$.ajax({
url: "test.html",
context: document.body
}).done(function() {
//show data
});
how to save data from ajax request into a variable and then show it? and how to show a message "loading..." while it is proccessing?
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function(data) { ... }); ??MightyPork– MightyPork2013-09-29 18:33:37 +00:00Commented Sep 29, 2013 at 18:33
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possible duplicate of stackoverflow.com/questions/905298/…Cole Tobin– Cole Tobin2013-09-29 18:35:07 +00:00Commented Sep 29, 2013 at 18:35
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3 Answers
$.ajax({
url: "test.html",
context: document.body
}).done(function(data) {
alert(data);
});
UPDATED:
$.ajax({
url: "test.html",
context: document.body
}).done(function(data) {
$('#loading').hide();
// alert(data);
});
markup:
<div id='loading'></div>
1 Comment
M1X
thanks how to show a "loading..." message when a request is send ?
For loading message, use beforeSend() in ajax.
beforeSend : function () {
$('body').html('Loading...');
}
Comments
You can try something like this. In your function you can show loading div
function your_ajax() {
$('#loading').show();
$.ajax({
url: "test.html",
context: document.body
}).done(function(data) {
$('#loading').hide();
alert(data);
});
}
Please add this part in your html
<div id="loading" style="display:none;">Loading </div>
1 Comment
M1X
how to wrap this in a timeout request for example refresh every 5 sec?