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Why does the function getSafestCountriesNames() return promise{pending} when i call it and when i use async/await it returns undefined?

This is the function:
export const getSafestCountriesNames = async() => {
try{
const names = await getCountriesData();
names.forEach((country) => {
const {score} = country.advisory;
if(score == 0){
const liEl = createDOMElement('li', {className: 'name'});
liEl.innerHTML = `${country.name}`;
return liEl;
}
});
}
// …
}
When I am calling it in here I get undefined:
async function func(){
const names = await getSafestCountriesNames();
console.log(names);
}
func();

It returns Promise{pending} because when a function has the async keyword, it has to return a Promise. See here.
Nonetheless there's a deeper issue in your code. Your return statement is inside the callback of the forEach. That's the return statement of the callback, not the one of getSafestCountriesNames.
Since you need to return a Promise (for you're using await), you can either wrap your logic inside something like:
return new Promise((resolve, reject) => {
// your logic here, resolving or rejecting instead of just returning
})
or keeping things simple: return Promise.resolve("something")
When you call the function with await, it resolves undefined because you're not resolving anything in your Promise.

Conditionally chaining promises inside and outside of a loop in javascript

In the following code, when the condition within the loop is satisfied we can't really be sure in which order are both asyncFunction called.
siblings.forEach(function(sibling, index) {
// This condition may only be satisfied once (only one sibling may have data-selected true)
if (sibling.getAttribute('data-selected') == 'true') {
asyncFunction(sibling);
}
})
asyncFunction(element);
Since I know asyncFunction() returns a promise I could chain those to ensure the order:
asyncFunction(sibling).then(asyncFunction(element));
But how to do this taking the condition into account ?
I considered wrapping the loop in a promise that resolves when the condition is satisfied or after the loop ends, but it seems a tad convoluted.
// Untested
function checkSiblings(siblings) {
let promise = new Promise(function(resolve, reject) {
siblings.forEach(function(sibling, index) {
if (sibling.getAttribute('data-selected') == 'true') {
resolve(asyncFunction(sibling));
}
})
resolve();
})
}
checkSiblings(siblings).then(asyncFunction(element));
I am sure this has been addressed before, but I can't find the right search keyboards.
Thank you

If only one of the siblings can meet the criteria, using find will simplify your code:
const selected = [...siblings].find( elem => elem.getAttribute('data-selected') == 'true' );
Mark checkSiblings as async so it always returns a Promise, and have it return the asyncFunction Promise when appropriate:
return selected ? asyncFunction(selected) : null;
Then you can chain them as before:
checkSiblings(siblings).then(() => asyncFunction(element));
All together you’d have something like this:
async function checkSiblings (siblings) {
const selected = [...siblings].find( elem => elem.getAttribute('data-selected') == 'true' );
return selected ? asyncFunction(selected) : null
}
checkSiblings(siblings).then(() => asyncFunction(element));

There are several ways. One of them is using Array.reduce to resolve promises sequentially:
[...siblings].reduce(
async (prev, sibling)) => {
// Await initial/previous promise (discard results)
await prev;
// This condition may only be satisfied once
// (only one sibling may have data-selected true)
if (sibling.getAttribute('data-selected') == 'true') {
// Return the async function promise
return asyncFunction(sibling);
}
// async functions always returns a promise.
// No need to return anything
// Initial promise autoresolved
}, Promise.resolve())
// Once all have been executed
.then( () => asyncFunction(element));

How to make async/await actually wait [duplicate]

I read that async functions marked by the async keyword implicitly return a promise:
async function getVal(){
return await doSomethingAync();
}
var ret = getVal();
console.log(ret);
but that is not coherent...assuming doSomethingAsync() returns a promise, and the await keyword will return the value from the promise, not the promise itsef, then my getVal function should return that value, not an implicit promise.
So what exactly is the case? Do functions marked by the async keyword implicitly return promises or do we control what they return?
Perhaps if we don't explicitly return something, then they implicitly return a promise...?
To be more clear, there is a difference between the above and
function doSomethingAync(charlie) {
return new Promise(function (resolve) {
setTimeout(function () {
resolve(charlie || 'yikes');
}, 100);
})
}
async function getVal(){
var val = await doSomethingAync(); // val is not a promise
console.log(val); // logs 'yikes' or whatever
return val; // but this returns a promise
}
var ret = getVal();
console.log(ret); //logs a promise
In my synopsis the behavior is indeed inconsistent with traditional return statements. It appears that when you explicitly return a non-promise value from an async function, it will force wrap it in a promise.
I don't have a big problem with it, but it does defy normal JS.

