我正在构建一个简单的障碍游戏,我现在遇到的问题是,我可以用箭头键左右移动,但不能上下移动。我不明白为什么上下不工作,当他们有相同的逻辑,左和右。我是一个人开发东西的新手,所以如果很明显的话,那就让我轻松一点吧。:)
document.addEventListener("DOMContentLoaded", () => {
// Grab the elements from the HTML
const floppy = document.getElementById("floppyDisk");
const gameBoard = document.querySelector(".gameBoard");
let userScore = document.getElementById("userScore");
let highscore = document.getElementById("highscore");
// Set up variables to be used later
let floppyPosX = 0;
let floppyPosY = 0;
let left = 20;
let top = 190;
// Use the arrows to move the floppy disk
function moveFloppy(e) {
if(e.keyCode == 39) {
left += 2;
floppy.style.left = floppyPosX + left + "px";
}
if(e.keyCode == 37) {
left -= 2;
floppy.style.left = floppyPosX + left + "px";
}
if(e.keycode == 38) {
top += 2;
floppy.style.top = floppyPosY + top + "px";
}
if(e.keycode == 40) {
top -= 2;
floppy.style.top = floppyPosY + top + "px";
}
}
// Invoke the moveFloppy function
document.onkeydown = moveFloppy;
// Generate Obstacles
// Function to move the obstacles
// Set function to repeat
// Call the function to generate the obstacles
})发布于 2020-10-23 15:34:52
您的问题在于您使用的是the (and ill-defined) KeyboardEvent.keyCode property而不是标准化的code属性。
将您的JavaScript更改为此,它应该可以工作:
left/floppyPosX逻辑的正确性。if语句更改为单个switch语句。as it has fallthrough unless you use break.
switch function moveFloppy(e) {
switch(e.code) {
case 'ArrowLeft':
left += 2;
floppy.style.left = floppyPosX + left + "px";
break;
case 'ArrowRight':
left -= 2;
floppy.style.left = floppyPosX + left + "px";
break;
case 'ArrowUp':
top += 2;
floppy.style.top = floppyPosY + top+ "px";
break;
case 'ArrowDown':
top -= 2;
floppy.style.top = floppyPosY + top + "px";
break;
default:
// TODO: Play a fart sound.
break;
}
}
document.addEventListener( 'keydown', moveFloppy );https://stackoverflow.com/questions/64502959
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