jQuery中的效率问题

I have this HTML structure in a page and an AJAX request returns the same structure:

<div id="feeds">
    <div class="items">
        <div class="feedClass">
            content
        </div>
        <div class="feedClass">
            content
        </div>
        <div class="feedClass">
            content
        </div>
    </div>
    <div class="pager">
        content
    </div>
    <div class="keys">
        content
    </div>
</div>

Contents of data are the same as above.

How can I replace this JavaScript code with a more efficent code to append the stuff in items to the current items, and replace the content in pager and keys. I have this so far, data is what is returned by the AJAX request:

var itemsFirstChildTag = $(id + " .items:first-child").get(0).nodeName;

var id = "#feeds";

$(".items", id).append($(id + " .items > "+itemsFirstChildTag, "<div>"+data+"</div>"));

$(".keys", id).replaceWith($(id + ".keys", "<div>"+data+"</div>"));

$(".pager", id).replaceWith($(id + ".pager", "<div>"+data+"</div>"));

I would try looking into using a better communication method, such as JSON or XML. That alone would make this whole ajax request a LOT more efficient. The idea being you can append the data you are returning to the appropriate location, you do not need to be returning any HTML.

$.ajax({
  type: 'post',
  url: 'some-script.php',
  success: function( response ){
      //neat jquery element creation
      var newFeed = $("div", {
         "class" : "feedClass",
         "text" : response.newText
      });

      $(".items").append(newFeed);
      $(".keys").text(response.newText);
  }
});

If you need to modify the text of an element you can use .text().

Get:

$(".some-class").text();

Set:

$(".some-class").text("Some Text");

If you need to modify the HTML of an element, use .html()

GET/SET same as $.text()

In short I'd definitely recommend reading as much of the jQuery documentation as you possibly can, there are a lot of neat tricks/demo's there.

var feeds = document.getElementById("feeds"); var feeddivs = feeds.getElementsByTagName("div"); var len = feeddivs.length; for(var i=0;i

}