Javascript 使用jquery的html画布不工作 你好 var CANVAS_WIDTH=400; var CANVAS_高度=200; var canvasElem=$(“”); $(canvasElem).appendTo('body'); var canvas=canvasElem.get(0); var context=canvas.getContext(“2d”); context.fillRect(0,0400200); 测试。。。
上面的代码不是绘制画布。但是如果我在Javascript 使用jquery的html画布不工作 你好 var CANVAS_WIDTH=400; var CANVAS_高度=200; var canvasElem=$(“”); $(canvasElem).appendTo('body'); var canvas=canvasElem.get(0); var context=canvas.getContext(“2d”); context.fillRect(0,0400200); 测试。。。,javascript,jquery,html5-canvas,Javascript,Jquery,Html5 Canvas,上面的代码不是绘制画布。但是如果我在之后写一些东西,例如用下面的内容替换第6行 此处 然后画画布。我无法找出上面代码中的错误。执行脚本时,DOM尚未就绪。您可以将脚本放在正文中,或者添加一个在DOM准备就绪后立即执行的事件侦听器: <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <ht
之后写一些东西,例如用下面的内容替换第6行此处
然后画画布。我无法找出上面代码中的错误。执行脚本时,DOM尚未就绪。您可以将脚本放在
正文中
,或者添加一个在DOM准备就绪后立即执行的事件侦听器:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html>
<head>
<title>Hello ...</title>
</head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.1/jquery.min.js"></script>
<script>
var CANVAS_WIDTH = 400;
var CANVAS_HEIGHT = 200;
var canvasElem = $("<canvas id='my-canvas' width='" + CANVAS_WIDTH + "' height='" + CANVAS_HEIGHT + "'></canvas>");
$(canvasElem).appendTo('body') ;
var canvas = canvasElem.get(0);
var context = canvas.getContext("2d");
context.fillRect(0,0,400,200);
</script>
<body>
Testing ...
</body>
</html>
window.onload=function(){
var CANVAS_WIDTH=400;
var CANVAS_高度=200;
var canvasElem=$(“”);
$(canvasElem).appendTo('body');
var canvas=canvasElem.get(0);
var context=canvas.getContext(“2d”);
context.fillRect(0,0400200);
}
window.onload = function() {
var CANVAS_WIDTH = 400;
var CANVAS_HEIGHT = 200;
var canvasElem = $("<canvas id='my-canvas' width='" + CANVAS_WIDTH + "' height='" + CANVAS_HEIGHT + "'></canvas>");
$(canvasElem).appendTo('body') ;
var canvas = canvasElem.get(0);
var context = canvas.getContext("2d");
context.fillRect(0,0,400,200);
}