是否可以在Javascript中链接setTimeout函数？

Question

是否可以链接setTimout函数以确保它们相互运行？

提前致谢

Answer 1

当然。当第一个发射时，只需设置下一个。

setTimeout(function() {
    // do something
    setTimeout(function() {
        // do second thing
    }, 1000);
}, 1000);

---

你也可以让自己成为一个小实用程序对象，它可以让你真正地链接那些可以让你像这样链接调用的东西：

delay(fn1, 400).delay(fn2, 500).delay(fn3, 800);

function delay(fn, t) {
    // private instance variables
    var queue = [], self, timer;
    
    function schedule(fn, t) {
        timer = setTimeout(function() {
            timer = null;
            fn();
            if (queue.length) {
                var item = queue.shift();
                schedule(item.fn, item.t);
            }
        }, t);            
    }
    self = {
        delay: function(fn, t) {
            // if already queuing things or running a timer, 
            //   then just add to the queue
        	  if (queue.length || timer) {
                queue.push({fn: fn, t: t});
            } else {
                // no queue or timer yet, so schedule the timer
                schedule(fn, t);
            }
            return self;
        },
        cancel: function() {
            clearTimeout(timer);
            queue = [];
            return self;
        }
    };
    return self.delay(fn, t);
}

function log(args) {
    var str = "";
    for (var i = 0; i < arguments.length; i++) {
        if (typeof arguments[i] === "object") {
            str += JSON.stringify(arguments[i]);
        } else {
            str += arguments[i];
        }
    }
    var div = document.createElement("div");
    div.innerHTML = str;
    var target = log.id ? document.getElementById(log.id) : document.body;
    target.appendChild(div);
}


function log1() {
	  log("Message 1");
}
function log2() {
	  log("Message 2");
}
function log3() {
	  log("Message 3");
}

var d = delay(log1, 500)
    .delay(log2, 700)
    .delay(log3, 600)

或者，既然它现在是ES6 +中承诺的时代，这里有类似的代码使用promises，我们让promise基础设施为我们排队和排序：

// utility function for returning a promise that resolves after a delay
function delay(t) {
    return new Promise(function (resolve) {
        setTimeout(resolve, t);
    });
}

Promise.delay = function (fn, t) {
    // fn is an optional argument
    if (!t) {
        t = fn;
        fn = function () {};
    }
    return delay(t).then(fn);
}

Promise.prototype.delay = function (fn, t) {
    // return chained promise
    return this.then(function () {
        return Promise.delay(fn, t);
    });

}

function log(args) {
    var str = "";
    for (var i = 0; i < arguments.length; i++) {
        if (typeof arguments[i] === "object") {
            str += JSON.stringify(arguments[i]);
        } else {
            str += arguments[i];
        }
    }
    var div = document.createElement("div");
    div.innerHTML = str;
    var target = log.id ? document.getElementById(log.id) : document.body;
    target.appendChild(div);
}

function log1() {
    log("Message 1");
}

function log2() {
    log("Message 2");
}

function log3() {
    log("Message 3");
}

Promise.delay(log1, 500).delay(log2, 700).delay(log3, 600);

Answer 2

如果你使用的是针对ES6的Typescript，那么使用Async Await非常简单。这也是非常容易阅读和对承诺答案的一点升级。

//WARNING: this is Typescript source code
//expect to be async
async function timePush(...arr){
    function delay(t){
        return new Promise((resolve,reject)=>{
            setTimeout(()=>{
                resolve();
            },t)
        })
    }
    //for the length of this array run a delay
    //then log, you could always use a callback here
    for(let i of arr){
        //pass the items delay to delay function
        await delay(i.time);
        console.log(i.text)
    }
}


timePush(
    {time:1000,text:'hey'},
    {time:5000,text:'you'},
    {time:1000,text:'guys'}
);

Answer 3

受@jfriend00的启发，我展示了一个较短的版本：

Promise.resolve()
  .then(() => delay(400))
  .then(() => log1())
  .then(() => delay(500))
  .then(() => log2())
  .then(() => delay(800))
  .then(() => log3());

function delay(duration) {
  return new Promise((resolve) => {
    setTimeout(() => resolve(), duration);
  });
}

function log1() {
  console.log("Message 1");
}

function log2() {
  console.log("Message 2");
}

function log3() {
  console.log("Message 3");
}

Answer 4

我遇到了同样的问题。我的解决方案是通过setTimeout调用self，它可以工作。

＆＃13;

＆＃13;

let a = [[20,1000],[25,5000],[30,2000],[35,4000]];

function test(){
  let b = a.shift();
  console.log(b[0]);
  if(a.length == 0) return;
  setTimeout(test,b[1]);
}

＆＃13;

＆＃13;

＆＃13;

数组a中的第二个元素是延迟时间

Answer 5

使用 async / await 和 @Penny Liu 示例：

(async() => {
  await delay(400)
  log1()
  await delay(500)
  log2()
  await delay(800)
  log3()
})()

async function delay(duration) {
  return new Promise((resolve) => {
    setTimeout(() => resolve(), duration);
  });
}

function log1() {
  console.log("Message 1");
}

function log2() {
  console.log("Message 2");
}

function log3() {
  console.log("Message 3");
}