重新加载内容后如何使JQuery工作？

Question

I have a block(div) where i use to put information. I have a hidden form there? and it shows on click. But If I reload data in block(div) it works like link .

This is my script:

var comment_block ;
var show_comment_block;
var show;
$(function(){
    $('A.reply').on('click',ShowCommentBlock);
});

function ShowCommentBlock(){
    $(this).parent().next('.answer-form').slideToggle('slow');
    return false;
}

And it is the code on my button:

<a class="reply" href="#comment<?php echo $comment['comment_id']?>">Ответить</a>

As i understand after content reload in block(div) i have to put this actions on new forms? I dont understand how to do that? or maybe there is some other kind of solution. Thanks.

Answer 1

You have to use delegation, eg:

$(document.body).on('click','A.reply', ShowCommentBlock);

A. Wolff

Thank U!

Answer 2

也许将其更改为：

 $('A.reply').live("click", ShowCommentBlock);