use*_*991 19 javascript css jquery html5
当我在滚动之外时,我试图激活当前滚动,特别是在 #DivDet
这是我试过的:
$("div#DivDet").scroll(function () {
// I don't know what i should have here
// something like $("div#scrlDiv").scroll();
});
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T.J*_*der 35
听起来你想div通过滚动另一个来回应滚动.
您已经确定了如何挂钩scroll事件.要设置元素的滚动位置(另一个div),请设置元素scrollTop和scrollLeft值(以像素为单位).例如,如果你想要两个div以近似的方式滚动,你可以将源div scrollTop和scrollLeft目标分配给目标div.
相关的JavaScript:
(function() {
var target = $("#target");
$("#source").scroll(function() {
target.prop("scrollTop", this.scrollTop)
.prop("scrollLeft", this.scrollLeft);
});
})();
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或者(来源):
(function() {
var target = $("#target")[0]; // <== Getting raw element
$("#source").scroll(function() {
target.scrollTop = this.scrollTop;
target.scrollLeft = this.scrollLeft;
});
})();
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完整页面:
<!DOCTYPE html>
<html>
<head>
<script src="http://code.jquery.com/jquery-1.9.1.min.js"></script>
<meta charset=utf-8 />
<title>Scroll Example</title>
<style>
.scroll-example {
display: inline-block;
width: 40%;
border: 1px solid black;
margin-right: 20px;
height: 100px;
overflow: scroll;
}
</style>
</head>
<body>
<p>Scroll the left div, watch the right one.</p>
<div id="source" class="scroll-example">
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</div>
<div id="target" class="scroll-example">
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</div>
<script>
(function() {
var target = $("#target");
$("#source").scroll(function() {
target.prop("scrollTop", this.scrollTop)
.prop("scrollLeft", this.scrollLeft);
});
})();
</script>
</body>
</html>
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