Bluebird Promises Join behaviour

问题:

If I execute the following code with Node.js

var Promise = require('bluebird');

Promise.join(
    function A() { console.log("A"); },
    function B() { console.log("B"); }
).done(
    function done() { console.log("done");}
);

The console will log

B
done

However I would expect

A
B
done

or

B
A
done

If it set a break point in function A it is never reached. Why is it that it processes B but not A?

回答1:

Promise.join takes promises as all its arguments but its last one, which is a function.

Promise.join(Promise.delay(100), request("http://...com"), function(_, res){
       // 100 ms have passed and the request has returned.
});

You're feeding it two functions so it does the following:

• Make a promise over function A() { ... } - basically returning a promise over it
• When it's done (immediately) execute the last argument, function B() {... } logging it.

See the docs:

Promise.join(Promise|Thenable|value promises..., Function handler) -> Promise

For coordinating multiple concurrent discrete promises. While .all() is good for handling a dynamically sized list of uniform promises, Promise.join is much easier (and more performant) to use when you have a fixed amount of discrete promises that you want to coordinate concurrently, for example:

var Promise = require("bluebird");

var join = Promise.join;

join(getPictures(), getComments(), getTweets(),

     function(pictures, comments, tweets) {

     console.log("in total: " + pictures.length + comments.length + tweets.length);

});

---

Update:

JSRishe came up with another clever way to solve this sort of pattern in this answer which looks something like:

Promise.delay(100).return(request("http://...com").then(function(res){
    // 100 ms have passed and the request has returned 
});

This works because the request already happens before the delay returns since the function is called in the same scope.