通过正则表达式查找特定的url <a>

Question

我需要一个特定的正则表达式模式来查找HTML网页中的URL

例如，我想搜索这个url：domainurl.com

这些是带有标签的URL

<a href="https://www.domainurl.com/refer/google-adsense/">fsdf</a>
<a title="Google Adsense" href="https://www.domainurl.com/refer/google-adsense/" target="_blank" rel="nofollow noopener">fgddf</a>
<a href="https://www.domainurl.com/page/pago">domain </a>

使用下面的代码regex

<a.*?[^>]* href="((https?:\/\/)?([\w\-])+\.{1}domainurl\.([a-z]{2,6})([\/\w\.-]*)*\/?)"

想要得到这个标签，我想目标应该是"_ blank" rel = "nofollow noopener"

<a title="Google Adsense" href="https://www.domainurl.com/refer/google-adsense/" target="_blank" rel="nofollow noopener">fgddf</a>

是否有任何正则表达式代码的目标= "_空白“和rel = "nofollow”?？

这就是我的https://regexr.com/49hne

Answer 1

对于使用positive lookbehind的完整URL

(?<=\<a.*?href=\")(.*?\..*?\.[a-z]+)

DEMO

仅限使用positive lookbehind的domainurl.com

(?<=\<a.*?www\.)([a-z]+\.[a-z]+)

DEMO2

对于target = "_ blank" and rel = "nofollow noopener"：

DEMO3

target.*?\".*\"

对于domainurl.com和target = "_ blank" and rel = "nofollow noopener"：

DEMO4