JQuery等價於原型代碼

Question

與此原型代碼等效的jQuery會是什麼？

如果你不認爲這是相關的並關閉它，你會介意直接用解決方案給我發消息嗎？

var comment_link = new Element('a', { 
    'title': "View snapshot", 
    'id': snapshot.id + '_comment', 
    'href': "#", 
    'class': "" 
}).update(snapshot.name.truncate(13)); 

comment_link.observe('click', respondToClick); 
comment_link.observe('mouseover', respondToMouseOver); 
comment_link.observe('mouseout', respondToMouseOut); 

我想這無濟於事：

var comment_link = new $j('<a/>', { 
    title: "View snapshot", 
    id: snapshot.id + '_comment', 
    href: "#", 
    class: "" 
}).html(snapshot.name.truncate(13)); 

$j(comment_link).click(respondToClick); 
$j(comment_link).mouseover(respondToMouseOver); 
$j(comment_link).mouseout(respondToMouseOut); 

Answer 1

var comment_link = $j('<a/>') 
    .attr('title', 'View snapshot') 
    .attr('id', snapshot.id + '_comment') 
    .attr('href', '#') 
    .html(snapshot.name.truncate(13)) 
    .click(respondToClick) 
    .mouseover(respondToMouseOver) 
    .mouseout(respondToMouseOut); 

應該這樣做