包括超級鏈接<a> jQuery中。內容

Question

我玩一起在這個片段中，以去除斷點（<br/>），並纏上了我的文字段落：

$('.articleText').contents().filter(function() { 
    return this.nodeType == 3; 
}).wrap('<p></p>').end().filter('br').remove(); 

來源：http://api.jquery.com/contents/

的問題是我的文本包含一個超鏈接（），這是不包括在本段 - 它只是被排除在外...

這是我的textsnippet：

Lorem Ipsum is simply dummy text of the printing and typesetting industry. <br/><br/>Lorem Ipsum has been the industry's standard dummy text ever since the 1500s, when an unknown printer took a <a href="#">galley</a> of type and scrambled it to make a type specimen book.

這是它是如何轉化：

<p>Lorem Ipsum is simply dummy text of the printing and typesetting industry.</p> 
<p>Lorem Ipsum has been the industry's standard dummy text ever since the 1500s, when an unknown printer took a</p> 
<a href="#">galley</a> 
<p>of type and scrambled it to make a type specimen book.</p> 

，這是應該的：

<p>Lorem Ipsum is simply dummy text of the printing and typesetting industry.</p> 
<p>Lorem Ipsum has been the industry's standard dummy text ever since the 1500s, when an unknown printer took a <a href="#">galley</a> of type and scrambled it to make a type specimen book.</p> 

這怎麼可能......？

Answer 1

方法.contents()只返回子女的內容，不是所有的後代。由於您的<a>標籤嵌套在子標籤中，因此.contents()方法不會採用該標籤。

試試這個：

$('.articleText').contents().add($(this).find("*").contents())... 

注：我還沒有測試上面，所以它可能需要一些小的調整。

更新：如果我誤解你，是不應該的<a>標籤被嵌入，那麼你必須在你的標記錯誤：

嘗試改變：

<p>Lorem Ipsum is simply dummy text of the printing and typesetting industry.<p>

到

<p>Lorem Ipsum is simply dummy text of the printing and typesetting industry.</p>

讓我知道你是否對此有任何疑問。祝你好運！ :)