为了清晰起见,编辑此问题。
我正在尝试根据mousemove
制作视差效果,但我遇到了一些问题。
1)我无法获得正确的window
偏移量。如果你看到JSFiddle,你会注意到偏移量与指针进入window
的位置有关。我希望asset-layer
偏移始终基于window
的中间位置。我该怎么做才能解决这个问题?
2)你会注意到视差会在重复时改变强度。我之前尝试过循环来迭代它们但我没有成功。为什么会发生这种情况?我该如何预防?
HTML
<section class="one">
<div class="parallax">
<div class="asset asset-layer4">4</div>
<div class="asset asset-layer3">3</div>
<div class="asset asset-layer2">2</div>
<div class="asset asset-layer1">1</div>
</div>
</section>
<section class="two">
<div class="parallax">
<div class="asset asset-layer4">4</div>
<div class="asset asset-layer3">3</div>
<div class="asset asset-layer2">2</div>
<div class="asset asset-layer1">1</div>
</div>
</section>
<section class="three">
<div class="parallax">
<div class="asset asset-layer4">4</div>
<div class="asset asset-layer3">3</div>
<div class="asset asset-layer2">2</div>
<div class="asset asset-layer1">1</div>
</div>
</section>
JS
var currentX = '';
var currentY = '';
var movementConstant = .015;
$(document).mousemove(function(e) {
if (currentX == '')
currentX = e.pageX;
var xdiff = e.pageX - currentX;
currentX = e.pageX;
if (currentY == '')
currentY = e.pageY;
var ydiff = e.pageY - currentY;
currentY = e.pageY;
$('.parallax div').each( function(i) {
var $el = $(this);
var movementx = (i + 1) * (xdiff * movementConstant);
var movementy = (i + 1) * (ydiff * movementConstant);
var newX = $el.position().left + movementx;
var newY = $el.position().top + movementy;
$el.css({left: newX + 'px', top: newY + 'px'});
});
});
CSS
.one,
.two,
.three {
position: relative;
width: 100%;
height: 200px;
}
.one { background-color: pink; }
.two { background-color: lightgray; }
.three { background-color: orange; }
.parallax {
position: absolute;
left: 50%;
top: 50%;
bottom: 50%;
right: 50%;
overflow: visible;
}
.asset {
position: absolute;
}
.asset-layer1 {
background-color: yellow;
}
.asset-layer2 {
background-color: green;
}
.asset-layer3 {
background-color: blue;
}
.asset-layer4 {
background-color: red;
}
提前谢谢。
答案 0 :(得分:0)
我猜,但你的问题可能是那个
var movement = (i + 1) * (xdiff * movementConstant);
应该是
var movement = (z + 1) * (xdiff * movementConstant);
实际上内部each
是无关紧要的,所以你可以写:
for (var z = 0; z < 5; z++) {
console.log(z);
var $el = $('.parallax-' + z);
var movement = (z + 1) * (xdiff * movementConstant);
var movementy = (z + 1) * (ydiff * movementConstant);
var newX = $el.position().left + movement;
var newY = $el.position().top + movementy;
$el.css('left', newX + 'px');
$el.css('top', newY + 'px');
}
但是,不是对每个项目应用单独的类,而是对所有类似项目使用单个类,然后执行以下操作:
for (var z = 0; z < 5; z++) {
console.log(z);
var $el = $('.parallax').eq(z);
或者,imo,甚至更好:
$('.parallax').each( function(i) {
var $el = $(this);
var movementx = (i + 1) * (xdiff * movementConstant);
var movementy = (i + 1) * (ydiff * movementConstant);
var newX = $el.position().left + movementx;
var newY = $el.position().top + movementy;
$el.css({left: newX + 'px', top: newY + 'px'});
}
答案 1 :(得分:0)
你真的让这个问题复杂化了。这应该是一个简单的逻辑。这里的输入只是从鼠标指针到窗口中心点的距离。计算刚刚完成输入。不需要计算任何鼠标移动差异。更确切地说,我们需要鼠标指针相对于中心点的坐标(因此它可以有负值,在这种情况下它应该在中心点的左侧)。
基于这个简单的逻辑,代码可以更加简化,如下所示:
var movementConstant = .05;
$(document).mousemove(function(e) {
var xToCenter = e.pageX - window.innerWidth/2;
var yToCenter = e.pageY - window.innerHeight/2;
$('.parallax div').each( function(i) {
var $el = $(this);
var newX = (i + 1) * (xToCenter * movementConstant);
var newY = (i + 1) * (yToCenter * movementConstant);
$el.css({left: newX + 'px', top: newY + 'px'});
});
});
我希望我能理解你想要什么。