我有一个简单的淡入淡出,我想无限期地进出。我发现插件可以做到这一点但是好奇,如果jquery已经有一个loop()api所以我可以在脚本中处理它。
<script type="text/javascript">
$(document).ready(function(){
$('.bottom-left').delay(1000).fadeIn(900);
$('.bottom-right').delay(3000).fadeIn(700);
});
</script>
答案 0 :(得分:7)
如果你想变得复杂,那么这可能变成很多代码,但简单的实现只会是几行。基本上你想要递归地调用一个隐藏或显示动画函数的回调函数中的元素的函数:
$(function () {
//declare a function that can fade in/out any element with a specified delay and duration
function run_animation($element, delay, duration) {
//animate fade in/out after delay
$element.delay(delay).fadeToggle(duration, function () {
//after fade in/out is done, recursively call this function again with the same information
//(if faded-out this time then next-time it will be faded-in)
run_animation($element, delay, duration);
});
}
//initialize the animations for each element, specifying a delay and duration as well
run_animation($('.bottom-left'), 1000, 900);
run_animation($('.bottom-right'), 3000, 700);
});
以下是演示:http://jsfiddle.net/xpw4D/
.fadeToggle()
的文档:http://api.jquery.com/fadeToggle
你可以稍微加强一下这个代码,然后动画两个以上的步骤:
$(function () {
function run_animation(options) {
//initialize the count variable if this is the first time running and reset it to zero if there are no more steps
if (typeof options.count == 'undefined' || options.count >= options.steps.length) {
options.count = 0;
}
options.element.delay(options.steps[options.count].delay).fadeToggle(options.steps[options.count].duration, function () {
options.count++;
run_animation(options);
});
}
run_animation({
element : $('.bottom-left'),
steps : [
{ delay : 1000, duration : 100 },
{ delay : 500, duration : 900 },
{ delay : 3000, duration : 500 }
]
});
run_animation({
element : $('.bottom-right'),
steps : [
{ delay : 2000, duration : 200 },
{ delay : 1000, duration : 1800 },
{ delay : 6000, duration : 1000 }
]
});
});