【发布时间】：2019-12-07 16:56:26
【问题描述】：
问一个简单的问题。这是在我的一个网站中使用的分页代码。我只是想让这更简单。删除所有页码，只保留“下一个”和“上一个”链接。
谁能指出我需要删除脚本的哪一部分以摆脱页码。非常感谢您的帮助。
<?php
error_reporting(E_ALL ^ E_NOTICE);
// How many adjacent pages should be shown on each side?
$adjacents = 5;
$query = $mysqli->query("select COUNT(*) as num from posts order by id desc");
$total_pages = mysqli_fetch_array($query);
$total_pages = $total_pages['num'];
$limit = 12; //how many items to show per page
$page = $_GET['page'];
if($page)
$start = ($page - 1) * $limit; //first item to display on this page
else
$start = 0; //if no page var is given, set start to 0
/* Get data. */
$result = $mysqli->query("select * from posts order by id desc LIMIT $start, $limit");
/* Setup page vars for display. */
if ($page == 0) $page = 1; //if no page var is given, default to 1.
$prev = $page - 1; //previous page is page - 1
$next = $page + 1; //next page is page + 1
$lastpage = ceil($total_pages/$limit); //lastpage is = total pages / items per page, rounded up.
$lpm1 = $lastpage - 1; //last page minus 1
$pagination = "";
if($lastpage > 1)
{
$pagination .= "<div class=\"pagination\">";
//previous button
if ($page > 1)
$pagination.= "<a href=\"videos-$prev.html\">« previous</a>";
else
$pagination.= "<span class=\"disabled\">« previous</span>";
//pages
if ($lastpage < 7 + ($adjacents * 2)) //not enough pages to bother breaking it up
{
for ($counter = 1; $counter <= $lastpage; $counter++)
{
if ($counter == $page)
$pagination.= "<span class=\"current\">$counter</span>";
else
$pagination.= "<a href=\"videos-$counter.html\">$counter</a>";
}
}
elseif($lastpage > 5 + ($adjacents * 2)) //enough pages to hide some
{
//close to beginning; only hide later pages
if($page < 1 + ($adjacents * 2))
{
for ($counter = 1; $counter < 4 + ($adjacents * 2); $counter++)
{
if ($counter == $page)
$pagination.= "<span class=\"current\">$counter</span>";
else
$pagination.= "<a href=\"videos-$counter.html\">$counter</a>";
}
$pagination.= "...";
$pagination.= "<a href=\"videos-$lpm1.html\">$lpm1</a>";
$pagination.= "<a href=\"videos-$lastpage.html\">$lastpage</a>";
}
//in middle; hide some front and some back
elseif($lastpage - ($adjacents * 2) > $page && $page > ($adjacents * 2))
{
$pagination.= "<a href=\"videos-1.html\">1</a>";
$pagination.= "<a href=\"videos-2.html\">2</a>";
$pagination.= "...";
for ($counter = $page - $adjacents; $counter <= $page + $adjacents; $counter++)
{
if ($counter == $page)
$pagination.= "<span class=\"current\">$counter</span>";
else
$pagination.= "<a href=\"videos-$counter.html\">$counter</a>";
}
$pagination.= "...";
$pagination.= "<a href=\"videos-$lpm1.html\">$lpm1</a>";
$pagination.= "<a href=\"videos-$lastpage.html\">$lastpage</a>";
}
//close to end; only hide early pages
else
{
$pagination.= "<a href=\"videos-1.html\">1</a>";
$pagination.= "<a href=\"videos-2.html\">2</a>";
$pagination.= "...";
for ($counter = $lastpage - (2 + ($adjacents * 2)); $counter <= $lastpage; $counter++)
{
if ($counter == $page)
$pagination.= "<span class=\"current\">$counter</span>";
else
$pagination.= "<a href=\"videos-$counter.html\">$counter</a>";
}
}
}
//next button
if ($page < $counter - 1)
$pagination.= "<a href=\"videos-$next.html\">next »</a>";
else
$pagination.= "<span class=\"disabled\">next »</span>";
$pagination.= "</div>\n";
}
$q = $mysqli->query("select * from posts order by id desc limit $start,$limit");
while($row=mysqli_fetch_assoc($q)){
// LOOP Code
} echo $pagination;?>
【问题讨论】：
标签： php html mysqli pagination