The return value will always be a promise. If you don't explicitly return a promise, the value you return will automatically be wrapped in a promise.
async function increment(num) {
return num + 1;
}
// Even though you returned a number, the value is
// automatically wrapped in a promise, so we call
// `then` on it to access the returned value.
//
// Logs: 4
increment(3).then(num => console.log(num));
Same thing even if there's no return! (Promise { undefined } is returned)
async function increment(num) {}
Same thing even if there's an await.
function defer(callback) {
return new Promise(function(resolve) {
setTimeout(function() {
resolve(callback());
}, 1000);
});
}
async function incrementTwice(num) {
const numPlus1 = await defer(() => num + 1);
return numPlus1 + 1;
}
// Logs: 5
incrementTwice(3).then(num => console.log(num));
Promises auto-unwrap, so if you do return a promise for a value from within an async function, you will receive a promise for the value (not a promise for a promise for the value).
function defer(callback) {
return new Promise(function(resolve) {
setTimeout(function() {
resolve(callback());
}, 1000);
});
}
async function increment(num) {
// It doesn't matter whether you put an `await` here.
return defer(() => num + 1);
}
// Logs: 4
increment(3).then(num => console.log(num));
In my synopsis the behavior is indeed inconsistent with traditional
return statements. It appears that when you explicitly return a
non-promise value from an async function, it will force wrap it in a
promise. I don't have a big problem with it, but it does defy normal
JS.
ES6 has functions which don't return exactly the same value as the return. These functions are called generators.
function* foo() {
return 'test';
}
// Logs an object.
console.log(foo());
// Logs 'test'.
console.log(foo().next().value);

Yes, an async function will always return a promise.
According to the tc39 spec, an async function desugars to a generator which yields Promises.
Specifically:
async function <name>?<argumentlist><body>
Desugars to:
function <name>?<argumentlist>{ return spawn(function*() <body>, this); }
Where spawn "is a call to the following algorithm":
function spawn(genF, self) {
return new Promise(function(resolve, reject) {
var gen = genF.call(self);
function step(nextF) {
var next;
try {
next = nextF();
} catch(e) {
// finished with failure, reject the promise
reject(e);
return;
}
if(next.done) {
// finished with success, resolve the promise
resolve(next.value);
return;
}
// not finished, chain off the yielded promise and `step` again
Promise.resolve(next.value).then(function(v) {
step(function() { return gen.next(v); });
}, function(e) {
step(function() { return gen.throw(e); });
});
}
step(function() { return gen.next(undefined); });
});
}

Your question is: If I create an async function should it return a promise or not? Answer: just do whatever you want and Javascript will fix it for you.
Suppose doSomethingAsync is a function that returns a promise. Then
async function getVal(){
return await doSomethingAsync();
}
is exactly the same as
async function getVal(){
return doSomethingAsync();
}
You probably are thinking "WTF, how can these be the same?" and you are right. The async will magically wrap a value with a Promise if necessary.
Even stranger, the doSomethingAsync can be written to sometimes return a promise and sometimes NOT return a promise. Still both functions are exactly the same, because the await is also magic. It will unwrap a Promise if necessary but it will have no effect on things that are not Promises.

Just add await before your function when you call it :
var ret = await getVal();
console.log(ret);

