Return promise from promise

Question

I wonder how to return promise from promise. E.g.

I have such construction:

doAsyncStuff()  // a promise
  .then( function(res) {
    doAnotherAsyncStuff(res)  // another promise
      .then( makeSomeThings )
      .then( function(anotherRes, opts) {
        ...
      })
    })
  .then( ... )

I want to write it like this:

doAsyncStuff()  // a promise
  .then( function(res) {
    doAnotherAsyncStuff(res)  // another promise
      .then( makeSomeThings )
      // and somehow push-out promise
  })
  .then( function(anotherRes) {
     ...
  })
  .then( ... )

How can I achieve such result?

the problem thing

var Promise = require('bluebird');
//noinspection JSUnresolvedFunction
var bcrypt = Promise.promisifyAll(require('bcrypt'));
var Sequelize = require('sequelize');
var config = require('config');

var sequelize = new Sequelize(config.get('db.connstring'));


//noinspection JSUnresolvedFunction
var User = sequelize.define('user', {
  name: {
    type: Sequelize.STRING
  },
  email: {
    type: Sequelize.STRING,
    validate: {
      isEmail: true
    }
  },
  passwordHash: {
    type: Sequelize.STRING
  },
  isConfirmed: {
    type: Sequelize.BOOLEAN,
    allowNull: false,
    defaultValue: false
  }
}, {
  freezeTableName: true,
  classMethods: {
    login: Promise.method(function (email, password) {
      if (!email || !password) throw new Error('Email and password are both required');
      var rv = this
        .find({where: {email: email.toLowerCase().trim()}})
        .then(function (user) {

          return bcrypt.compareAsync(password, user.passwordHash).then(function (res) {
            console.log(email, password, res);
          });
          // if i dont use pacthed compare here, i have no problem ..
          // return bcrypt.compare(password, user.passwordHash, function(err, res) {
          //    console.log(email, password, res);
          //  });
        });
      console.log('B', rv);
      return rv;
    })
  }
});

sequelize.sync({force: true}).then(function () {
  var pwd = 'pwd';
  //noinspection JSUnresolvedFunction
  bcrypt.hashAsync(pwd, 4).then(function (salt) {
    var u1 = User.create({
      name: 'u1',
      email: 'u1@ex.com',
      passwordHash: salt
    }).then(function (result) {
      User.login('u1@ex.com', pwd).then(function (res) {
        console.log('A', res)
      })
    });
  });
});

Here is a good article about promises, if you want to figure out: http://pouchdb.com/2015/05/18/we-have-a-problem-with-promises.html — Andrew, Feb 06 '16 at 12:22

score 5 · Answer 1 · answered May 04 '15 at 00:33

5

Just return your other promise

doAsyncStuff()  // a promise
  .then( function(res) {
    return doAnotherAsyncStuff(res)  // another promise
  })
  .then( function(anotherRes) {
     ...
  })
  .then( ... )

answered May 04 '15 at 00:33

Transcendence

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Asad Saeeduddin · Answer 2 · 2015-05-04T00:53:33.247

2

If a function represents the transformation of the previous (resolved) result to either the next result, or a promise which will resolve to the next result, you just need to pass it to then directly.

function return1() {
    return Promise.resolve(1);
}

function wait(ms) {
    return new Promise(function (resolve) {
        setTimeout(resolve, ms);
    });
}

function increment(val) {
    return wait(1000).then(function () {
        return val + 1;
    });
}

function square(val) {
    return wait(1000).then(function () {
        return val * val;
    });
}

var p = return1()
    .then(increment)
    .then(square); // Returns a promise that will resolve to 4 (eventually)

p.then(function (result) { console.log(result); });

Demo here: http://jsfiddle.net/Lzxtuu1b/

edited May 04 '15 at 00:53

answered May 04 '15 at 00:33

Asad Saeeduddin

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@akaRem I've tried to address this in my answer. In the `increment` function, I call `wait`, which is analogous to your `doAnotherAsyncStuff` function and returns a promise. I chain it with a function that returns an incremented value (analogously to your `makeSomeThings`). Then I just return this promise. The outer `then` takes care of resolving the function before passing it to the next callback. – Asad Saeeduddin May 04 '15 at 00:58
thank you very much for your edits. I've added example where ive stuck.. Could you help? – akaRem May 04 '15 at 01:03
@akaRem I see you've added another code snippet, but you haven't actually explained what the problem is. – Asad Saeeduddin May 04 '15 at 01:43

score 0 · Answer 3 · edited May 23 '17 at 11:54

As the other answers already suggested, you must return from your then callbacks. Whether you return a plain value or a promise for a value doesn't matter, both will work; but when you return nothing then the resulting promise will be resolved with undefined.

In your particular examples, it is here:

doAsyncStuff()  // a promise
  .then( function(res) {
    return doAnotherAsyncStuff(res)  // another promise
//  ^^^^^^
      .then( makeSomeThings )
   …

…
  .then(function (user) {
      return bcrypt.compareAsync(password, user.passwordHash).then(function (res) {
        console.log(email, password, res);
        return res;
//      ^^^^^^
      });

An in your last example, you can even unnest al lot calls when you use return:

var pwd = 'pwd';
sequelize.sync({force: true}).then(function () {
  //noinspection JSUnresolvedFunction
  return bcrypt.hashAsync(pwd, 4);
}).then(function (salt) {
  return User.create({
    name: 'u1',
    email: 'u1@ex.com',
    passwordHash: salt
  });
}).then(function (result) {
  return User.login('u1@ex.com', pwd);
}).then(function (res) {
   console.log('A', res);
});

Return promise from promise

3 Answers3

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