I'm confused about Promise!
I use Promise then without return like this:
new Promise((resolve, reject) => {
resolve("1");
}).then((v1) => {
console.log("v1");
new Promise((resolve, reject) => {
//Time-consuming operation, for example: get data from database;
setTimeout(() => {
resolve(2)
}, 3000);
}).then((v11) => {
console.log("v11");
})
}).then((v2) => {
console.log("v2")
});
I get this result v1 v2 v11.
Then, I use another way of writing, Like below:
new Promise((resolve, reject) => {
resolve("1");
}).then((v1) => {
console.log("v1");
return new Promise((resolve, reject) => {
setTimeout(() => {
resolve(2)
}, 3000)
}).then((v11) => {
console.log("v11");
})
}).then((v2) => {
console.log("v2")
});
I get another result v1 v11 v2.
Maybe, There is another case:
new Promise((resolve, reject) => {
resolve("1");
}).then((v1) => {
console.log("v1");
return new Promise((resolve, reject) => {
setTimeout(() => {resolve(2)}, 3000)
}).then((v11) => {
console.log("v11");
return new Promise((resolve, reject) => {
setTimeout(() => {resolve(2)}, 4000)
}).then(() => {
console.log("v12")
})
})
}).then((v2) => {
console.log("v2")
});
I get this result v1 v11 v12 v2
I can't understand the second return I want to know why I get this result?
It will be easier to understand the control flow if you actually print the values of the resolved promises and not only the names of the variables:
Version 1
new Promise((resolve, reject) => {
resolve("1");
}).then((v1) => {
console.log("v1:", v1);
new Promise((resolve, reject) => {
//Time-consuming operation, for example: get data from database;
setTimeout(() => {
resolve(2)
}, 3000);
}).then((v11) => {
console.log("v11:", v11);
})
}).then((v2) => {
console.log("v2:", v2)
});
Version 2
new Promise((resolve, reject) => {
resolve("1");
}).then((v1) => {
console.log("v1:", v1);
return new Promise((resolve, reject) => {
setTimeout(() => {
resolve(2)
}, 3000)
}).then((v11) => {
console.log("v11:", v11);
})
}).then((v2) => {
console.log("v2:", v2)
});
Version 3
new Promise((resolve, reject) => {
resolve("1");
}).then((v1) => {
console.log("v1:", v1);
return new Promise((resolve, reject) => {
setTimeout(() => {resolve(2)}, 3000)
}).then((v11) => {
console.log("v11:", v11);
return new Promise((resolve, reject) => {
setTimeout(() => {resolve(2)}, 4000)
}).then((v12) => {
console.log("v12:", v12)
})
})
}).then((v2) => {
console.log("v2:", v2)
});
Now you can see what gets passed to the callbacks:
Result 1
v1: 1
v2: undefined
v11: 2
Result 2
v1: 1
v11: 2
v2: undefined
Result 3
v1: 1
v11: 2
v12: 2
v2: undefined
Explanation
As you can see when in the .then() handlers you don't return a promise, it acts as if you returned an already resolved promise with value undefined - like if you did:
return Promise.resolve(undefined);
and thus the next .then() handler can be called immediately.
If, on the other hand, you return a promise that is not resolved yet, then the next .then() handler will not be invoked immediately but only after that returned promise gets resolved.
And that explains the order of execution that is different when you don't return a promise - and what happens is as if an already resolved promise got returned implicitly for you.
function one() {
return new Promise((resolve,reject) => {
setTimeout(function () {
console.log("one 1 ");
resolve('one one');
}, 2000);
});
}
function two() {
return new Promise((resolve,reject) => {
setTimeout(function () {
console.log("two 2 ");
resolve('two two');
}, 10000);
});
}
function three(){
setTimeout(function () {
console.log("three 3 ");
}, 5000);
}
one().then((msg) => {
console.log('one : ', msg);
return two();
}).then((msg) => {
console.log('two :', msg);
return three();
}).then((msg) => {
console.log('three :', msg);
})
.catch((error) => {
console.error('Something bad happened:', error.toString());
});
console three show undefined because three not parse resolve
Related
I have a similar problem in dialogflow fulfillment where I need to book any appointments on the Google calendar. I would reject all functions p1, p2 and p3 if only one of them is rejected. In the code below, although p2 is rejected, the others p1 and p3 are executed (I wish all functions p1, p2 and p3 were not performed).