async doesn't return the promise, the await keyword awaits the resolution of the promise. async is an enhanced generator function and await works a bit like yield
I think the syntax (I am not 100% sure) is
async function* getVal() {...}
ES2016 generator functions work a bit like this. I have made a database handler based in top of tedious which you program like this
db.exec(function*(connection) {
if (params.passwd1 === '') {
let sql = 'UPDATE People SET UserName = #username WHERE ClinicianID = #clinicianid';
let request = connection.request(sql);
request.addParameter('username',db.TYPES.VarChar,params.username);
request.addParameter('clinicianid',db.TYPES.Int,uid);
yield connection.execSql();
} else {
if (!/^\S{4,}$/.test(params.passwd1)) {
response.end(JSON.stringify(
{status: false, passwd1: false,passwd2: true}
));
return;
}
let request = connection.request('SetPassword');
request.addParameter('userID',db.TYPES.Int,uid);
request.addParameter('username',db.TYPES.NVarChar,params.username);
request.addParameter('password',db.TYPES.VarChar,params.passwd1);
yield connection.callProcedure();
}
response.end(JSON.stringify({status: true}));
}).catch(err => {
logger('database',err.message);
response.end(JSON.stringify({status: false,passwd1: false,passwd2: false}));
});
Notice how I just program it like normal synchronous particularly at
yield connection.execSql and at yield connection.callProcedure
The db.exec function is a fairly typical Promise based generator
exec(generator) {
var self = this;
var it;
return new Promise((accept,reject) => {
var myConnection;
var onResult = lastPromiseResult => {
var obj = it.next(lastPromiseResult);
if (!obj.done) {
obj.value.then(onResult,reject);
} else {
if (myConnection) {
myConnection.release();
}
accept(obj.value);
}
};
self._connection().then(connection => {
myConnection = connection;
it = generator(connection); //This passes it into the generator
onResult(); //starts the generator
}).catch(error => {
reject(error);
});
});
}

Is it possible to return a value from then()? [duplicate]

This question already has answers here:
How do I return the response from an asynchronous call?
(41 answers)
Closed 4 years ago.
I have read that a when you return a value from then() it gets wrapped in a Promise and gets returned as a promise. Is there a way to return just a value and NOT a promise object?
function TestPromise() {
//someAsyncFunction returns a Promise.
return someAsyncFunction();
}
//Assume valueFromPromise = 2;
//After the addition newValue becomes 3
//And we return 3 instead of 3 wrapped up in a Promise.
function someFunction() {
return latestValue = TestPromise().then(valueFromPromise => {
newValue = valueFromPromise + 1;
return newValue;
});
}
function main() {
let value = someFunction();
//Use value here to do something.
}
I know what I am asking can be achieved via async/await. But I want to properly understand Promises.
Thanks

It may help to realize what is happening under the hood. Promises are really just syntax sugar around the old callbacks.
function TestPromise() {
//someAsyncFunction returns a Promise.
return someAsyncFunction();
}
function someAsyncFunction() {
return new Promise(function(resolve, reject) {
fetchSomeValue()
.then(function(someValue) {
resolve(someValue);
});
}
Until the Promise has called the resolve() callback, there is no way to get that data. async/await really is quite clean, I'm not sure why you would want to avoid it.
function TestPromise() {
//someAsyncFunction returns a Promise.
return someAsyncFunction();
}
//Assume valueFromPromise = 2;
//After the addition newValue becomes 3
//And we return 3 instead of 3 wrapped up in a Promise.
async function someFunction() {
const valueFromPromise = await TestPromise();
return valueFromPromise + 1;
}
async function main() {
let value = await someFunction();
//Use value here to do something.
}

Short answer to the question in title:
no with vanilla promises, and yes with async await.
Long answer:
Value produced from TestPromise() can only be accessed inside the subsequent .then or .catch when working with vanilla Promises.
And then itself is creating and returning a Promise Object every time you call it, so you can have your Promise resolutions for every then separate.
Say:
const p = new Promise((resolve) => {
setTimeout(() => resolve(1), 2000);
}).then((val) => {
console.log(val);
return new Promise((resolve) => {
setTimeout(() => resolve(2), 1000);
})
}).then((val) => {
console.log(val);
return 3
});
p.then(console.log);
Async/Await is just a sugar over Promises and Generators which does the aforementioned for you and unwraps the promise value and passes it to you like here:
async function someFunction() {
const valueFromPromise = /* here it passes the unwrapped value back after await */await TestPromise();
return valueFromPromise + 1;
}
But still async function itself is a promise, so you have to receive its value either as attaching another then to async function like this:
someFunction().then(finalValue => console.log(finalValue)
or by using it within another async function again with unwrapping through await like this:
async function outerAsyncFunction() {
const finalValue = await someFunction();
console.log(finalValue);
}

