有一个简单的绘图应用程序。问题出在坐标(redraw
函数)中:它们必须是鼠标,但接近2x鼠标。代码中的问题是什么?
<html>
<head>
<style type="text/css" media="screen">
body{
background-color: green;
}
#workspace{
width: 700px;
height:420px;
margin: 40px auto 20px auto;
border: black dashed 1px;
}
#canvas{
background: white;
width: 100%;
height: 100%;
}
</style>
<script type="text/javascript" src="jquery-1.7.1.js"></script>
<script type="text/javascript">
$(document).ready(
function() {
var context = document.getElementById('canvas').getContext("2d");
var clickX = new Array();
var clickY = new Array();
var clickDrag = new Array();
var paint;
$('#canvas').mousedown(function(e){
var mouseX = e.pageX - this.offsetLeft;
var mouseY = e.pageY - this.offsetTop;
paint = true;
addClick(mouseX,mouseY, false);
redraw();
});
$('#canvas').mousemove(function(e){
if(paint){
addClick(e.pageX - this.offsetLeft, e.pageY - this.offsetTop, true);
redraw();
}
});
$('#canvas').mouseup(function(e){
paint = false;
});
$('#canvas').mouseleave(function(e){
paint = false;
});
function addClick(x, y, dragging)
{
clickX.push(x);
clickY.push(y);
clickDrag.push(dragging);
}
function redraw(){
canvas.width = canvas.width; // Clears the canvas
context.strokeStyle = "#df4b26";
context.lineJoin = "round";
context.lineWidth = 5;
for(var i=1; i < clickX.length; i++)
{
context.beginPath();
if(clickDrag[i] && i){
context.moveTo(clickX[i-1], clickY[i-1]);
}else{
context.moveTo(clickX[i], clickY[i]);
}
context.lineTo(clickX[i], clickY[i]);
context.closePath();
context.stroke();
}
console.log(clickX[clickX.length-1]+" "+clickY[clickX.length-1]);
}
});
</script>
</head>
<body>
<div id="workspace">
<canvas id="canvas"/>
</div>
</body>
</html>
答案 0 :(得分:12)
您不应该通过CSS设置画布的宽度/高度。它使画布伸展而不是提高分辨率。这意味着虽然坐标是正确的,但它们在视觉上会在其他地方结束。
例如,100
的x坐标将被拉伸到视觉上是200
的x坐标(或其他东西;至少它会大于100
,因为它已被拉伸)。
相反,请使用:
<canvas id="canvas" width="700" height="420" />
并删除CSS中的width: 100%
。