今天我想知道如何使一个 Canvas 元素顺利地跟随另一个 Canvas 元素。例如，我正在尝试制作一款 Canvas 元素连续平滑地跟随玩家(可以使用 W、A、S 和 D 移动)的游戏。我有一个想法，使用毕达哥拉斯定理来检查从 A 点( Canvas 元素)移动到 B 点(玩家)的最接近且最快的方法。但是，我没有物理方法可以做到这一点。有谁知道如何让 Canvas 元素尽可能顺利地持续跟随玩家，以便以最快的速度到达玩家手中？
这里我有一个非常糟糕的遵循方式的例子:
<!DOCTYPE html>
<html>
<head>
<title>Target Following Test</title>
</head>
<body>
<script src="https://code.jquery.com/jquery-2.1.0.js"></script>
<center>
<canvas id="canvas" width="800" height="500"></canvas>
</center>
<script>
var canvas = document.getElementById("canvas");
var ctx = canvas.getContext("2d");
var width = canvas.width;
var height = canvas.height;
var circle = function(x, y, radius, fillCircle, color) {
ctx.beginPath();
ctx.fillStyle = color;
ctx.arc(x, y, radius, 0, Math.PI * 2, false);
if (fillCircle) {
ctx.fill();
} else {
ctx.stroke();
}
};
var drawRect = function(x, y, color) {
ctx.fillStyle = color;
ctx.fillRect(x, y, 20, 20)
}
//Moving Obstacle
var Obstacle = function(x, y) {
this.x = x;
this.y = y;
this.vSpeed = 0;
this.hSpeed = 0;
}
Obstacle.prototype.drawOb = function(color) {
drawRect(this.x, this.y, "Red")
}
Obstacle.prototype.follow = function() {
this.y += this.vSpeed
this.x += this.hSpeed
if (this.x < ball.x - 9) {
this.hSpeed = 1;
}
if (this.x > ball.x - 10) {
this.hSpeed = -1;
}
if (this.y > ball.y - 10) {
this.vSpeed = -1;
}
if (this.y < ball.y - 9) {
this.vSpeed = 1;
}
}
Obstacle.prototype.checkCollision = function(direction) {
return (ball.x - ball.radius < this.x + 20) &&
(ball.x + ball.radius > this.x) &&
(ball.y - ball.radius < this.y + 20) &&
(ball.y + ball.radius > this.y);
}
// The Ball constructor
var Ball = function() {
this.x = 20
this.y = 20
this.xSpeed = 0;
this.ySpeed = 0;
this.radius = 10;
};
// Draw the ball at its current position
Ball.prototype.draw = function() {
circle(this.x, this.y, 10, true, "Black");
};
Ball.prototype.reposition = function(reX, reY) {
this.x = reX;
this.y = reY;
}
// Update the ball's position based on its speed
Ball.prototype.move = function() {
this.x += this.xSpeed;
this.y += this.ySpeed;
if (this.x < 11) {
this.x = 11;
} else if (this.x > width - 11) {
this.x = width - 11;
} else if (this.y < 11) {
this.y = 11;
} else if (this.y > height - 11) {
this.y = height - 11;
}
};
// Set the ball's direction based on a string
Ball.prototype.setDirection = function(direction) {
if (direction === "up") {
this.xSpeed = 0;
this.ySpeed = -2;
} else if (direction === "down") {
this.xSpeed = 0;
this.ySpeed = 2;
} else if (direction === "left") {
this.xSpeed = -2;
this.ySpeed = 0;
} else if (direction === "right") {
this.xSpeed = 2;
this.ySpeed = 0;
} else if (direction === "stop") {
this.xSpeed = 0;
this.ySpeed = 0;
}
};
function simulate() {
var prev_ball_x = ball.x;
var prev_ball_y = ball.y;
var prev_fol_x = follower.x;
var prev_fol_y = follower.y;
ball.move();
follower.follow()
if (follower.checkCollision()) {
ball.setDirection('stop');
follower.vSpeed = 0;
follower.hSpeed = 0;
follower.x = prev_fol_x;
follower.y = prev_fol_y;
ball.x = prev_ball_x;
ball.y = prev_ball_y;
}
}
function draw() {
ctx.clearRect(0, 0, width, height);
ball.draw();
follower.drawOb();
ctx.strokeRect(0, 0, width, height);
}
// An object to convert keycodes into action names
var keyActions = {
37: "left",
38: "up",
39: "right",
40: "down"
};
// The keydown handler that will be called for every keypress
$("body").keydown(function(event) {
var direction = keyActions[event.keyCode];
ball.setDirection(direction);
});
$("body").keyup(function(event) {
ball.setDirection('stop');
})
setInterval(function() {
// separate drawing and simulating phases
simulate();
draw();
}, 10);
// Create all the Objects!
var ball = new Ball();
var follower = new Obstacle(400, 100);
</script>
</body>
</html>
最佳答案
注意:我还没有真正检查过你的代码...... 但希望我正确理解你的问题。如果我这样做，解决方案可能会非常简单。
最简单、快速的方法是将 Canvas 元素沿直线移动到播放器，无需毕达哥拉斯先生的帮助。为此，您需要知道玩家的位置 (x, y)，您确实这样做了。
我从 AS3 问题中获取了缓动函数，但对于 JS 来说是相同的:AS 3 simple ease
每次更新时，将跟随者移动到玩家的位置:
follower.x += (player.x - follower.x) / delay;
follower.y += (player.y - follower.y) / delay;
示例:Fiddle
这不是对您的脚本的直接修复，但希望它有帮助
关于javascript - 目标平稳跟随，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/37548099/