这是布雷登斯答案的一个实现。

Var键= {} handleKeyPress(evt) { let {keyCode, type} = evt ||事件;//处理IE let isKeyDown = (type == 'keydown'); keys[keyCode] = isKeyDown; // test: enter键被按下，shift键没有被按下 if(isKeyDown && keys[13] && !keys[16]){ Console.log('用户按下enter没有shift') ｝ }; 窗口。addEventListener(“弹起”,handleKeyPress); 窗口。addEventListener(“keydown”,handleKeyPress);

只是让某些东西更稳定:

var keys = [];
$(document).keydown(function (e) { 
    if(e.which == 32 || e.which == 70){
    keys.push(e.which);
    if(keys.length == 2 && keys.indexOf(32) != -1 && keys.indexOf(70) != -1){
        
        
        
        alert("it WORKS !!"); //MAKE SOMETHING HERE---------------->
        
        
        
        keys.length = 0;
    }else if((keys.indexOf(32) == -1 && keys.indexOf(70) != -1) || (keys.indexOf(32) != -1 && keys.indexOf(70) == -1) && (keys.indexOf(32) > 1 || keys.indexOf(70) > 1)){
    }else{
        keys.length = 0;
    }
}else{
    keys.length = 0;
}
});

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;
    }
});

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);

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。