我正在尝试按照播放器制作一个屏幕,以便播放器位于屏幕中间。我已经在另一个游戏中制作了它,但在这里它不起作用。这是我的代码:
var c = document.getElementById("main");
var ctx = c.getContext("2d");
var screen = document.getElementById("screen").getContext("2d");
var WhatUSeeWidth = document.getElementById("screen").width;
var WhatUSeeHeight = document.getElementById("screen").height;
ctx.beginPath();
for (i = 0; i < 100; i ++) {
if (i % 2) {
ctx.fillStyle = "red";
}
else {
ctx.fillStyle = "blue";
}
ctx.fillRect(0, i * 100, 500, 100);
}
var player = {
x : 700,
y : 800
}
setInterval(tick, 100);
function tick() {
screen.beginPath();
screen.drawImage(c, player.x - WhatUSeeWidth / 2, player.y - WhatUSeeHeight / 2, WhatUSeeWidth, WhatUSeeHeight, 0, 0, WhatUSeeWidth, WhatUSeeHeight);
}
canvas {
border: 2px solid black;
}
<canvas id="main" width="500" height="500"h></canvas>
<canvas id="screen" width="500" height="500"></canvas>
我想在“屏幕”画布中使用drawImage
绘制蓝色和红色画布答案 0 :(得分:0)
好的,从你的评论我明白你在找什么。但问题是你可能从没有理解的例子开始。我试着向你解释你做了什么,但是你应该寻找一个好的指南,从基础开始并加深动画(例如:http://www.html5canvastutorials.com/)。
HTML
<canvas id="canvasLayer" width="500" height="500"></canvas>
的Javascript
var canvas = document.getElementById("canvasLayer");
var context = canvas.getContext("2d");
var WhatUSeeWidth = document.getElementById("canvasLayer").width;
var WhatUSeeHeight = document.getElementById("canvasLayer").height;
var player = {
x : 0,
y : 0
}
function drawBackground() {
for (i = 0; i < 100; i ++) {
if (i % 2) {
context.fillStyle = "red";
}
else {
context.fillStyle = "blue";
}
context.fillRect(0, i * 100, 500, 100);
}
}
function playerMove() {
context.beginPath();
var radius = 5;
context.arc(player.x, player.y, radius, 0, 2 * Math.PI, false);
context.fillStyle = 'green';
context.fill();
context.lineWidth = 1;
context.strokeStyle = '#003300';
context.stroke();
}
setInterval(tick, 100);
function tick() {
context.clearRect(0, 0, canvas.width, canvas.height);
drawBackground();
player.x++;
player.y++;
playerMove();
}
使用正确的答案进行编辑
错误位于对象“播放器”的位置。它位于画布外面,宽度:500高度:500,“播放器”位于x:700 y:800。
更改播放器的位置。
var player = {
x : 50,
y : 50
}