如何從jQuery段落中刪除不需要的元素

Question

我正在解析網頁中的html內容。

我需要這段轉換：

<p>Bryan Cranston was a guest on the third episode of <a href="/wiki/ALF%27s_Hit_Talk_Show" title="ALF's Hit Talk Show">ALF's Hit Talk Show</a>. 
Bryan Cranston is an American actor best known for his roles as Walter White in <i>Breaking Bad</i>, for which he won three consecutive Outstanding Lead Actor in a Drama Series Emmy Awards; and as Hal, the father in <i>Malcolm in the Middle</i>.</p> 

到這一點：

<p>Bryan Cranston was a guest on the third episode of ALF's Hit Talk Show. 
Bryan Cranston is an American actor best known for his roles as Walter White in <i>Breaking Bad</i>, for which he won three consecutive Outstanding Lead Actor in a Drama Series Emmy Awards; and as Hal, the father in <i>Malcolm in the Middle</i>.</p> 

正如你所看到的，我只是保持我的元素，並刪除所有其他元素。

我被卡住，這是我到目前爲止有：提前 http://jsfiddle.net/JyxL4/1/

謝謝！

Answer 1

我不知道你爲什麼使用每一個，但我認爲是必要的，那麼你可以使用方法.unwrap()刪除鏈接標籤，但保持內部的HTML。
試試這個：

$('p').each(function(){ 
    $(this).find('a').contents().unwrap(); 
}); 

或更容易

$('p a').contents().unwrap(); 

DEMO

Answer 2

$('p *').not('i').contents().unwrap(); 

或

$('p *').not('i').each(function(){ 
    $(this).replaceWith(this.childNodes); 
}); 

JSFIDDLE