打破分页页数

Question

I am using this pagination script given below. The problem is that it gives the continuous count of the pages. Say for example if there are 10 pages it gives out numbers 1 2 3 4 5 6 7 8 9 10 (all clickable to respective pages). What I want is to break pagination count after 5. It should display 1 2 3 4 5.......9 10. Then when I click on page number 5 it should hide 1 and display 6 as 2 3 4 5 6....9 10. Please help me.

$start = 0;
$limit = 1;

if(isset($_GET['id'])){
  $id = (!empty($_GET['id']))?$_GET['id']:null;
  $start = ($id-1)*$limit;
}else{
  $id = null;
}

$stmt = $pdo->prepare("SELECT ad_id, ad_description, ad_url, ad_memclicks, ad_visclicks, ad_status, avc_clicks FROM ads
                       LEFT JOIN ads_viewed_cost ON ads.ad_clicks = ads_viewed_cost.avc_id
                       WHERE ad_user = :user LIMIT $start, $limit");
$stmt-> bindValue(':user', $sessionid);
$stmt-> execute();

$count = $pdo->prepare("SELECT COUNT(*) as cnt FROM ads LEFT JOIN ads_viewed_cost ON ads.ad_clicks = ads_viewed_cost.avc_id WHERE ad_user = :user");
$count-> bindValue(':user', $sessionid);
$count-> execute();
$rf = $count->fetch();
$rows = $rf['cnt'];
$total = ceil($rows/$limit);

if($id>1){
  echo "<span class='pre'><a href='?id=".($id-1)."' class='button'>PREVIOUS</a></span>";
}
if($id!=$total){
  echo "<span class='next'><a href='?id=".($id+1)."' class='button'>NEXT</a></span>";
}

echo "<ul class='pagination'>";
for($i=1; $i<=$total; $i++){
  if($i == $id){
    echo "<li class='current'>".$i."</li>";
  }else{
    echo "<li><a href='?id=".$i."'>".$i."</a></li>";
  }
}
echo "</ul>";

Answer 1

I don't think what you're describing is the best way to do pagination as you're losing the first page as you go, but this does what you are asking for.

$dots = false;
for ( $i=1; $i<=$total; $i++ ) {
    if( $i == $id ) {
        echo "<li class='current'>{$i}</li>";
    } else {
        if ( ( $id < 5 && $i < 5 ) || ( ( $i > $id-4 && $i < $id ) || $i > $total-2 || $i == $id+1 ) ) {
            echo "<li><a href='?id=".$i."'>{$i}</a></li>";
        } else {
            if ( $i > $id && !$dots ) {
                echo "...";
                $dots = true;
            }
        }
    }
}

If the current $id is less than 5, it shows the first pages. After that, it shows the current page, three before it, and one after it (so when id=5, you see 2, 3, 4, 5, 6).

The dots are shown if there is a gap between the current page + 1 and the last two pages, but are only ever shown once (what the boolean is for).

The last two pages are always shown.

Tested with a larger page set (up to 100) and works as expected.

---

A more common way to do pagination, as StackOverflow itself does it, would be once you are outside the extremes, you show the first page, dots, current page and a couple either side, dots, and last page.

1 ... 12 13 [14] 15 16 ... 100

Without the first page anchor, with what I've provided for you here, it could take a LOT of clicking to get back to page one with a large page set.

To achieve this method, you can use this code instead. The changes are in the if logic and a second boolean with handling for both sets of dots.

$dotsLow = false;
$dotsHi = false;
for ( $i=1; $i<=$total; $i++ ) {
    if( $i == $id ) {
        echo "<li class='current'>{$i}</li>";
    } else {
        if (
            ( $i == 1 || $i == $total ) ||
            ( $id < 5 && $i < 5 ) ||
            ( $id > $total-4 && $i > $total-4 ) ||
            ( $i >= $id-2 && $i <= $id+2)
        ) {
            echo "<li><a href='?id=".$i."'>{$i}</a></li>";
        } else {
            if ( $i > $id && !$dotsHi ) {
                echo "...";
                $dotsHi = true;
            }
            if ( $i < $id && !$dotsLow ) {
                echo "...";
                $dotsLow = true;
            }
        }
    }
}