Fra*_*ckl 9 javascript jquery scroll infinite-scroll
我有一个类似于的代码:
<div id='right-column'>
<div id='results'>
<div id='result1>
<div class='main'></div>
<div class='details'></div>
</div>
<!-- ... -->
<div id='result50>
<div class='main'></div>
<div class='details'></div>
</div>
</div>
</div>
Run Code Online (Sandbox Code Playgroud)
div.main当用户点击a时,总是可见(固定高度)和div.details"展开/折叠" .div.mainresult div如果#results scrollHeight大于#right-column height,我想创建一个连续的滚动循环.
在这种情况下,滚动过去#result50会显示#result1,滚动之前#result1会显示#result50.
我不能.append()将第一个孩子放在底部,因为在某些情况下,result可以在列的顶部和底部看到一部分a .除非我检测到是否展开/折叠,否则我
无法复制a .当用户展开div 时,a
的高度会发生变化这一事实使得它变得更加复杂......result.detailsresult.details
以下是连续滚动循环(2列)的示例:
$(document).ready(function() {
var num_children = $('#up-left').children().length;
var child_height = $('#up-left').height() / num_children;
var half_way = num_children * child_height / 2;
$(window).scrollTop(half_way);
function crisscross() {
$('#up-left').css('bottom', '-' + window.scrollY + 'px');
$('#down-right').css('bottom', '-' + window.scrollY + 'px');
var firstLeft = $('#up-left').children().first();
var lastLeft = $('#up-left').children().last();
var lastRight = $('#down-right').children().last();
var firstRight = $('#down-right').children().first();
if (window.scrollY > half_way ) {
$(window).scrollTop(half_way - child_height);
lastRight.appendTo('#up-left');
firstLeft.prependTo('#down-right');
} else if (window.scrollY < half_way - child_height) {
$(window).scrollTop(half_way);
lastLeft.appendTo('#down-right');
firstRight.prependTo('#up-left');
}
}
$(window).scroll(crisscross);
});Run Code Online (Sandbox Code Playgroud)
div#content {
width: 100%;
height: 100%;
position: absolute;
top:0;
right:0;
bottom:0;
left:0;
}
#box {
position: relative;
vertical-align:top;
width: 100%;
height: 200px;
margin: 0;
padding: 0;
}
#up-left {
position:absolute;
z-index:4px;
left: 0;
top: 0px;
width: 50%;
margin: 0;
padding: 0;
}
#down-right {
position:fixed;
bottom: 0px;
z-index: 5px;
left: 50%;
width: 50%;
margin: 0;
padding: 0;
}
h1 {margin: 0;padding: 0;color:#fff}
.black {background: black;}
.white {background: grey;}
.green {background: green;}
.brown {background: brown;}Run Code Online (Sandbox Code Playgroud)
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<script src="http://cdnjs.cloudflare.com/ajax/libs/jquery/1.9.0/jquery.min.js"></script>
<div id="content">
<div id="up-left">
<div id="box" class="brown">
<h1>ONE</h1>
</div>
<div id="box" class="black">
<h1>TWO</h1>
</div>
<div id="box" class="white">
<h1>THREE</h1>
</div>
<div id="box" class="black">
<h1>FOUR</h1>
</div>
<div id="box" class="white">
<h1>FIVE</h1>
</div>
<div id="box" class="black">
<h1>SIX</h1>
</div>
</div><!-- #up-left -->
<div id="down-right">
<div id="box" class="white">
<h1>SIX</h1>
</div>
<div id="box" class="black">
<h1>FIVE</h1>
</div>
<div id="box" class="white">
<h1>FOUR</h1>
</div>
<div id="box" class="black">
<h1>THREE</h1>
</div>
<div id="box" class="white">
<h1>TWO</h1>
</div>
<div id="box" class="green">
<h1>ONE</h1>
</div>
</div><!-- #down-right -->
</div><!-- .content -->Run Code Online (Sandbox Code Playgroud)
关于如何做到这一点的任何提示/想法?