function f1() {
return Promise.all([p1(), p2(), p3()])
.then(value => {
alert('ok');
})
.catch(err => {
console.log('err: ' + err);
});
}
function p1() {
new Promise((resolve, reject) => {
setTimeout(resolve, 1000, alert("one"));
});
}
function p2() {
new Promise((resolve, reject) => {
reject('reject');
});
}
function p3() {
new Promise((resolve, reject) => {
setTimeout(resolve, 3000, alert("three"));
});
}
f1();
Well it contains a lot of code to implement it so I would give short instructions.
You need to have a way to reject your request or whatever you are doing. For examples with axious we can use CancelToken to cancel HTTP request.
Now you need to subscribe on each request and cancel the request or whatever you are using.
It is not clear what exactly you need. Functions can not be rejected or not executed if you run them in parallel. You can only notify the internal function code that the cancel operation was requested from the outside. We don't know what async operations are performed inside your async functions. Demo
import { CPromise } from "c-promise2";
function makePromise(ms, label) {
return new CPromise((resolve, reject, { onCancel }) => {
setTimeout(resolve, ms);
onCancel(() => console.log(`onCancel(${label})`));
});
}
CPromise.all([
makePromise(2000, "one"),
makePromise(1500, "two").then(() => {
throw Error("Oops");
}),
makePromise(4000, "three")
]).then(
(v) => console.log(`Done: ${v}`),
(e) => console.warn(`Fail: ${e}`)
);
onCancel(one)
onCancel(three)
Fail: Error: Oops
The problem is you're not returning anything from p1, p2, p3. So when you call Promise.all([p1(), p2(), p3()]) which is actually calling with Promise.all([undefined, undefined, undefined])(resolves anyway) which does not have a rejection. That's why you're not seeing the error.
Add return in your functions.
function f1() {
return Promise.all([p1(), p2(), p3()])
.then(value => {
alert('ok');
})
.catch(err => {
console.log('err: ' + err);
});
}
function p1() {
return new Promise((resolve, reject) => {
setTimeout(resolve, 1000);
});
}
function p2() {
return new Promise((resolve, reject) => {
reject('reject');
});
}
function p3() {
return new Promise((resolve, reject) => {
setTimeout(resolve, 3000);
});
}
f1();
Remember Promises are not cancellable, if you really want cancel execution part, you can try something like this. I don't guarantee this works all the time and I don't think good idea to do in this way.
const timers = []
function f1() {
return Promise.all([p1(), p2(), p3()])
.then(value => {
alert('ok');
})
.catch(err => {
console.log('err: ' + err);
timers.forEach((timerId) => clearTimeout(timerId))
});
}
function p1() {
return new Promise((resolve, reject) => {
const timerId = setTimeout(() => {
alert(1)
resolve()
}, 1000);
timers.push(timerId)
});
}
function p2() {
return new Promise((resolve, reject) => {
reject('reject');
});
}
function p3() {
return new Promise((resolve, reject) => {
const timerId = setTimeout(() => {
alert(2)
resolve()
}, 3000);
timers.push(timerId)
});
}
f1();
This question already has answers here:
When should I use a return statement in ES6 arrow functions
(6 answers)
Closed 2 years ago.
