问题
I want to do an AJAX call via jQuery load() and only once it returns, then fadeOut the old content and fadeIn the new content. I want to old content to remain showing until the new content is retrieved, at which point the fade Out/In is triggered.
Using:
$('#data').fadeOut('slow').load('/url/').fadeIn('slow');
the content fades in and out and a few moments the later the load() call returns, and the data updates, but the fade has already completed.
回答1:
Use callbacks to the control the order of the calls.
var $data = $('#data');
$data.fadeOut('slow', function() {
$data.load('/url/', function() {
$data.fadeIn('slow');
});
});
(Note: I'm not 100% sure about if using var $data = ...
and $data.doStuff()
will actually work - if it does, it saves you from having to look up the div in the DOM tree every time, but if it doesn't, just remove the first line and use $('#data')
everywhere...
回答2:
The problem is related to the fact that all three functions, fadeOut, load and fadeIn are asynchronous. Each of the above functions accept a callback argument (a function) which will run when the function has finished execution. E.g.
$('#data').fadeOut(functionToRunWhenFadeOutIsComplete);
//If you have defined 'functionToRunWhenFadeOutIsComplete' it will run after fadeOut is over.
Armed with this knowledge, you can now solve your problem.
var fadeInData = function fadeInData() { $('#data').fadeIn(); }
var loadData = function loadData() { $('#data').load('url', fadeInData); }
$('#data').fadeOut(loadData);
Alternatively, you can define loadData, fadeInData as an inline anonymous functions.
来源:https://stackoverflow.com/questions/1550642/how-can-i-get-jquery-load-to-complete-before-fadeout-fadein