您好我正在使用Ajax和Php来显示分页
在我的代码中第一个和上一个链接正在运行,但下一个和上一个链接无法正常工作我的意思此下一步和上一次链接将转到最后一页。当我按上一个链接时,应该转到最后一页但是如果按下一步链接,它也会转到最后一页。 下一步链接应转到下一页。你能告诉我为什么吗?
这是我的PHP代码:
function paginate_function($item_per_page, $current_page, $total_records, $total_pages)
{
$pagination = '';
if($total_pages > 0 && $total_pages != 1 && $current_page <= $total_pages){ //verify total pages and current page number
$pagination .= '<ul class="pagination">';
$right_links = $current_page + 15;
$previous = $current_page - 1; //previous link
$next = $current_page + 1; //next link
$first_link = true; //boolean var to decide our first link
if($current_page > 1){
$previous_link = ($previous==0)?1:$previous;
$pagination .= '<li class="first"><a href="#" data-page="1" title="First">first</a></li>'; //first link
$pagination .= '<li><a href="#" data-page="'.$previous_link.'" title="Previous">prev</a></li>'; //previous link
for($i = ($current_page-2); $i < $current_page; $i++){ //Create left-hand side links
if($i > 0){
$pagination .= '<li><a href="#" data-page="'.$i.'" title="Page'.$i.'">'.$i.'</a></li>';
}
}
$first_link = false; //set first link to false
}
if($first_link){ //if current active page is first link
$pagination .= '<li class="first active">'.$current_page.'</li>';
}elseif($current_page == $total_pages){ //if it's the last active link
$pagination .= '<li class="last active">'.$current_page.'</li>';
}else{ //regular current link
$pagination .= '<li class="active">'.$current_page.'</li>';
}
for($i = $current_page+1; $i < $right_links ; $i++){ //create right-hand side links
if($i<=$total_pages){
$pagination .= '<li><a href="#" data-page="'.$i.'" title="Page '.$i.'">'.$i.'</a></li>';
}
}
if($current_page < $total_pages){
$next_link = ($i > $total_pages)? $total_pages : $i;
$pagination .= '<li><a href="#" data-page="'.$next_link.'" title="Next">Next</a></li>'; //next link
$pagination .= '<li class="last"><a href="#" data-page="'.$total_pages.'" title="Last">Last</a></li>'; //last link
}
$pagination .= '</ul>';
}
return $pagination; //return pagination links
}
非常感谢您的帮助:)
答案 0 :(得分:0)
哇。我刚刚解决了我的问题。它应该是变量$ next而不是Next Link上的$ next_link。谢谢。