无法使用JQUERY删除html内容
[英]Unable to remove html content using JQUERY
问题描述
我无法删除ID为"Third" div并在ID为"Fourth" div中显示结果内容。 请注意：-我需要将#Third内容附加到#Fourth ，然后在以后删除。
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title></title>
<script src="../jquery-ui-1.10.4.custom/js/jquery-1.10.2.js" type="text/javascript"></script>
<script src="../jquery-ui-1.10.4.custom/js/jquery-ui-1.10.4.custom.min.js" type="text/javascript"></script>
<link href="../jquery-ui-1.10.4.custom/css/ui-lightness/jquery-ui-1.10.4.custom.min.css"
rel="stylesheet" type="text/css" />
<script type="text/javascript">
$(function() {
var widget = {}
$("#Vshowdialog").click(function() {
widget.doInitialize();
});
widget.doInitialize = function() {
$("#Fourth").append($("#First").html());
$("#Fourth").append($("#Second").html());
$("#Fourth").append($("#Third").html());
widget.showLightBox();
};
widget.showLightBox = function() {
text = $('#Fourth').html();
$(text).find("#Third").remove();
$("#Fourth").dialog();
};
});
</script>
</head>
<body>
<input type="button" id="Vshowdialog" value="ShowDialog" />
<div id="First">
<div id="content">
This is my main Firstcontent.</div>
<div id="sidebar">
This is a FirstSideBar</div>
</div>
<div id="Second">
<div id="secondOne">
This is my main Second content.</div>
<div id="secondTwo">
This is a second sidebar</div>
</div>
<div id="Third">
hello
<div id="thirdOne">
This is my main third content.</div>
<div id="thirdTwo">
This is a third sidebar</div>
</div>
<div id="Fourth">
</div>
</body>
</html>
5 个解决方案
解决方案1
1 2014-06-26 11:32:42
解决方案2
1 2014-06-26 11:44:33
$(function() {
var widget = {}
$("#Vshowdialog").click(function() {
widget.doInitialize();
});
widget.doInitialize = function() {
$("#Fourth").append($("#First").html());
$("#Fourth").append($("#Second").html());
$("#Fourth").append($("#Third").html());
widget.showLightBox();
};
widget.showLightBox = function() {
$('#Fourth').find("#thirdOne").remove();
$('#Fourth').find("#thirdTwo").remove();
$("#Fourth").dialog();
};
});
试试这个，它将正常工作。
解决方案3
0 2014-06-26 11:39:47
您要将其他元素的“内部html”附加到#Fourth ； 为了保持标识符完整，您应该这样做：
$('#First,#Second,#Third').appendTo('#Fourth');
然后，在稍后阶段，您只需删除#Third ：
$('#Third').remove();
解决方案4
0 2014-06-26 11:49:22
您将要初始化此变量
text = $('#Fourth').html();
但是，当您仅包含内部html时，您会在#Fourth中寻找#Third
var text = $('#Fourth').html(); //contents of fourth
$(text).find("#Third").remove(); //third was added as inner Html without the wrapping container.
你应该能够改变
$("#Fourth").append($("#Third").html());
至
$("#Fourth").append($("#Third"));
真正的问题是，为什么要在第三位到第四位添加然后删除它？ 假设代码是固定的，则在执行时，您的代码与此等效。
$("#Fourth").append($("#Third"));
$("#Fourth").find("#Third").remove();
解决方案5
0 2014-06-26 12:04:05
//cache the third div content
var tempThird = $('#Third');
//add your result to Fourth div
$('#Fourth').html("resultcontent");
//remove the third div
$('#Third').remove();
//tempThird will still have the essence of #Third div which you can use anywhere.
您可以在http://api.jquery.com/remove/上查看删除API文档。
希望这能解决您的问题。
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