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")
} }

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。

使keydown甚至调用多个函数，每个函数检查特定的键并适当地响应。

document.keydown = function (key) {

    checkKey("x");
    checkKey("y");
};

我会试着在按键时添加一个按键事件处理程序。例句:

window.onkeydown = function() {
    // evaluate key and call respective handler
    window.onkeypress = function() {
       // evaluate key and call respective handler
    }
}

window.onkeyup = function() {
    window.onkeypress = void(0) ;
}

这只是为了说明一个模式;这里我就不详细介绍了(尤其是浏览器特定的level2+事件注册)。

请回复这是否有帮助。

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);

Easiest, and most Effective Method

//check key press
    function loop(){
        //>>key<< can be any string representing a letter eg: "a", "b", "ctrl",
        if(map[*key*]==true){
         //do something
        }
        //multiple keys
        if(map["x"]==true&&map["ctrl"]==true){
         console.log("x, and ctrl are being held down together")
        }
    }

//>>>variable which will hold all key information<<
    var map={}

//Key Event Listeners
    window.addEventListener("keydown", btnd, true);
    window.addEventListener("keyup", btnu, true);

    //Handle button down
      function btnd(e) {
      map[e.key] = true;
      }

    //Handle Button up
      function btnu(e) {
      map[e.key] = false;
      }

//>>>If you want to see the state of every Key on the Keybaord<<<
    setInterval(() => {
                for (var x in map) {
                    log += "|" + x + "=" + map[x];
                }
                console.log(log);
                log = "";
            }, 300);