$(document).ready(function () {
        // using ascii 17 for ctrl, 18 for alt and 83 for "S"
        // ctr+alt+S
        var map = { 17: false, 18: false, 83: false };
        $(document).keyup(function (e) {
            if (e.keyCode in map) {
                map[e.keyCode] = true;
                if (map[17] && map[18] && map[83]) {
                    // Write your own code here, what  you want to do
                    map[17] = false;
                    map[18] = false;
                    map[83] = false;
                }
            }
            else {
                // if u press any other key apart from that "map" will reset.
                map[17] = false;
                map[18] = false;
                map[83] = false;
            }
        });

    });

case 65: //A
jp = 1;
setTimeout("jp = 0;", 100);

if(pj > 0) {
ABFunction();
pj = 0;
}
break;

case 66: //B
pj = 1;
setTimeout("pj = 0;", 100);

if(jp > 0) {
ABFunction();
jp = 0;
}
break;

这不是最好的方法，我知道。

这不是一个通用的方法，但在某些情况下是有用的。这是有用的组合，如CTRL + something或Shift + something或CTRL + Shift + something，等。

示例:当您想使用CTRL + P打印页面时，第一个按下的键总是CTRL + P, CTRL + S, CTRL + U和其他组合也是如此。

document.addEventListener (keydown,函数(e) { //SHIFT +某个值 如果(e.shiftKey) { 开关(e.code) { 例“钥匙”: console.log('Shift + S'); 打破; ｝ ｝ //CTRL + SHIFT +一些 如果(e。&& e.shiftKey){ 开关(e.code) { 例“钥匙”: console.log('CTRL + Shift + S'); 打破; ｝ ｝ })；

对于任何使用React的人，这里是我的解决方案:

import { useEffect, useState } from "react";
import Backdrop from '@mui/material/Backdrop';

export const Example = () => {
  const [backdropOpen, setBackdropOpen] = useState(false);

  useEffect(() => {
    // Keys that need to be pressed at the same time in order for
    // the 'backdropOpen' variable to be 'true'
    const keysArr = ['ControlLeft', 'ShiftLeft', 'AltLeft'];
    const keysMap = {};
    let backdropOpenLocal = false;

    const keydownEvent = 'keydown';
    const keyupEvent = 'keyup';

    const checkKeys = () => {
      const keysArePressed = keysArr.every((value) => keysMap[value] === keydownEvent);
      if (keysArePressed !== backdropOpenLocal) {
        backdropOpenLocal = keysArePressed;
        setBackdropOpen(keysArePressed);
      }
    }

    const handleKeyDown = (event) => {
      const keyCode = event.code;
      if (keysArr.includes(keyCode) && keysMap[keyCode] !== keydownEvent) {
        keysMap[keyCode] = keydownEvent;
      }
      checkKeys();
    }

    const handleKeyUp = (event) => {
      const keyCode = event.code;
      if (keysArr.includes(keyCode) && keysMap[keyCode] !== keyupEvent) {
        keysMap[keyCode] = keyupEvent;
      }
      checkKeys();
    }

    document.addEventListener('keydown', handleKeyDown);
    document.addEventListener('keyup', handleKeyUp);

    return () => {
      document.removeEventListener('keydown', handleKeyDown);
      document.removeEventListener('keyup', handleKeyUp);
    }
  }, []);

  return (
    <React.Fragmemnt>
      <div>
        <Backdrop
          open={backdropOpen}
        >
          <span>
            It worked!
          </span>
        </Backdrop>
      </div>
    </React.Fragmemnt>
  );
}

请记住，我们需要在useEffect函数中使用backdropOpenLocal而不是backdropOpen，因为我们只想更新局部作用域变量并保持作用域的状态。

如果我们更新Example组件的状态并尝试访问backdropOpen，我们将得到与之前相同的值，除非我们在useEffect的依赖数组中传递backdropOpen;这将导致useEffect内的作用域变量被重置，我们不希望这样。

我使用case, if和bool。我有一个项目，这对我很有效

window.addEventListener("keydown", onKeyDown, false);
window.addEventListener("keyup", onKeyUp, false);

function onKeyDown(event) {
  var keyCode = event.keyCode;
  switch (keyCode) {
    case 68: //D
      keyd = true;
      break;
    case 32: //spaaaaaaaaaaaaaaace
      keyspace = true;
      break;
    case 65: //A
      keya = true;
      break;
    case 37:
      keya = true;
      break;
    case 38:
      keyspace = true;
      break;
    case 39:
      keyd = true;
      break;
  }
}

function onKeyUp(event) {
  var keyCode = event.keyCode;

  switch (keyCode) {
    case 68: //dddddd
      keyd = false;
      break;
    case 32: //spaaaaaaaaaaaaaaaaaaaaaace
      keyspace = false;
      break;
    case 65: //aaaaa
      keya = false;
      break;
    case 37:
      keya = false;
      break;
    case 38:
      keyspace = false;
      break;
    case 39:
      keyd = false;
      break;
  }
}

如果有人需要简单的解决方案。

let keys = [];
document.addEventListener("keydown", (e) => {
  keys.push(e.key);
  if (keys.includes("Control") && keys.includes("o")) {
    console.log("open");
  }

  if (keys.includes("Control") && keys.includes("s")) {
    console.log("save");
  }
});

// clear the keys array
document.addEventListener("keyup", () => {
  keys = [];
});