【发布时间】:2018-11-11 12:41:56
【问题描述】:
我有自定义 HTML 分页模板:
<ul class="number-list">
<li><a href="/article/" class="page-number js-page-filter " data-page="1">1</a></li>
// Current page with class [.red]
<li><a href="/article/?page=2" class="page-number js-page-filter red" data-page="2">2</a></li>
<li><a href="/article/?page=3" class="page-number js-page-filter " data-page="3">3</a></li>
<li><a href="/article/?page=4" class="page-number js-page-filter " data-page="4">4</a></li>
<li><a href="/article/?page=5" class="page-number js-page-filter " data-page="5">5</a></li>
<li><span class="page-number">...</span></li>
<li><a href="/article/?page=31" class="page-number js-page-filter " data-page="31">31</a></li>
<li><a href="/article/?page=32" class="page-number js-page-filter " data-page="32">32</a></li>
</ul>
我可以使用此代码获取当前页面网址:
$current = $doc->find('.number-list li a.red')->attr('href');
在我的情况下如何获取下一页网址?
我累了:
$next = $doc->find('.number-list li a.red')->next('a')->attr('href');
$next = $doc->find('.number-list li a.red')->next('li a')->attr('href');
$next = $doc->find('.number-list li a.red')->next('.page-number')->attr('href');
$next = $doc->find('li a.red')->next('li a')->attr('href');
$next = $doc->find('li a.red')->next('li .page-number')->attr('href');
【问题讨论】: