Level 50
why would you hide them?
try using ->simplePaginate($numberPerPage) maybe?
Nov 16, 2018
4
Level 1
hide fields in paginator
Hi everyone. I've implemented the default paginator of laravel.
collectionName->paginate($numberPerPage).
All is good. I receive the response:
"current_page": 1, "data":[ all data here], "first_page_url": "http://homestead.test/api/user/orders?page=1", "from": 1, "last_page": 2, "last_page_url": "http://homestead.test/api/user/orders?page=2", "next_page_url": "http://homestead.test/api/user/orders?page=2", "path": "http://homestead.test/api/user/orders", "per_page": 5, "prev_page_url": null, "to": 5, "total": 10
Now i want to hide some fields which i not use, such us first_page_url and others. How can i do ?? I want to do this every time i call the function paginate().
Thank you for your time.
Level 1
I don't need them in the front end. I want to optimize to the best the system. I know that are only bytes but i want to save space as much as possible.
Level 75
First, I have no idea why you would want to hide a pagination link. But you can write any pagination template you need, you are free to do that. The docs under pagination explains it,
As example here is a custom template I use:
@if ($paginator->hasPages())
<ul class="pagination pagination">
{{-- Previous Page Link --}}
@if ($paginator->onFirstPage())
<li class="disabled"><span>«</span></li>
@else
<li><a href="{{ $paginator->previousPageUrl() }}" rel="prev">«</a></li>
@endif
@if($paginator->currentPage() > 3)
<li class="hidden-xs"><a href="{{ $paginator->url(1) }}">1</a></li>
@endif
@if($paginator->currentPage() > 4)
<li><span>...</span></li>
@endif
@foreach(range(1, $paginator->lastPage()) as $i)
@if($i >= $paginator->currentPage() - 2 && $i <= $paginator->currentPage() + 2)
@if ($i == $paginator->currentPage())
<li class="active"><span>{{ $i }}</span></li>
@else
<li><a href="{{ $paginator->url($i) }}">{{ $i }}</a></li>
@endif
@endif
@endforeach
@if($paginator->currentPage() < $paginator->lastPage() - 3)
<li><span>...</span></li>
@endif
@if($paginator->currentPage() < $paginator->lastPage() - 2)
<li class="hidden-xs"><a href="{{ $paginator->url($paginator->lastPage()) }}">{{ $paginator->lastPage() }}</a></li>
@endif
{{-- Next Page Link --}}
@if ($paginator->hasMorePages())
<li><a href="{{ $paginator->nextPageUrl() }}" rel="next">»</a></li>
@else
<li class="disabled"><span>»</span></li>
@endif
</ul>
@endif
But you need to on your own learn css (not a library) and you will discover how to handle things like you ask. A pagination link is just that, a link with css style applied.
If you don't want extra links, as suggested use simple paginate.
Level 23
If you really just want the list of items back you can:
collectionName->paginate($numberPerPage)->items()
that will get you only the array of the collection items
1 like
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