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我試圖創建一個視差效果,即絕對定位的子元素應的速度移動速度慢於它的父上滾動子抵消母公司。視差效果 - 計算上滾動
孩子永遠是父母的130%的高度,但家長可以是任何高度:
HTML:
<div class="parallax-window lg">
<div class="parallax-image image-1"></div>
<div class="parallax-content">Hello World</div>
</div>
<div class="parallax-window">
<div class="parallax-image image-2"></div>
<div class="parallax-content">Hello World</div>
</div>
CSS:
.parallax-window {
min-height: 300px;
position: relative;
overflow: hidden;
}
.parallax-window.lg {
min-height: 600px;
}
.parallax-image {
position: absolute;
top: 0;
left: 0;
width: 100%;
height: 130%;
background-size: cover;
background-repeat: no-repeat;
background-position: 50% 50%;
transform: translate3d(0, 0, 0);
z-index: -1;
}
.image-1 {
background-image: url(https://i.ytimg.com/vi/TbC-vUPMR7k/maxresdefault.jpg);
}
.image-2 {
background-image: url(https://i.ytimg.com/vi/xi5-YrAEChc/maxresdefault.jpg);
}
我有一個公式來移動圖像,但我的數學是明確的路要走:
var win = $(window),
win_h = win.height(),
parallaxers = $('.parallax-window');
function scroll_events() {
var win_top = win.scrollTop(),
win_btm = win_top + win_h;
parallaxers.each(function() {
var cont = $(this),
cont_top = cont.offset().top,
cont_h = cont.height(),
cont_btm = cont_top + cont_h,
para = cont.find('.parallax-image'),
para_h = Math.round(cont_h * 1.3);
if (cont_btm > win_top && cont_top <= win_btm) {
var diff = (win_h - cont_h)/(win_h - para_h),
value = -Math.round((win_top * diff));
// para.css('transform', 'translate3d(0,' + value*-1 + 'px, 0)');
para.css('top', value + 'px');
}
});
}
的圖像移動,但不是在正確率。
圖像應是與母體的頂部線的時候,元件首先進入視區。然後在滾動後,當圖像到達視口頂部時,圖像的底部應該與父視圖的底部一致。
任何幫助將大規模感激!
FIDDLE(https://jsfiddle.net/8dwLwgy7/1/)