我想做这样的事情https://muzzleapp.com/。我希望右侧的移动项目可以向上滑动并从下到上逐渐消失。项目应连续移动,因此在最后一个div之后,它将重新开始。我已经尝试过自己的代码。但是从起点开始就卡住了。它从一开始就无法正常启动。
html
function moveItems(el) {
var x = 1;
var flag = 0;
var elems = $(el).nextAll();
count = elems.length;
elems.each (function (i) {
setTimeout(function() {
if (x>1) {
y = x-1;
$('#slider_'+y).show().delay(2000).slideUp().fadeOut();
}
$('#slider_'+x).show().delay(2000).slideUp();
x++;
if (!--count) {
setTimeout(function() {
moveItems('.panel');
}, 6000)
}
}, i*2000);
})
}
moveItems('.panel');
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="panel" id='slider_1'>
<div class="panel-body">
<div class="row">
slider 1
</div>
</div>
</div>
<div class="panel" id='slider_2'>
<div class="panel-body">
<div class="row">
slider 2
</div>
</div>
</div>
<div class="panel" id='slider_3'>
<div class="panel-body">
<div class="row">
slider 3
</div>
</div>
</div>
答案 0 :(得分:1)
我稍微修改了您的逻辑以处理隐藏和显示。这是更新的预览https://jsfiddle.net/Aravi/hd465gpc/13/
function moveItems(el) {
var elems = document.querySelectorAll(el);
Array.from(elems).forEach((item,index) => {
setTimeout(function() {
index+=1;
$('#slider_'+index).show().delay(2000).slideUp().fadeOut();
if (index == elems.length) {
setTimeout(function() {
for(j=1;j<= elems.length;j++){ $('#slider_'+j).show()}
moveItems('.panel');
}, 6000)
}
}, index*2000);
})
}
moveItems('.panel');
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="panel" id='slider_1'>
<div class="panel-body">
<div class="row">
slider 1
</div>
</div>
</div>
<div class="panel" id='slider_2'>
<div class="panel-body">
<div class="row">
slider 2
</div>
</div>
</div>
<div class="panel" id='slider_3'>
<div class="panel-body">
<div class="row">
slider 3
</div>
</div>
</div>
<div class="panel" id='slider_4'>
<div class="panel-body">
<div class="row">
slider 4
</div>
</div>
</div>
<div class="panel" id='slider_5'>
<div class="panel-body">
<div class="row">
slider 5
</div>
</div>
</div>