我正在制作一个页面,在向下滚动的情况下,页面的左侧将向上滑动,页面的右侧将向下滑动。 向上滚动时,页面左侧将向下滑动,页面右侧将向上滑动。
如果用户只滚动一次(使用鼠标滚轮),我有这个工作,当你滚动多次时(在功能完成之前,左右两侧将继续滚动并弄乱。有没有办法禁用多个滚动?
我添加了我的代码和下面的小提琴。
<div class="left">
<div id="left1" class="onLeft Side">
</div>
<div id="left2" class="onLeft Side">
</div>
<div id="left3" class="onLeft Side">
</div>
</div>
<div class="right">
<div id="right3" class="onRight Side">
</div>
<div id="right2" class="onRight Side">
</div>
<div id="right1" class="onRight Side">
</div>
</div>
$(document).ready(function(){
var wholeHeight = jQuery('.right').height();
var rightLength = jQuery('.onRight').length;
jQuery('.right').css({top:-((wholeHeight*rightLength)-wholeHeight)});
var leftHeight = jQuery('.left').height();
var leftLength = jQuery('.onLeft').length;
var tot = (leftHeight * leftLength) - leftHeight;
console.log('tot', tot)
$('body').bind('mousewheel', function(e){
var height = jQuery('.left').height();
var leftTop = jQuery('.left').position().top;
var rightTop = jQuery('.right').position().top;
if(e.originalEvent.wheelDelta /120 > 0) {
if (leftTop != 0) {
console.log('scrolling up !');
jQuery('.left').animate({top:leftTop + height});
jQuery('.right').animate({top:rightTop - height});
} else {
console.log('The up end');
}
} else {
if (leftTop != -tot) {
console.log('scrolling down !');
jQuery('.left').animate({top:leftTop - height});
jQuery('.right').animate({top:rightTop + height});
} else {
console.log('the down end')
}
}
});
});
https://jsfiddle.net/11pftj26/
谢谢
答案 0 :(得分:2)
试试这个:
$(document).ready(function(){
var wholeHeight = jQuery('.right').height();
var rightLength = jQuery('.onRight').length;
jQuery('.right').css({top:-((wholeHeight*rightLength)-wholeHeight)});scrolling=false;
var leftHeight = jQuery('.left').height();
var leftLength = jQuery('.onLeft').length;
var tot = (leftHeight * leftLength) - leftHeight;
console.log('tot', tot)
$('body').bind('mousewheel', function(e){
if(scrolling)
return;
scrolling=true;
var height = jQuery('.left').height();
var leftTop = jQuery('.left').position().top;
var rightTop = jQuery('.right').position().top;
if(e.originalEvent.wheelDelta /120 > 0) {
if (leftTop != 0) {
console.log('scrolling up !');
jQuery('.left').animate({top:leftTop + height},{done:function(){scrolling=false}});
jQuery('.right').animate({top:rightTop - height},{done:function(){scrolling=false}});
} else {
console.log('The up end');
scrolling=false;
}
} else {
if (leftTop != -tot) {
console.log('scrolling down !');
jQuery('.left').animate({top:leftTop - height},{done:function(){scrolling=false}});
jQuery('.right').animate({top:rightTop + height},{done:function(){scrolling=false}});
} else {
console.log('the down end');
scrolling=false;
}
}
});
});
答案 1 :(得分:0)
您可以使用Jquery的.one()
功能,阅读官方指南。
答案 2 :(得分:0)
当您快速滚动两次时,leftTop
和leftRight
的计算将会关闭。您可以做的是从DOM状态中分离顶部计算:
https://jsfiddle.net/11pftj26/2/
$(document).ready(function(){
var wholeHeight = jQuery('.right').height();
var rightLength = jQuery('.onRight').length;
jQuery('.right').css({top:-((wholeHeight*rightLength)-wholeHeight)});
var leftHeight = jQuery('.left').height();
var leftLength = jQuery('.onLeft').length;
var tot = (leftHeight * leftLength) - leftHeight;
var index = 0;
function animate(index) {
jQuery('.left').stop().animate({top: -1 * index * leftHeight});
jQuery('.right').stop().animate({top: -1 * (rightLength - index - 1) * wholeHeight });
}
$('body').bind('mousewheel', function(e) {
if (e.originalEvent.wheelDelta > 0) {
index = Math.max(0, index - 1);
} else {
index = Math.min(index + 1, rightLength - 1);
}
animate(index);
});
});