CODEWITHSUNDEEP
phppagination
Douglas Roos
Douglas Roos

Reputation: 613

Issue with pagination results in PHP

I'm developing a search function for a web, until now works like a charm but i have an issue with the pagination. I have the following URL format:

https://www.server.com/search?w=something&re=34&page=1

I need to change specifically the page number in the URL, i have a function that retrieves the full URL but is not useful because when i click on the page number it does this:

https://www.server.com/search?w=something&re=34&page=1&page=2&page=3

I need to replace specifically the page= to the number i want. How can i achieve this?

The full code of the function is this:

function PAGINAR($page,$totalPages){
        global $link;

        $nextpage= $page + 1;
        $prevpage= $page - 1;

        $paginado = '<ul id="pagination-digg">';

        if ($page == 1){
            $paginado .= '<li class="previous-off">&laquo; Previous</li>';
        }else{
            $paginado .= '<li class="previous"><a href="'.getUrl().'&page='.$prevpage.'">&laquo; Previous</a></li>';
        }

        for($i = max(1, $page - 5); $i <= min($page + 5, $totalPages); $i++){
            if ($page == $i){
                $paginado .= '<li class="active">'.$i.'</li>';
            }else{
                $paginado .= '<li><a href="'.getUrl().'&page='.$i.'">'.$i.'</a></li>';
            }
        }

        if($totalPages > $page ){      
               $paginado .= '<li class="next"><a href="'.getUrl().'&page='.$nextpage.'">Next &raquo;</a></li>';
            }else{
                $paginado .= '<li class="next-off">Next &raquo;</li>';
            }

        $paginado .= '</ul>';

        return $paginado;
    }

And this retrieves the URL

function getUrl() {
  $url  = @( $_SERVER["HTTPS"] != 'on' ) ? 'http://'.$_SERVER["SERVER_NAME"] :  'https://'.$_SERVER["SERVER_NAME"];
  $url .= $_SERVER["REQUEST_URI"];
  return $url;
}

Thanks in advance

Upvotes: 0

Views: 74

Answers (2)

Danijel
Danijel

Reputation: 12689

Here is one example that might help you:

$url = 'https://www.server.com/search?w=something&re=34&page=1';

parse_str( parse_url( $url, PHP_URL_QUERY ), $params );
$params['page'] = 2;

echo http_build_query( $params );   // outputs: w=something&re=34&page=2

parse_url
parse_str
http_build_query

Upvotes: 1

tazer84
tazer84

Reputation: 1823

This is a great example of why you don't want your back-end dictating your front-end. Ideally you should be passing back only the result set with the current page number and letting the front-end generate the first and last page links.

But since you're doing it this way, the problem is REQUEST_URI already contains the variable page. So you have two options:

1. Parse REQUEST_URI to either get the page parameter out or replace the number of the page with the number you want

preg_replace('/page=\d+/', 'page=' . $nextPage, $_SERVER['REQUEST_URI'])

2. Parse out the script variables (everything after '?') and re-attach the vars you want from $_GET

$baseUrl = explode('?', $_SERVER['REQUEST_URI'])[0];

$varArray = $_GET;
unset($varArray['page']);
$url = $baseUrl . '?' . implode('&', $varArray);

If this seems hacky, its because it is. Use JSON and let the front-end deal with setting up the front-end.

Upvotes: 1

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