perPage = 20;

$currentPage = LengthAwarePaginator::resolveCurrentPage();

if ($currentPage == 1) {
    $start = 0;
}
else {
    $start = ($currentPage - 1) * $perPage;
}

$currentPageCollection = $collection->slice($start, $perPage)->all();

$paginatedTop100 = new LengthAwarePaginator($currentPageCollection, count($collection), $perPage);

$paginatedTop100->setPath(LengthAwarePaginator::resolveCurrentPath());


return view('top100videos', ['top100' => $paginatedTop100 ]);

You didn't read the whole issue. The person did not know what they were doing who created the issue.

How come? What did I miss?

late now I know, but I would probably run 2 queries, one to take(100) entries and then pluck() the id's then do a second wherein($ids)->paginate(20);

Not a bad idea at all! I'll try it out tomorrow... Thanks for the input...

Wow that is something I used appends often in version 5.1 but there is a problem with it in 5.4 I hope they fix it.

Followup, I created an issue, I use skip and take from time to time, I have been on 5.1, just switched to 5.4.

https://github.com/laravel/framework/issues/20218

I closed above issue and am just using a Lengthaware now.

@Snapey - tried your way... works like charm and it's super clean and readable... Thank you for your input!

@Snapey Your are genius, thank you for the idea it really solved a big issue for me.