Is it possible to chain setTimeout functions in JavaScript?

Question

Is it possible to chain setTimout functions to ensure they run after one another?

Answer 1

Three separate approaches listed here:

1. Manually nest setTimeout() callbacks.
2. Use a chainable timer object.
3. Wrap setTimeout() in a promise and chain promises.

Manually Nest setTimeout callbacks

Of course. When the first one fires, just set the next one.

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

Chainable Timer Object

You can also make yourself a little utility object that will let you literally chain things which would let you chain calls like this:

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)

Wrap setTimeout in a Promise and Chain Promises

Or, since it's now the age of promises in ES6+, here's similar code using promises where we let the promise infrastructure do the queuing and sequencing for us. You can end up with a usage like this:

Promise.delay(fn1, 500).delay(fn2, 700).delay(fn3, 600);

Here's the code behind that:

// 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);

The functions you supply to this version can either by synchonrous or asynchronous (returning a promise).

Answer 2

Inspired by the Promise-based solution in jfriend00's answer, I demonstrated a shorter version:

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 3

With ES6, this is pretty simple using async/await. This is also very easy to read and a little upgrade to the promises answer.

// 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 4

I have encountered the same issue. My solution was to call self by setTimeout, it works.

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]);
}

the second element in array a is time to be delayed

Answer 5

Using async / await with @Penny Liu example:

(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");
}