console.log("Before");
getUser(1)
.then((user) => getRepositories(user.gitHubUsername))
.then((repos) => {
getCommits(repos[0]);
})
.then((commits) => {
console.log("commits: ", commits);
})
.catch((err) => console.log("Error: ", err.message));
console.log("After");
function getUser(id) {
return new Promise((resolve, reject) => {
setTimeout(() => {
console.log("Reading a user from a database...");
resolve({
id: id,
gitHubUsername: "mosh"
});
}, 2000);
});
}
function getRepositories(username) {
return new Promise((resolve, reject) => {
setTimeout(() => {
console.log("Calling GitHub API...");
resolve(["repo1", "repo2", "repo3"]);
}, 2000);
});
}
function getCommits(repo) {
return new Promise((resolve, reject) => {
setTimeout(() => {
console.log("Calling GitHub API...");
resolve("commit");
}, 2000);
});
}
now here evey function which is chained is contains a promise which is resolving something i.e username, array etc. but when i run this code on the console it displays commnits : undefined
you can see it in this attached image
You missed the return:
console.log("Before");
getUser(1)
.then((user) => getRepositories(user.gitHubUsername))
.then((repos) => {
return getCommits(repos[0]);
})
.then((commits) => {
console.log("commits: ", commits);
})
.catch((err) => console.log("Error: ", err.message));
console.log("After");
function getUser(id) {
return new Promise((resolve, reject) => {
setTimeout(() => {
console.log("Reading a user from a database...");
resolve({
id: id,
gitHubUsername: "mosh"
});
}, 2000);
});
}
function getRepositories(username) {
return new Promise((resolve, reject) => {
setTimeout(() => {
console.log("Calling GitHub API...");
resolve(["repo1", "repo2", "repo3"]);
}, 2000);
});
}
function getCommits(repo) {
return new Promise((resolve, reject) => {
setTimeout(() => {
console.log("Calling GitHub API...");
resolve("commit");
}, 2000);
});
}
I'm new to js and I'm learning promises. I came up with this code which will print the resolved values from every function and will call new functions using .then
function login() {
return new Promise((resolve, reject) => {
setTimeout(() => {
resolve({username : 'default'})
}, 1000)
})
}
function getVideos() {
return new Promise((resolve, reject) => {
setTimeout(() => {
resolve(['vid1', 'vid2'])
},1000)
})
}
function getDesc() {
return new Promise((resolve, reject) => {
setTimeout(() => {
resolve('desc')
}, 1000)
})
}
const test = login()
test.then(res => {
console.log(res)
getVideos()
})
.then(res => {
console.log(res)
getDesc()
})
.then(res => console.log(res))
But, I'm not getting the expected result, I thought all the statements in .then() needed to be executed and resolved to continue to the next .then() statement. But clearly this is not the case here as I'm getting the following as the output -
{ username: 'default' }
undefined
undefined
But I expected the output to be similar to this -
{username : 'default}
['vid1', 'vid2']
desc
pls, point where I'm going wrong here. Any help is appreciated, thanks a lot
You need to add return to inside your chains when you're calling those functions. For example:
function login() {
return new Promise((resolve, reject) => {
setTimeout(() => {
resolve({username : 'default'})
}, 1000)
})
}
function getVideos() {
return new Promise((resolve, reject) => {
setTimeout(() => {
resolve(['vid1', 'vid2'])
},1000)
})
}
function getDesc() {
return new Promise((resolve, reject) => {
setTimeout(() => {
resolve('desc')
}, 1000)
})
}
const test = login()
test.then(res => {
console.log(res)
return getVideos()
})
.then(res => {
console.log(res)
return getDesc()
})
.then(res => console.log(res))
This is a good guide: https://javascript.info/promise-chaining
Make a couple of very minor changes:
const test = login()
test.then(res => {
console.log(res)
return getVideos()
})
.then(res => {
console.log(res)
return getDesc()
})
.then(res => console.log(res))
To chain promises together, you need to return the promises at each step. As it is you're implicitly returning undefined which is not a promise.
I need to call multiple promises inside a for loop, but it gives me unhandled promise exception during each run. The way to solve this is to return the second promise, this way the last catch would be executed and it would work without errors, but - second and more iterations would not be executed.
const doFirstThing = () => {
return new Promise((resolve, reject) => {
setTimeout(() => (resolve(['a', 'b', 'c'])), 1000);
});
}
const doSecondThing = () => {
return new Promise((resolve, reject) => {
setTimeout(() => (reject('test')), 500); // reject here
});
}
doFirstThing()
.then(results => {
results.forEach(result => {
doSecondThing(result)
.then(() => console.log(result, 'done'));
});
})
.catch(error => console.log(error));
How can I deal with this?
