jQuery完全用另一个DOM替换元素的DOM - 更快的方法？

Question

I'm using jQuery's AJAX for getting new content from server. Data is loaded in JSON:

$.ajax({
    url: url,
    data: {
        'ajax': '1',
    },
    dataType: 'json',
    success: somefunction
});

For server-side application limitation, I'm not able to setup more JSON variables inside so I have to load everything into content. That is why I have to load result into jQuery, than search and replace some elements on page, like this (used in somefunction ):

var somefunction = function(data) {
    var con = $('<div></div>').html(data.content); // just $(data.content) is not working
    $('div#mainContent').html(con.find('div#ajax-content').html());
    ... // same process with three more divs
}

EDIT: Please, note that I have to do same process to replace three divs!

There is more about that, but as example, it's enough I hope. My question: For some logic way, I expect that loading result into DOM ( $(data.content) ), parsing to html ( con.find('dix#ajax-content').html() ) and back to DOM ( $('div#mainContent').html() ) seems to me like loosing some resources and decreasing the perfomance so I would like to know if there is any faster way to do it and load DOM directly, like:

$('div#mainContent').dom(con.find('div#ajax-content').dom());

I tried to google it but maybe I don't know what to type in. Also jQuery documentation does not helped me a lot.

Some facts:

• jQuery 1.9.1
• jQuery UI 1.10.3 available

Finally, I know that it would be much more better to do something with server-side app to provide more JSON variables, however, I need to write not-so-easy peace of code which is requiring longer time to develop which I don't have right now. Doing it on client side would be temporary solution for now, however, I don't want to decrease performace a lot.

Side-question:

is it correct to use find() function in this case or there is any better one?

EDIT 2 (not working parsing string) I'm expecting this working but it's not:

content = '<div id="ajax-title">Pečivo běžné, sladké, slané</div>
<div id="ajax-whereami"><a href="/category/4">Chléba a pečivo</a> » Pečivo běžné, sladké, slané</div>';
$(content);

Answer 1

I'm not sure it this will help:

$('div#mainContent').replaceWith(con.find('div#ajax-content'))
 .attr("id","mainContent")

Now you don't have to set the html of the element and get the html of the Element you just created from JSON. Not sure if this actually is faster but it does skip the 2 html() steps.

Answer 2

Actually, $(data.content) should work just fine, but you have to keep in mind that the top level elements can only be reached via .filter() instead of .find() . If the elements you wish to target are at least one level deeper than the root you should use .find() though; in the examples below you can replace .filter() with .find() where appropriate.

var $con = $(data.content);
$('div#mainContent')
  .empty()
  .append($con.filter('div#ajax-content'))
  .append($con.filter('div#another-id'))
  .append($con.filter('div#and-another-id'));

You can also combine the selectors together:

  .append($con.filter('div#ajax-content, div#another-id, div#and-another-id'));

Lastly, since identifiers should only appear once inside a document, you can drop the div part:

  .append($con.filter('#ajax-content, #another-id, #and-another-id'));

Update

Okay, it seems that jQuery doesn't evaluate data.content properly when there are newlines in the wrong places; this should work in all cases:

var wrapper = document.createElement('div');
wrapper.innerHTML = data.content;

var $con = $(wrapper);

Answer 3

You could use .load , although I believe it is essentially just a wrapper for the same thing:

$("#mainContent").load(url + " #ajax-content", data);

You can increase performance on the server side by sending only the specific ajax content (less to download and parse) although that may be difficult.

Other than that, your best bet is to use vanilla JS over jQuery, at least for the appending (using innerHTML directly).