How to execute promises in series?

var promiseReturningFuncs = [];
for(var i = 0; i < 5; i++){
promiseReturningFuncs.push(askQuestion);
}
var programmers = [];
Promise.reduce(promiseReturningFuncs, function(resp, x) {
console.log(typeof resp);
if(typeof resp != "function") {
programmers.push(resp);
}
return x();
})
.then(function(resp) {
programmers.push(resp);
console.log(programmers);
});
My goal: execute the askQuestion function in series and resolve an array of objects created by that function. (this function must execute in series so that it can respond to user input)
So imagine that the askQuestion function returns a promise that resolves a object I want to add to an array.
This is my messy way of doing it.
I am looking to find a cleaner way of doing it, ideally, i wouldn't even need to push to an array, I would just have a final .then, where the response is an array.

Since you appear to be using the Bluebird promise library, you have a number of built-in options for sequencing your promise returning functions. You can use Promise.reduce(), Promise.map() with a concurrency value of 1, Promise.mapSeries or Promise.each(). If the iterator function returns a promise, all of these will wait for the next iteration until that promise resolves. Which to use depends more upon the mechanics of how your data is structured and what result you want (neither of which you actually show or describe).
Let's suppose you have an array of promise returning functions and you want to call them one at a time, waiting for the one to resolve before calling the next one. If you want all the results, then I'd suggest Promise.mapSeries():
let arrayOfPromiseReturningFunctions = [...];
// call all the promise returning functions in the array, one at a time
// wait for one to resolve before calling the next
Promise.mapSeries(arrayOfPromiseReturningFunctions, function(fn) {
return fn();
}).then(function(results) {
// results is an array of resolved results from all the promises
}).catch(function(err) {
// process error here
});
Promise.reduce() could also be used, but it would accumulate a single result, passing it from one to the next and end with one final result (like Array.prototype.reduce() does).
Promise.map() is a more general version of Promise.mapSeries() that lets you control the concurrency number (the number of async operations in flight at the same time).
Promise.each() will also sequence your functions, but does not accumulate a result. It assumes you either don't have a result or you are accumulating the result out-of-band or via side effects. I tend to not like to use Promise.each() because I don't like side effect programming.

You could solve this in pure JS using ES6 (ES2015) features:
function processArray(arr, fn) {
return arr.reduce(
(p, v) => p.then((a) => fn(v).then(r => a.concat([r]))),
Promise.resolve([])
);
}
It applies the function given to the array in series and resolves to an array of the results.
Usage:
const numbers = [0, 4, 20, 100];
const multiplyBy3 = (x) => new Promise(res => res(x * 3));
// Prints [ 0, 12, 60, 300 ]
processArray(numbers, multiplyBy3).then(console.log);
You'll want to double check browser compatibility but this works on reasonably current Chrome (v59), NodeJS (v8.1.2) and probably most others.

You can use recursion so that you can move to the next iteration in a then block.
function promiseToExecuteAllInOrder(promiseReturningFunctions /* array of functions */) {
var resolvedValues = [];
return new Promise(function(resolve, reject) {
function executeNextFunction() {
var nextFunction = promiseReturningFunctions.pop();
if(nextFunction) {
nextFunction().then(function(result) {
resolvedValues.push(result);
executeNextFunction();
});
} else {
resolve(resolvedValues);
}
}
executeNextFunction();
}
}

Executing one after another using a recursive function( in a non promise way):
(function iterate(i,result,callback){
if( i>5 ) callback(result);askQuestion().then(res=>iterate(i+1,result.concat([res]),callback);
})(0,[],console.log);
For shure this can be wrapped in a promise:
function askFive(){
return new Promise(function(callback){
(function iterate(i,result){
if( i>5 ) callback(result);askQuestion().then(res=>iterate(i+1,result.concat([res]),callback);
})(0,[],console.log);
});
}
askFive().then(console.log);
Or:
function afteranother(i,promise){
return new Promise(function(resolve){
if(!i) return resolve([]);
afteranother(i-1,promise).then(val=>promise().then(val2=>resolve(val.concat([val2])));
});
}
afteranother(5,askQuestion).then(console.log);