js实现放大镜效果

Posted 2020-10-30 aSnow

html部分

<div id="box">
        <img width="100%" height="100%" src="images/1.png" alt="#">
        <div id="move"></div>
      </div>
        <div id="bimg">
              <img id="b_bimg" src="images/1.png" alt="#">
        </div>

css部分

            *{
  margin:0px;
  padding:0px;
}
#box{
  width:330px;
  height:330px;
 float: left;
  margin-left:17%;
  position:relative;
}
#move{
  background-image:url(images/move.png);
  width:100px;
  height:100px;
  position:absolute;
  left:0px;
  top:0px;
  display:none;
}
#bimg{
  width:400px;
  height:400px;
  overflow:hidden;
 float: left;
 margin-left: 20px;
  position: relative;
  display:none;
}
#bimg img{
  width:1000px;
  height:1000px;
  position:absolute;
  top:0px;
  left:0px;
}

js部分

function bigimg(){
  var bbox = document.getElementById("box");
  var bmove = document.getElementById("move");
  var bbimg = document.getElementById("bimg");
  var b_bimg = document.getElementById("b_bimg");
  bbox.onmouseover = function(){//鼠标移动到box上显示大图片和选框
    bbimg.style.display = "block";
    bmove.style.display="block";
  }
  bbox.onmouseout = function(){//鼠标移开box不显示大图片和选框
    bbimg.style.display = "none";
    bmove.style.display="none";
  }
  bbox.onmousemove = function(e){//获取鼠标位置
      var e=e=e||window.event;
    var x = e.clientX;//鼠标相对于视口的位置
    var y = e.clientY;
    var t = bbox.offsetTop;//box相对于视口的位置
    var l = bbox.offsetLeft;
    var _left = x - l - bmove.offsetWidth/2;//计算move的位置left值
    var _top = y - t -bmove.offsetHeight/2;//计算move的位置top值
    if(_top<=0){//滑到box的最顶部
      _top = 0;
    }else if(_top>=bbox.offsetHeight-bmove.offsetHeight){//滑到box的最底部
      _top = bbox.offsetHeight-bmove.offsetHeight ;//bbox.offsetHeight-bmove.offsetHeight这是top的最大值
    }
    if(_left<=0){//滑到box的最左边
      _left=0;
   } else if(_left>=bbox.offsetWidth-bmove.offsetWidth){//滑到box的最右边
      _left=bbox.offsetWidth-bmove.offsetWidth ;//bbox.offsetWidth-bmove.offsetWidth这是left的最大值
      }
    bmove.style.top = _top +"px";//设置move的位置
    bmove.style.left = _left + "px";
     var w =b_bimg.offsetWidth/bbox.offsetWidth;//计算移动的比例大图宽除以小图宽
     var h =b_bimg.offsetHeight/bbox.offsetHeight;
    var b_bimg_top = _top*h;//计算大图的位置
    var b_bimg_left = _left*w;
    b_bimg.style.top = -b_bimg_top + "px";//设置大图的位置信息
    b_bimg.style.left = -b_bimg_left + "px";
  }
    
}
bigimg()

效果如下