如果你想找到任何按键事件与控制键，你可以这样做

onkeypress = (e) =>{
console.log(e);
if(e.ctrlKey && e.code == "KeyZ"){
    document.write("do somthing")
} }

我使用这种方式(必须检查按下Shift + Ctrl的地方):

// create some object to save all pressed keys
var keys = {
    shift: false,
    ctrl: false
};

$(document.body).keydown(function(event) {
// save status of the button 'pressed' == 'true'
    if (event.keyCode == 16) {
        keys["shift"] = true;
    } else if (event.keyCode == 17) {
        keys["ctrl"] = true;
    }
    if (keys["shift"] && keys["ctrl"]) {
        $("#convert").trigger("click"); // or do anything else
    }
});

$(document.body).keyup(function(event) {
    // reset status of the button 'released' == 'false'
    if (event.keyCode == 16) {
        keys["shift"] = false;
    } else if (event.keyCode == 17) {
        keys["ctrl"] = 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。

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。

谁需要完整的示例代码。左+右补充道

var keyPressed = {};
document.addEventListener('keydown', function(e) {

   keyPressed[e.key + e.location] = true;

    if(keyPressed.Shift1 == true && keyPressed.Control1 == true){
        // Left shift+CONTROL pressed!
        keyPressed = {}; // reset key map
    }
    if(keyPressed.Shift2 == true && keyPressed.Control2 == true){
        // Right shift+CONTROL pressed!
        keyPressed = {};
    }

}, false);

document.addEventListener('keyup', function(e) {
   keyPressed[e.key + e.location] = false;

   keyPressed = {};
}, false);

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 = [];
});