Delay chained promise

Question

I\'m trying to do the following but it isn\'t working. How can I adjust the code to have a delay between .then and .done?

myServi         


        

Answer 1

You can create a simple delay function that returns a promise and use that in your promise chain:

function delay(t, val) {
   return new Promise(function(resolve) {
       setTimeout(function() {
           resolve(val);
       }, t);
   });
}

myService.new(a).then(function(temp) {
    return delay(60000, temp);
}).then(function(temp) {
    return myService.get(a, temp);
}).then(function (b) {
    console.log(b);
});

---

You could also augment the Promise prototype with a .delay() method (which some promise libraries like Bluebird already have built-in). Note, this version of delay passes on the value that it is given to the next link in the chain:

Promise.prototype.delay = function(t) {
    return this.then(function(val) {
        return delay(t, val);
    });
}

Then, you could just do this:

myService.new(a).delay(60000).then(function(temp) {
    return myService.get(a, temp);
}).then(function (b) {
    console.log(b);
});

Answer 2

You can create a .delay() function set at Promise.prototype which returns a new Promise having value of original promise.

Promise.prototype.delay = function(t) {
  return this.then(function(data) {
    return new Promise(function(resolve) {
      setTimeout(resolve, t || 0, data);
    })
  })
};

Promise.resolve(123)
.delay(6000)
.then(function(data) {
  console.log(data)
});

Answer 3

A slightly improved version of jfriend00's approach avoids needing the pass the resolved value like this:

function delay(t) {
  return (val) => {
    return new Promise((resolve, reject) => {
      setTimeout(() => resolve(val), t)
    })
  }
}

can be used like this:

myService.new(a)
  .then(delay(60000))
  .then(function(temp) {
    return myService.get(a, temp);
  }).then(function (b) {
    console.log(b);
  });