使用Q库时Node.js承诺不起作用

Question

I am trying to learn promises on my own. This is the code I wrote -

var Q = require('q');

var promise = Q.fcall(function() {
  // I expect this time out to delay returning the response 7.
  setTimeout( console.log('hi'), 1000 );
  return 7;
});

promise.then(function(contents) {
  console.log(contents);
});
// Added this timeout so that the javascript execution context(node ex.js) remains alive before the code in the then block is resolved.
setTimeout( function(){console.log('bye');}, 1000 );

Now this is not printing the contents. I just get C:\\node\\Ex_Files_UaR_Node\\first>node example3.js hi bye

I was expecting to get - hi 7 bye

Please let me know if there is anything very much evident which I am missing.

EDIT:

PS The below code resolves the promise -

var Q = require('q');

var promise = Q.fcall(function() {
  setTimeout( function(){console.log('hi');}, 1000 );
  return 7;
});

promise.then(function(contents) {
  console.log(contents);
});

setTimeout( function(){console.log('bye');}, 1000 );

However the basic idea behind adding setTimeOut in Q.fcall was to delay the execution of the promise. Any idea how to do that ?

Answer 1

There you go

let p = new Promise((resolve, reject) => {  
   setTimeout(() => resolve(4), 2000);
});

p.then((res) => {  
  console.log(res);
});

ES6 Promises are simpler. p is resolved only after 2 seconds. Till then, It is not resolved, neither rejected. So then executes only after 2 seconds

Answer 2

There is a problem with your 1st log statement.

I've modified your code to work properly.

var Q=require('q');
var promise = Q.fcall(function() {
    setTimeout( function(){console.log('hi')}, 1000 );
    return 7;
})
promise.then(function(contents) {
    console.log(contents);
});

setTimeout(function(){console.log('bye');}, 1000 );

But this code will print 7 , bye and hi in respective order because, the following actions happens syncronously:

• Promise gets registered with the call to the function fcall() .
• The promise variable registers a handler using then() method.
• Your timeout function, which is aiming at logging 'bye', is registered.

Now the following things happens asynchronously in the following order:

• Your registered promise callback, where you are registering a timeout function to print 'hi' and returning value 7, is called as callback from fcall() .
• Now the setTimeout registers the function aiming to log 'hi' and returns value 7.
• Promise gets resolved with value 7 and it logs the value to console.
• The timeout for 'bye' gets expired and logs it to the console.
• The timeout for 'hi' gets expired and logs it to the console.

Hope it makes the flow clear.

Answer 3

I could not reproduce the problem of not getting the 7 in the output (even though your code has several logical mistakes), as this snippet (without any of the corrections) does produce it:

 // This is browser version, so `require` is replaced by `script src` tag. //var Q = require('q'); var promise = Q.fcall(function() { // I expect this time out to delay returning the response 7. setTimeout( console.log('hi'), 1000 ); return 7; }); promise.then(function(contents) { console.log(contents); }); // Added this timeout so that the javascript execution context(node ex.js) remains alive before the code in the then block is resolved. setTimeout( function(){console.log('bye');}, 1000 ); 

 <script src="https://cdnjs.cloudflare.com/ajax/libs/q.js/1.5.0/q.js"></script> 

There are however several issues:

• The first argument to the first setTimeout is not a function: you pass it console.log('hi') instead of function(){console.log('hi');} .
• return 7 will be executed immediately. If you want a delay for the initial promise to settle, then Q.fcall is not the method you need: you need Q.Promise instead.

Here is how your intended behaviour would be coded then:

 var promise = Q.Promise(function (resolve, reject) { setTimeout( function () { console.log('hi'); resolve(7); }, 1000 ); }); promise.then(function(contents) { console.log(contents); }); setTimeout( function(){console.log('bye');}, 1000 ); 

 <script src="https://cdnjs.cloudflare.com/ajax/libs/q.js/1.5.0/q.js"></script> 

Note that Node has a native Promise library with which you can do the same. There is really no need to include Q. Just replace Q.Promise(...) with new Promise(...) :

 var promise = new Promise(function (resolve, reject) { setTimeout( function () { console.log('hi'); resolve(7); }, 1000 ); }); promise.then(function(contents) { console.log(contents); }); setTimeout( function(){console.log('bye');}, 1000 ); 

Note that there is a slight difference in the output order for the "7". This is because the native implementation of Promises uses a microtask to schedule the execution of the then callback, while Q uses an event on the main queue to schedule that execution, which comes later. Many will consider the native behaviour "more correct".