To prevent unhandled promise exception chain .catch() to lat .then(); substitute using Promise.all() and .map() for .forEach()
const doFirstThing = () => {
return new Promise((resolve, reject) => {
setTimeout(() => (resolve(['a', 'b', 'c'])), 1000);
})
}
const doSecondThing = (r) => {
return new Promise((resolve, reject) => {
setTimeout(() => (reject('test')), 500); // reject here
})
}
doFirstThing()
.then(results => {
return Promise.all(results.map(result => {
return doSecondThing(result)
}));
})
.then((result) => console.log(result, 'done'))
.catch(error => console.log(`err:${error}`));
Here is my situation:
fetchData(foo).then(result => {
return new Promise((resolve, reject) => {
setTimeout(() => {
resolve(result + bar);
}, 0)
});
}).then(result => {
return new Promise((resolve, reject) => {
setTimeout(() => {
resolve( result + woo);
}, 0)
});
}).then(result => {
setTimeout(() => {
doSomething(result);
}, 0)
});
Where each setTimeout is a different async operation using the callback pattern.
It is really painfull to wrap each function inside a promise, I feel like the code should look more like this:
fetchData(foo).then(result => {
setTimeout(() => {
return result + bar;
}, 0)
}).then(result => {
setTimeout(() => {
return result + woo;
}, 0)
}).then(result => {
setTimeout(() => {
doSomething(result);
}, 0)
});
Obviously this doesn't work.
Am I using Promises right? Do I really have to wrap all existing async function in promises?
EDIT:
Actually I realize my example was not totally reprensentative of my situation, I did not make it clear that the setTimeout in my example is meant to reprensent en async operation. This situation is more representative of my situation:
fetchData(foo).then(result => {
return new Promise((resolve, reject) => {
asyncOperation(result, operationResult => {
resolve(operationResult);
}, 0)
});
}).then(result => {
return new Promise((resolve, reject) => {
otherAsyncOperation(result, otherAsyncResult => {
resolve( otherAsyncResult);
}, 0)
});
}).then(result => {
doSomething(result);
});
Am I using Promises right? Do I really have to wrap all existing async function in promises?
Yes. Yes.
I feel like the code should look more like this
No, it shouldn't. It rather should look like this:
function promiseOperation(result) {
return new Promise((resolve, reject) => {
asyncOperation(result, resolve, 0)
});
}
function otherPromiseOperation(result) {
return new Promise((resolve, reject) => {
otherAsyncOperation(result, resolve, 0)
});
}
fetchData(foo).then(promiseOperation).then(otherPromiseOperation).then(doSomething);
It is really painfull to wrap each function inside a promise
Well, don't repeatedly write it out every time. You can abstract this wrapping into a function!
function promisify(fn) {
return value => new Promise(resolve => {
fn(value, resolve, 0)
});
}
const promiseOperation = promisify(asyncOperation);
const otherPromiseOperation = promisify(otherAsyncOperation);
fetchData(foo).then(promiseOperation).then(otherPromiseOperation).then(doSomething);
Notice that most promise libraries come with a such a promisification function included, so your whole code reduces to these three lines.
You are using promise right. Just a small note on the first snippet of code: you are not returning a promise from the last then() callback:
...
}).then(result => {
setTimeout(() => {
doSomething(result);
}, 0)
});
This is correct if you need simply to do an async operation without returning to the caller of fetchData the value of the last async operation. If you need to return this value, you need to convert to promise this operation too:
fetchData(foo).then(result => {
return new Promise((resolve, reject) => {
setTimeout(() => {
resolve(result + bar);
}, 0)
});
}).then(result => {
return new Promise((resolve, reject) => {
setTimeout(() => {
resolve(result + woo);
}, 0)
});
}).then(result => {
return new Promise((resolve, reject) => {
setTimeout(() => {
resolve(doSomething(result));
}, 0)
});
});
Here I suppose doSomething is a sync function returning a value.
Said so, if you want to reduce the noise of create the promise to wrap setTimeout every time, you can create a utility function setTimeoutWithPromise:
function setTimeoutWithPromise(operation, millisec) {
return new Promise((resolve, reject) => {
setTimeout(() => {
resolve(operation());
}, millisec)
});
}
And clean your code:
fetchData(foo)
.then(result => setTimeoutWithPromise(() => result + bar, 0))
.then(result => setTimeoutWithPromise(() => result + woo, 0))
.then(result => setTimeoutWithPromise(() => doSomething(result), 0));