[英]Limit pages numbers on PHP pagination
我有一個 PHP 分頁腳本。 當頁面記錄有幾百條時,分頁結果太大。 如何限制頁碼/鏈接?
示例:< 1 | 2 ... 37 | 38 | 39 | 40 | 41 | 42 ... 82 | 83 >
這是我的 PHP 腳本
<?php
$ppp = 10;
$rows = mysql_num_rows($query);
$nmpages = ceil($rows/$ppp);
// if current page is not 1, draw PREVIOUS link
if ($pg > 1 && $nmpages != 0) {
echo "<a href=\"?pg=".($pg-1)."\"><</a> ";
}
For($i = 1 ; $i <= $nmpages ; $i++) {
If($i == $pg) {
echo "<a href=\"#\" class=\"selected\"><b>".$i."</b></a> ";
} else {
echo "<a href=\"?pg=".$i."\">".$i."</a> ";
}
}
// if current page less than max pages, draw NEXT link
if ($pg < $nmpages && $nmpages != 0) {
echo "<a href=\"?pg=".($pg+1)."\">></a>";
}
?>
您知道如何使用我擁有的特定 PHP 腳本執行此操作嗎?
嘗試這個 :
<?php
$link = "";
$page = $_GET['pg']; // your current page
// $pages=20; // Total number of pages
$limit=5 ; // May be what you are looking for
if ($pages >=1 && $page <= $pages)
{
$counter = 1;
$link = "";
if ($page > ($limit/2))
{ $link .= "<a href=\"?page=1\">1 </a> ... ";}
for ($x=$page; $x<=$pages;$x++)
{
if($counter < $limit)
$link .= "<a href=\"?page=" .$x."\">".$x." </a>";
$counter++;
}
if ($page < $pages - ($limit/2))
{ $link .= "... " . "<a href=\"?page=" .$pages."\">".$pages." </a>"; }
}
echo $link;
?>
輸出 :
//At page=1
1 2 3 4 ... 20
//At page=12
1 ... 12 13 14 15 ... 20
//At page=18
1 ... 18 19 20
基於@Makesh 的代碼的改進或更確切地說是重寫。
function get_pagination_links($current_page, $total_pages, $url)
{
$links = "";
if ($total_pages >= 1 && $current_page <= $total_pages) {
$links .= "<a href=\"{$url}?page=1\">1</a>";
$i = max(2, $current_page - 5);
if ($i > 2)
$links .= " ... ";
for (; $i < min($current_page + 6, $total_pages); $i++) {
$links .= "<a href=\"{$url}?page={$i}\">{$i}</a>";
}
if ($i != $total_pages)
$links .= " ... ";
$links .= "<a href=\"{$url}?page={$total_pages}\">{$total_pages}</a>";
}
return $links;
}
輸出:
page = 1
1 2 3 4 5 6 ... 20
page = 10
1 ... 5 6 7 8 9 10 11 12 13 14 15 ... 20
page = 19
1 ... 14 15 16 17 18 19 20
嘗試制作一個頁面括號,例如比實際頁面少 10 和 10,例如更改 for 語句:
For($i = $pg-10 ; $i <= $pg+10 ; $i++)
我想通過當前頁面和總頁面獲取一組數字。 所以這就是我想出的。 我希望這對其他人有幫助-
注意 - 如果總頁數超過 10,則此項工作。我覺得如果總頁數少於 10,則只需將其顯示在單個 for 循環中。
function getSmartPageNumbers($currentPage, $totalPage)
{
$pageNumbers = [];
$diff = 2;
$firstChunk = [1, 2, 3];
$lastChunk = [$totalPage - 2, $totalPage - 1, $totalPage];
if ($currentPage < $totalPage) {
$loopStartAt = $currentPage - $diff;
if ($loopStartAt < 1) {
$loopStartAt = 1;
}
$loopEndAt = $loopStartAt + ($diff * 2);
if ($loopEndAt > $totalPage) {
$loopEndAt = $totalPage;
$loopStartAt = $loopEndAt - ($diff * 2);
}
if (!in_array($loopStartAt, $firstChunk)) {
foreach ($firstChunk as $i) {
$pageNumbers[] = $i;
}
$pageNumbers[] = '.';
}
for ($i = $loopStartAt; $i <= $loopEndAt; $i++) {
$pageNumbers[] = $i;
}
if (!in_array($loopEndAt, $lastChunk)) {
$pageNumbers[] = '.';
foreach ($lastChunk as $i) {
$pageNumbers[] = $i;
}
}
}
return $pageNumbers;
}
測試:
getSmartPageNumbers(8, 20);
Array
(
[0] => 1
[1] => 2
[2] => 3
[3] => .
[4] => 6
[5] => 7
[6] => 8
[7] => 9
[8] => 10
[9] => .
[10] => 18
[11] => 19
[12] => 20
)
getSmartPageNumbers(1, 20);
Array
(
[0] => 1
[1] => 2
[2] => 3
[3] => 4
[4] => 5
[5] => .
[6] => 18
[7] => 19
[8] => 20
)
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.