Pagination links() with request that are only built with key

I have a set of records that needs to be paginated. Based on a Laracast's lesson I added a popular filter that will sort by "most viewed" restaurants. This query doesn't have any value like popular=true, only a key.

If my request is ?cuisine=pasta&popular&page=2 and I try to create pagination using

{{ $restaurants->appends(request()->input())->links() }}

the url for page 3 will remove popular key and only show ?cuisine=pasta&page=3.

How can I keep this popular query across navigation?

Thanks!

MichalOravec

5 years ago

Level 75

Use this

{{ $restaurants->withQueryString()->links() }}

From docs: If you wish to append all current query string values to the pagination links you may use the withQueryString method.

https://laravel.com/docs/7.x/pagination#displaying-pagination-results

jdebuchy

5 years ago

Level 14

Thanks for your answer @michaloravec. Unfortunately this is not working. I guess it has to do with the fact that popular value is null.

#query: array:2 [
    "cuisine" => "pasta"
    "popular" => null
 ]

jlrdw

5 years ago

Level 75

Both sides of = needs something like

popular=true

see https://stackoverflow.com/questions/6461668/boolean-in-a-uri-query

Compare:

tjhs/dog/indexadmin?p=2&psch=&aval=n

tjhs/dog/indexadmin?p=2&psch=c&aval=y

aval is a boolean for is the dog available for adoption or not.

First line search name field is blank to show all, therefore

psch=

Query string works the same in all languages, php, java, asp.net, etc.

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