Level 75
Use a lengthaware, and first pass store count in session. Then page 2, 3, etc you don't need to count again.
1 like
Feb 17, 2022
12
Level 1
Method paginate - Laravel
Good morning guys, I have a complex query that takes about 3 seconds to execute returning 113 thousand records with about 10 joins, etc, etc ...
So far ok, but because of paging it is as if we generated 2 queries one to know the total of records and the other to bring the records that would take 6 seconds ... After all the pagination does a count with the * of all tables which would be totally unnecessary because I use the paginate method ...
Would it be possible to make a custom pagination for this case? At first with some laravel feature even without using the traditional paginate ???
The intention would be to change the query that generates the count with * to count(0) because the * represents that I want to count all fields from all tables and would give this catastrophe, where if I pass a single field ready already solves, it drops from 3 seconds to 1 second ...
Does anyone have any suggestions ??? Thanks for now ...
Level 51
@Dárcio Júnior If that solves your problem, please mark his answer as "Best answer" to close the discussion.
Level 1
@jlrdw good afternoon friend ...
I'm having a single problem only, the paging is failing to retrieve the filtered parameters, follows my code below, where am I going wrong ????
public function paginate($items, $perPage = 5, $page = null) { $items = $items instanceof Collection ? $items : Collection::make($items); $page = $page ?: (Paginator::resolveCurrentPage() ?: 1);
$options = [
'path' => request()->url(),
'query' => request()->query()
];
return new LengthAwarePaginator($items->forPage($page, $perPage)->values(), $items->count(), $perPage, $page, $options);
}
Level 75
@Dárcio Júnior an example https://laracasts.com/discuss/channels/guides/length-aware-paginator
You have to append the query parameters.
1 like
Level 1
@jlrdw I am working withQueryString in the pagination view -> {{ $tableData->withQueryString()->links() }}
Level 75
@Dárcio Júnior just pass the parameters as an array to appends. I have had no problems with the paginator doing this.
@php echo str_replace('/?', '?', $report->appends($array-of-parameters)->render(') @endphp
If you named it $myparameters, then:
@php echo str_replace('/?', '?', $report->appends($myparameters)->render()) @endphp
1 like
Level 1
@jlrdw I understood ... Perfectly in the view ... In the controller, I made this way:
public function paginate($items, $perPage = 5, $page = null) { $items = $items instanceof Collection ? $items : Collection::make($items); $page = $page ?: (Paginator::resolveCurrentPage() ?: 1);
$params = ['query' => request()->query(), 'page' => $page];
$options = [
'path' => request()->url(),
'params', $params
];
return new LengthAwarePaginator($items->forPage($page, $perPage)->values(), $items->count(), $perPage, $page, $options);
}
Level 75
@Dárcio Júnior You are using collection? Why? Just paginate database results.
But for collections: https://laracasts.com/discuss/channels/guides/paginate-collection-simple-example-guide
Or write you own custom. A paginator is just basic math and links to the pages.
Level 1
@jlrdw Good afternoon friend, just this code here already solved, it was a problem in an object of my form with respect to the parameters ...
public function paginateLengthAwarePaginator($items, $perPage = 5, $page = null, $options = []) { $items = $items instanceof Collection ? $items : Collection::make($items);
$page = $page ?: (Paginator::resolveCurrentPage() ?: 1);
$options = [
'path' => request()->url(),
'query' => request()->query()
];
return new LengthAwarePaginator($items->forPage($page, $perPage), $items->count(), $perPage, $page, $options);
}
Your answer was the best ... Thanks
Translated with www.DeepL.com/Translator (free version)
Level 1
@MohamedTammam Close the question please ... Thanks, my problem was resolved ...
Level 51
Level 75
Wow.
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