您可以使用 jQuery.append()来移动.prepend()项目,而无需克隆它们。
您将使用与延迟加载 (AJAX) 无限滚动类似的技术,但在这种情况下,您想要处理向上和向下滚动,并且不是从服务器加载新内容,而是只是回收现有的 DOM 元素列表。
下面我演示一种技术。我将滚动位置存储在元素的.data缓存中,以便在检测滚动方向时轻松检索。我选择检测滚动方向以避免预先进行不必要的变量分配以提高性能。否则,您将获取元素并对不会在该方向发生的滚动事件进行数学计算。
滚动处理程序:
$('#right-column').on('scroll', function (e) {
var $this = $(this),
$results = $("#results"),
scrollPosition = $this.scrollTop();
if (scrollPosition > ($this.data('scroll-position') || 0)) {
// Scrolling down
var threshold = $results.height() - $this.height() - $('.result:last-child').height();
if (scrollPosition > threshold) {
var $firstResult = $('.result:first-child');
$results.append($firstResult);
scrollPosition -= $firstResult.height();
$this.scrollTop(scrollPosition);
}
} else {
// Scrolling up
var threshold = $('.result:first-child').height();
if (scrollPosition < threshold) {
var $lastResult = $('.result:last-child');
$results.prepend($lastResult);
scrollPosition += $lastResult.height();
$this.scrollTop(scrollPosition);
}
}
$this.data('scroll-position', scrollPosition)
});
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一个完整的工作示例:
$('#right-column').on('scroll', function (e) {
var $this = $(this),
$results = $("#results"),
scrollPosition = $this.scrollTop();
if (scrollPosition > ($this.data('scroll-position') || 0)) {
// Scrolling down
var threshold = $results.height() - $this.height() - $('.result:last-child').height();
if (scrollPosition > threshold) {
var $firstResult = $('.result:first-child');
$results.append($firstResult);
scrollPosition -= $firstResult.height();
$this.scrollTop(scrollPosition);
}
} else {
// Scrolling up
var threshold = $('.result:first-child').height();
if (scrollPosition < threshold) {
var $lastResult = $('.result:last-child');
$results.prepend($lastResult);
scrollPosition += $lastResult.height();
$this.scrollTop(scrollPosition);
}
}
$this.data('scroll-position', scrollPosition)
});
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$('#right-column').on('scroll', function (e) {
var $this = $(this),
$results = $("#results"),
scrollPosition = $this.scrollTop();
if (scrollPosition > ($this.data('scroll-position') || 0)) {
// Scrolling down
var threshold = $results.height() - $this.height() - $('.result:last-child').height();
if (scrollPosition > threshold) {
var $firstResult = $('.result:first-child');
$results.append($firstResult);
scrollPosition -= $firstResult.height();
$this.scrollTop(scrollPosition);
}
} else {
// Scrolling up
var threshold = $('.result:first-child').height();
if (scrollPosition < threshold) {
var $lastResult = $('.result:last-child');
$results.prepend($lastResult);
scrollPosition += $lastResult.height();
$this.scrollTop(scrollPosition);
}
}
$this.data('scroll-position', scrollPosition)
});
$('#results').on('click', '.result', function (e) {
$(this).find('.details').toggle();
});
$('#newNumber').on('input', function (e) {
var results = '';
for (var n = 1; n <= $(this).val(); n++) {
results +=
'<div class="result" id="result' + n + '">' +
' <div class="main">Result ' + n + '</div>' +
' <div class="details">Details for result ' + n + '</div>' +
'</div>';
}
$('#results').html(results);
});Run Code Online (Sandbox Code Playgroud)
body {
font-family: sans-serif;
}
h1 {
font: bold 2rem/1 Georgia, serif;
}
p {
line-height: 1.5;
margin-bottom: 1em;
}
label {
font-weight: bold;
margin-bottom: 1em;
}
.column {
box-sizing: border-box;
float: left;
width: 50%;
height: 100vh;
padding: 1em;
overflow: auto;
}
#right-column {
background-color: LemonChiffon;
}
.result {
padding: 1em;
cursor: pointer;
}
.result .main {
height: 2em;
font-weight: bold;
line-height: 2;
}
.result .details {
display: none;
}Run Code Online (Sandbox Code Playgroud)
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