无刷新更新页面上的数据 [英] Update data on a page without refreshing
问题描述
我有一个网站,我需要更新状态。 就像一个航班,你离开,巡航或降落。 我希望能够刷新状态,而无需我vieuwers拥有并重新加载整个页面。我知道有与AJAX和jQuery的一种方式,但我没有对如何工作的任何理解。我allso不想让他们拥有并点击一个按钮。 如果有任何身体知道如何将做我多少AP preciate了!
I have a website where I need to update a status. Like for a flight, you are departing, cruise or landed. I want to be able to refresh the status without having my vieuwers to have and reload the whole page. I know there is a way with AJAX and jQuery, but I don't have any understanding of how that works. I allso don't want them to have and click a button. If any body knows how that would be done I much appreciate it!
推荐答案
在一般情况下,如果你不知道如何运作的东西,寻找其中可以借鉴的例子。
In general, if you don't know how something works, look for an example which you can learn from.
对于这个问题,可以考虑<大骨节病> 这个DEMO 骨节病>
For this problem, consider this DEMO
您可以看到加载内容Ajax是很容易做到与jQuery:
You can see loading content with AJAX is very easily accomplished with jQuery:
$(function(){
// don't cache ajax or content won't be fresh
$.ajaxSetup ({
cache: false
});
var ajax_load = "<img src='http://automobiles.honda.com/images/current-offers/small-loading.gif' alt='loading...' />";
// load() functions
var loadUrl = "http://fiddle.jshell.net/deborah/pkmvD/show/";
$("#loadbasic").click(function(){
$("#result").html(ajax_load).load(loadUrl);
});
// end
});
试着去了解它是如何工作,然后尝试复制它。祝你好运。
Try to understand how this works and then try replicating it. Good luck.
您可以找到相应的教程<大骨节病> 这里
You can find the corresponding tutorial HERE
现在下面的事件开始了ajax 负荷
功能:
Right now the following event starts the ajax load
function:
$("#loadbasic").click(function(){
$("#result").html(ajax_load).load(loadUrl);
});
您还可以定期做:如何激发AJAX请求定期
(function worker() {
$.ajax({
url: 'ajax/test.html',
success: function(data) {
$('.result').html(data);
},
complete: function() {
// Schedule the next request when the current one's complete
setTimeout(worker, 5000);
}
});
})();
我做了这个实现您<大骨节病> 这里 <的演示/骨节病>。在本演示中,每2秒(的setTimeout(工人,2000年);
)的内容更新
您也可以只立即加载数据:
You can also just load the data immediately:
$("#result").html(ajax_load).load(loadUrl);
其中有<大骨节病> 本 骨节病>相应的演示。
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