如何用原生javascript实现放大镜效果_html/css_WEB-ITnose
随着科技的发展,网购已成为大家生活中必不可少的一个模式,各种电商平台也如雨后春笋般涌现出来,今天我们就来用原生js来实现类似淘宝选购物品时的放大镜效果.
这里要用到大小两张图片,我选取的是800x800和350x350大小的两张图片
图片来源于网络
首先写出html和css样式
html部分
<body> <div class="min"> <img src="/static/imghw/default1.png" data-src="img/1.jpg" class="lazy" / alt="如何用原生javascript实现放大镜效果_html/css_WEB-ITnose" > <div class="fd"></div> </div> <div class="max"><img src="/static/imghw/default1.png" data-src="img/2.jpg" class="lazy" / alt="如何用原生javascript实现放大镜效果_html/css_WEB-ITnose" ></div> </body>
原理是创建min和max两个区域,将小图img/1.jpg和创建的放大镜divfd放到min中,将大图img/2.jpg放到max中
css样式部分
<style type="text/css"> .min{ width: 350px; height: 350px; border: 1px solid black; float: left; position: relative; } .max{ width: 350px; height: 350px; border: 1px solid black; float: left; display: none; overflow: hidden; position: relative; } .max img{ position: absolute; margin: 0 auto; } .fd{ width: 153.125px; height: 153.125px; background-color: skyblue; opacity: 0.3; position: absolute; top: 0; left: 0; display: none; }这里需要强调的是 2.放大镜的宽高,如果按照我选的尺寸的两张图宽高必须为153.125px,否则会出现左侧所选区域和右侧显示区域不能完全吻合的情况; 3.大图的父级max定义的框尺寸为什么比里面的图片小?(这里的框相当于一个窗户,里面的图片相当于窗子一面的物体,无论里面的物体多大也只能显示出窗子的尺寸) 4.当鼠标放在小图外区域时,大图和放大镜无显示,所以开始置max和fd里display:none;
为了显示效果我们先把display:none注掉,此时效果如下
样式图
js部分
首先分析逻辑顺序
定义变量1.鼠标覆盖显示max和fd2.确定放大镜的移动范围(不能出min)3.max的对应显示然后按顺序书写代码
<script type="text/javascript"> var min = document.querySelector(".min"), max = document.querySelector(".max"), img = document.querySelector(".max img"), fd = document.querySelector(".fd"); min.onmouseover = function(){ //1.鼠标覆盖显示max和fd max.style.display = "block"; fd.style.display = "block"; } //离开时隐藏 min.onmouseout= function(){ max.style.display = "none"; fd.style.display = "none"; } //2.fd的移动范围 min.onmousemove = function(){ //鼠标触摸的点 var x = event.clientX-min.offsetLeft-fd.offsetWidth/2; var y = event.clientY-min.offsetTop-fd.offsetHeight/2; //最大移动距离 var maxX = min.clientWidth-fd.offsetWidth; var maxY = min.clientHeight-fd.offsetHeight; //边界判断 if(x<=0){ x=0; }else if(x>=maxX){ x=maxX; } if(y<=0){ y=0; }else if(y>=maxY){ y=maxY; } //fd的位置 fd.style.left = x+"px"; fd.style.top = y+"px"; //fd/min = max/img //移动比例 var yidongX = x/maxX; var yidongY = y/maxY; //移动 //3.max的对应显示 // 对于大图而言 放大镜在小图上移动的比例 相当于img在可显示区域上移动的比例 放大镜右移等于图片左移 // 也就是本质上为img - max 然而需要负值 则*-1 简化后 为max-img img.style.left = yidongX * (max.clientWidth - img.offsetWidth) + "px"; img.style.top = yidongY * (max.clientHeight - img.offsetHeight) + "px"; }</script>最后根据需求完善即可实现效果如下
1.gif
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