为啥document触发房子touchstart外，下次点击触发start，显示startTouchX 次数加1
<!DOCTYPE html>
<html>
<head>
<meta charset="UTF-8">
<title>Document</title>
<meta name="viewport" content="width=device-width,initial-scale=1.0,maximum-scale=1.0,user-scalable=0">
<style>
*{
margin:0;
padding: 0;
}
.wrapper{
overflow:hidden;
}
.swipArea{
height: 100px;
background-color: green;
text-align: center;
line-height: 100px;
position: absolute;
left: 0;
right: 0;
}
</style>
</head>
<body>
<div>
<div id="swipArea">swipArea</div>
</div>
<script>
function Swip(el){
this.el = el;
this.init();
}
Swip.prototype.init = function(){
var startTouchX = 0,
startTouchY = 0,
endTouchX = 0,
endTouchY = 0;
var deltaX = 0, deltaY = 0;
var direction = "left";
var d = "";
var that = this;
this.el.addEventListener("touchstart",function(evt){
evt.preventDefault();
var touches = evt.touches[0];
startTouchX = touches.pageX;
startTouchY = touches.pageY;
// console.log("startTouchX : " + startTouchX)
},false);
document.addEventListener("touchmove",function(evt){
evt.preventDefault();
var touches = evt.touches[0];
endTouchX = touches.pageX;
endTouchY = touches.pageY;
deltaY = endTouchY - startTouchY;
if(deltaY > 0) that.el.style.top = deltaY + "px";
// console.log("endTouchX : " + endTouchX + ",endTouchY : " + endTouchY)
},false);
document.addEventListener("touchend",function(evt){
evt.preventDefault();
document.ontouchmove = null;
document.ontouchstart = null;
deltaX = endTouchX - startTouchX;
deltaY = endTouchY - startTouchY;
d = Math.abs(deltaX) > Math.abs(deltaY) ? "h" : "v";
if(d == "h") direction = deltaX > 0 ? "right" : "left";
else if(d == "v") direction = deltaY > 0 ? "down" : "top";
that.el.style.top = "0";
// console.log("deltaX : " + deltaX + ",deltaY : " + deltaY) ;
},false);
}
var swipArea = document.querySelector("#swipArea");
new Swip(swipArea);
</script>
</body>
</html>
我是蜗牛1回答
-
李晓健
这个应该不是加1，是你点的位置不一样吧。在屏幕上点，你是没法保证两次点的位置完全一样的，精确到一个像素
随时随地看视频慕课网APP
相关分类
JavaScript