是否有可能在JavaScript中检测“空闲”时间?
我的主要用例可能是预取或预加载内容。
我将空闲时间定义为用户不活动或没有任何CPU使用的时间段
是否有可能在JavaScript中检测“空闲”时间?
我的主要用例可能是预取或预加载内容。
我将空闲时间定义为用户不活动或没有任何CPU使用的时间段
当前回答
你也许可以使用上面列出的鼠标移动技巧来检测网页上的不活跃状态,但这并不能告诉你用户不在另一个窗口或选项卡的另一个页面上,或者用户在Word、Photoshop或WoW中,只是在这个时候没有在看你的页面。
一般来说,我只会预取,并依赖于客户端的多任务处理。如果你真的需要这个功能,你可以在Windows中使用ActiveX控件做一些事情,但它充其量是丑陋的。
其他回答
下面是tvanfosson的想法的粗略jQuery实现:
$(document).ready(function(){
idleTime = 0;
//Increment the idle time counter every second.
var idleInterval = setInterval(timerIncrement, 1000);
function timerIncrement()
{
idleTime++;
if (idleTime > 2)
{
doPreload();
}
}
//Zero the idle timer on mouse movement.
$(this).mousemove(function(e){
idleTime = 0;
});
function doPreload()
{
//Preload images, etc.
}
})
(部分灵感来自Equiman回答的良好核心逻辑。)
sessionExpiration.js
sessionExpiration.js是轻量级的,但有效和可定制的。一旦实现,只在一行中使用:
sessionExpiration(idleMinutes, warningMinutes, logoutUrl);
Affects all tabs of the browser, not just one. Written in pure JavaScript, with no dependencies. Fully client side. (If so wanted.) Has warning banner and countdown clock, that is cancelled by user interaction. Simply include the sessionExpiration.js, and call the function, with arguments [1] number of idle minutes (across all tabs) until user is logged out, [2] number of idle minutes until warning and countdown is displayed, and [3] logout url. Put the CSS in your stylesheet. Customize it if you like. (Or skip and delete banner if you don't want it.) If you do want the warning banner however, then you must put an empty div with ID sessExpirDiv on your page (a suggestion is putting it in the footer). Now the user will be logged out automatically if all tabs have been inactive for the given duration. Optional: You may provide a fourth argument (URL serverRefresh) to the function, so that a server side session timer is also refreshed when you interact with the page.
这是一个例子,如果你不改变CSS,它看起来是什么样子的。
我使用这种方法,因为您不需要在事件触发时不断重置时间。相反,我们只记录时间,这将生成空闲的起始点。
function idle(WAIT_FOR_MINS, cb_isIdle) {
var self = this,
idle,
ms = (WAIT_FOR_MINS || 1) * 60000,
lastDigest = new Date(),
watch;
//document.onmousemove = digest;
document.onkeypress = digest;
document.onclick = digest;
function digest() {
lastDigest = new Date();
}
// 1000 milisec = 1 sec
watch = setInterval(function() {
if (new Date() - lastDigest > ms && cb_isIdel) {
clearInterval(watch);
cb_isIdle();
}
}, 1000*60);
},
您可以在文档主体上附加一个单击或鼠标移动事件,以重置计时器。
有一个函数,你在定时间隔调用,检查定时器是否超过指定的时间(如1000毫秒),并开始你的预加载。
您要求优雅,我创建了一个简单的类来支持惰性检查(具有空闲状态),除了命令方式(带有回调)之外。此外,当违反空闲时间时,该类支持“backToActive”。
class Idle {
constructor(timeout = 10, idleCallback = null, backToActiveCallback = null, autoStart = true, backToActiveOnXHR = false) {
this.timeout = timeout
this.idleCallback = idleCallback
this.backToActiveCallback = backToActiveCallback
this.autoStart = autoStart // only F5
this.backToActiveOnXHR = backToActiveOnXHR
this.idle = false
this.timer = null
this.events = ['scroll', 'mousedown', 'mousemove', 'keypress', 'scroll', 'touchstart']
this.init()
}
init() {
if(this.backToActiveOnXHR) {
this.events.push('load')
}
this.events.forEach(name => {
window.addEventListener(name, this.backToActive, true)
})
if(this.autoStart) {
this.backToActive()
}
}
goIdle = () => {
this.idle = true
if(!!this.idleCallback) {
this.idleCallback(this.timeout)
}
}
backToActive = () => {
if(this.idle) {
this.backToActiveCallback()
}
this.idle = false
clearTimeout(this.timer)
this.timer = setTimeout(this.goIdle, this.timeout * 1000)
}
}
用法:
let idleCallback = timeout => { console.log(`Went idle after ${timeout} seconds`) }
let backToActiveCallback = () => { console.log('Back to active') }
let idle = new Idle(30, idleCallback, backToActiveCallback)
devtools的结果:
// Went idle after 30 seconds <--- goes idle when no activity is detected
// Back to active <--- when the user is detected again
支持懒惰的好处:
setInterval(() => {
common.fetchApi('/api/v1/list', { status: idle.idle ? 'away' : 'online' }).then(/* show a list of elements */)
}, 1000 * 5)
你为什么要一张惰性支票?有时我们使用周期性的XHR(带setInterval),即当用户观看航班、乘车、电影、订单等列表时。对于每个XHR,我们可以添加关于他/她的活动状态(在线/离开)的信息,这样我们就可以了解系统中的活跃用户。
我的课程是基于Equiman和Frank Conijn的回答。