Query HTML returned from AJAX_问答_开发者_运维开发者技术经验分享

开发者 https://www.devze.com 2023-02-18 09:02 出处：网络

How can I quer开发者_开发知识库y HTML thats returned from AJAX? I tried $.post(\"gethtml.php\", { url: $url.val() }, function(data) {

相关专题：jquery

How can I quer开发者_开发知识库y HTML thats returned from AJAX?

I tried

$.post("gethtml.php", { url: $url.val() }, function(data) {
    var $html = $(data),
            $links = $("link, script, style", $html),
            $body = $("body", $html);

    $links.each(function() { $(this).addClass("tmp"); });
    console.log($html);
    console.log($body.html());
});

$html looks like [meta, title, script, style#gstyle, script, textarea#csi, div#mngb, iframe, center, script, script] and $body.html() null

UPDATE

A simple setup at http://jsfiddle.net/uvnrJ/1/

$(function() {
  var html = "<!doctype html><html><head><title>title here ... </title></head><body>This is the body</body></html>",
      $html = $(html);
  console.log($html); // jQuery(title, <TextNode textContent="This is the body">)
  console.log($html.find("body")); // jQuery()
  console.log($html.find("title")); // jQuery()
  console.log($html.filter("title")); // jQuery(title)
});

Shows that jQuery seem to have problems parsing the HTML string?

Try this...

..........
var elements = $('<div></div>');
elements.html(data);
alert(elements.find('body').html());
...........

I think you should search for your elements directly in the data variable or use the find() method on $(data). Something like:

$.post("gethtml.php", { url: $url.val() }, function(data) {
    var $html = $(data).find('html'), // First way
        $links = $("link, script, style", data), // Second way
        $body = $("body", data); // Could be better written as $(data).find('body')

    $links.each(function() { $(this).addClass("tmp"); });
    console.log($html);
    console.log($body.html());
});

I ran into the same issue and was forced to resort to parsing the HTML string directly:

    var headStartIndex = htmlString.toLowerCase().indexOf("<head>") + "<head>".length;
    var headEndIndex = htmlString.toLowerCase().indexOf("</head>");
    var newHead = htmlString.substring(headStartIndex, headEndIndex);

    var bodyStartIndex = htmlString.toLowerCase().indexOf("<body>") + "<body>".length;
    var bodyEndIndex = htmlString.toLowerCase().indexOf("</body>");
    var newBody = htmlString.substring(bodyStartIndex, bodyEndIndex);

    console.log(newHead);
    console.log(newBody);

you can try this, This works

$.ajax({
    'url': url,
    'dataType': 'html',
    'success': function(e){
        var $div = $("<div id='_some_id'>" + e + "</div>");
        $div = $div.filter("_some_id");
        $div.find("span"); // you can search for anything here. it would work now
    }
});