问题: 当我旋转添加的图像时,图像不会像我希望的那样居中。我一直在画布上方的白色区域上方。
尝试使用16:9比例图像,旋转后你会看到我的问题。这将用于移动,因此图片的宽度可以是最大640。
Here is a fiddle 目前的代码:
HTML:
<div id="files">
<input type="file" id="imageLoader" name="imageLoader" />
</div>
<canvas id="imageCanvas"></canvas>
<canvas id="preview"></canvas>
<input type="button" id="saveButton" value="Save" />
<div id="rotate">
<button type="button" id="rotateLeft">Rotate -90°</button>
<button type="button" id="rotateRight">Rotate 90°</button>
</div>
JS:
var imageLoader = document.getElementById('imageLoader');
imageLoader.addEventListener('change', handleImage, false);
var image = "";
var canvas = document.getElementById('imageCanvas');
var ctx = canvas.getContext("2d");
var fileName = "";
var angleInDegrees = 0;
function handleImage(e) {
var reader = new FileReader();
reader.onload = function(event) {
var img = new Image();
img.onload = function() {
canvas.width = img.width;
canvas.height = img.height;
ctx.drawImage(img, 0,0);
fileName = img.name;
image = img;
};
img.src = event.target.result;
};
reader.readAsDataURL(e.target.files[0]);
}
function drawRotated(degrees) {
ctx.clearRect(0, 0, canvas.width, canvas.height);
ctx.save();
ctx.translate(canvas.width / 2, canvas.height / 2);
ctx.rotate(degrees * Math.PI / 180);
ctx.drawImage(image, -image.width / 2, -image.width / 2);
ctx.restore();
}
$("#rotateRight").click(function () {
angleInDegrees += 90;
drawRotated(angleInDegrees);
});
$("#rotateLeft").click(function () {
angleInDegrees -= 90;
drawRotated(angleInDegrees);
});
这是我的新小提琴:)如果我希望它最大640宽度怎么办?
答案 0 :(得分:2)
你的drawImage中有一个拼写错误:
ctx.drawImage(image, -image.width / 2, -image.height / 2);
除非您的图像始终是方形,否则您必须调整画布大小以允许旋转的图像适合。
演示如何调整画布大小:http://jsfiddle.net/m1erickson/3E99A/
如果您不想更改画布大小,则会剪裁图像。此外,如果图像是横向的,则必须在drawimage中翻转宽度/高度:
// flip width/height when the image is in landscape orientation
ctx.drawImage(image, -image.height/2, -image.width/2);
关于如何调整画布大小的代码:
<!doctype html>
<html>
<head>
<link rel="stylesheet" type="text/css" media="all" href="css/reset.css" /> <!-- reset css -->
<script type="text/javascript" src="http://code.jquery.com/jquery.min.js"></script>
<style>
#canvas{
border: 1px solid green;
}
</style>
<script>
$(function(){
var canvas=document.getElementById("canvas");
var ctx=canvas.getContext("2d");
var imgWidth=200;
var imgHeight=300;
var size={width:imgWidth, height:imgHeight};
var rotation=0;
var deg2Rad=Math.PI/180;
var count1=0;
var count2=0;
var img=new Image();
img.onload=function(){
imgWidth=img.width;
imgHeight=img.height;
size={width:imgWidth, height:imgHeight};
draw();
}
img.src="https://dl.dropboxusercontent.com/u/139992952/stackoverflow/card.png";
function draw(){
canvas.width=size.width;
canvas.height=size.height;
// calculate the centerpoint of the canvas
var cx=canvas.width/2;
var cy=canvas.height/2;
var info=document.getElementById("info");
info.innerHTML="canvas size: "+(count1++)+": "+cx+" / "+cy;
// draw the rect in the center of the newly sized canvas
ctx.clearRect(0,0,canvas.width,canvas.height);
ctx.fillStyle="rgba(216,216,150,1.0)";
ctx.translate(cx,cy);
ctx.rotate(rotation * deg2Rad);
ctx.drawImage(img,-imgWidth/2,-imgHeight/2);
}
document.getElementById("rotate").addEventListener("click", resizeClicked, false);
function resizeClicked(e){
rotation+=30;
newSize(imgWidth,imgHeight,rotation);
draw();
}
function newSize(w,h,a){
var rads=a*Math.PI/180;
var c = Math.abs(Math.cos(rads));
var s = Math.abs(Math.sin(rads));
size.width = h * s + w * c;
size.height = h * c + w * s ;
}
}); // end onload
</script>
</head>
<body>
<button id="rotate">Rotate with resize</button>
<p id=info></p>
<canvas id="canvas" width=400 height=400></canvas>
</body>
</html>
答案 1 :(得分:0)
尝试更改此行:
ctx.drawImage(image, -image.width / 2, -image.width / 2);
为:
ctx.drawImage(image, -image.width / 2, -image.height / 2);
^^^^^^
或宽度将用作旋转两个轴的基础。
另请参阅此答案,了解旋转的一种方法,并根据旋转将画布调整到图像:
Rotate original Image with jQuery or JavaScript