只要没有鼠标运动或按键，我使用此代码每x秒运行一次功能。只要没有用户活动，我才能以2倍的价值增加计时器，并将其限制为300000？

Question

使用超时，增加超时，然后像下面再次致电超时。

在下面的示例中，我们在500ms中调用超时，然后我们加倍，以1000ms的重新调用，将其翻倍至2000毫秒，等等。

eDit

我不确定您希望它如何工作，但是将您的jQuery添加到混合中会看起来像这样。一旦您停止了超时，一旦您停止移动超时就开始返回

mousemove

Answer 1

500ms

let timeout
let interval = 500
let maxTime = 10000

function myTimeout() {
      console.log("Delayed: " + interval + "ms")
      interval *= 2
      if(interval < maxTime) {
            timeout = setTimeout(myTimeout, interval)
      } else {
            console.log("I am exiting!")
      }
}

$(document).ready(function () {
    // Zero the idle timer on mouse movement.
    $(document).on('mousemove', function (e) {
        clearTimeout(timeout)
        interval = 500
        timeout = setTimeout(myTimeout, interval)
    });
    $(document).on('keypress', function (e) {
        clearTimeout(timeout)
        interval = 500
        timeout = setTimeout(myTimeout, interval)
    });
    
    // Start the Timeouts
    timeout = setTimeout(myTimeout, interval)
});

我已将您的代码修改为此，我认为这是您在运行后需要看到的控制台（不要移动鼠标会重置计时器），

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

var time = 5000;
var idleInterval;
$(document).ready(function () {
    
    idleInterval = setInterval(timerIncrement, time);
    
    $(this).on('mousemove', function (e) {
        idleTime = 0;
        time = 5000;
    	clearInterval(idleInterval);
    	idleInterval = setInterval(timerIncrement, time);
    });
    $(this).on('keypress', function (e) {
        idleTime = 0;
        time = 5000;
    	clearInterval(idleInterval);
    	idleInterval = setInterval(timerIncrement, time);
    });
});

function timerIncrement() {

    console.log('refreshContent is called after '+time+' milliseconds');
    refreshContent();

    time = time*2;
    if( time > 300000 ) { time = 300000; }
    clearInterval(idleInterval);
    idleInterval = setInterval(timerIncrement, time);
}	
function refreshContent()
{
    console.log('Stuffs you want to do when refresh content is called');
}

Answer 2

首先，您无法将Settimeout设置为变量。 。

在您的示例中，您可以使用以下方式：<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

一个简单的示例与

var idleTime = 0;
var time = 5000;
var currentActiveTimeoutId = 0;
$(document).ready(function () {
    // Zero the idle timer on mouse movement.
    $(this).on('mousemove', function (e) {
        idleTime = 0;
        time = 5000;
        //reset timeout counter
        setTimeoutCounter(time);
    });
    $(this).on('keypress', function (e) {
        idleTime = 0;
        time = 5000;
        //reset timeout counter
        setTimeoutCounter(time);
    });
    setTimeoutCounter(time);
});

//function to set the timeout and call the incrementor
function setTimeoutCounter(timeOut){
  //clear previous timeout created
  clearTimeout(currentActiveTimeoutId);
  currentActiveTimeoutId = setTimeout(timerIncrement, timeOut);
}

function timerIncrement() {
    idleTime = idleTime + 1;
    if (idleTime > 1) {
        time = time*2;
        if( time > 300000 ) { time = 300000; }
        refreshContent();
        //recall set timeout and apply the next value of time
        setTimeoutCounter(time);
    }
}

和

Answer 3

事件。

当触发（或达到最大时间）

时，

Answer 4

keypress

mousemove

2K25中的同样需求）keypress