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Can I make a synchronous promise in the JavaScript Q library?

rated 0 times [ 13] [ 1] / answers: 1 / hits: 18804 / 11 Years ago, thu, june 20, 2013, 12:00:00

I want to do something like the following:

delay( 2500 )

  .then( function () { console.log( Step 1 done ) } )

  .then( delay( 7500 ) )

  .then( function () { console.log( Step 2 done ) } );

So implementation of delay has been demonstrated many times before:

function delay( ms ) {

  var deferred = Q.defer();

  setTimeout( deferred.resolve, ms );

  return deferred.promise;

}

But if I run the above in node.js I get:

... delay of 2500ms

Step 1 done

Step 2 done

... delay of ~7500ms

rather than what I expect to see:

... delay of 2500ms

Step 1 done

... delay of 7500ms

Step 2 done

In the examples provided on https://github.com/kriskowal/q/wiki/Examples-Gallery I can't find any examples of synchronous functions (functions that return a value without any callbacks involved) chained with promise functions.

Any ideas how to mix in synchronous actions with asynchronous promises?

I've tried:

function synchronousPromise() {

  var deferred = Q.defer();

  console.log( Synchronous function call );

  deferred.resolve();

  return deferred.promise;

}



delay( 2500 )

  .then( function(){synchronousPromise()} )

  .then( function(){delay( 7500 )} )

  .then( function(){synchronousPromise()} );

And this outputs:

... delay of 2500ms

Time now is 2013-06-20

Time now is 2013-06-20

... delay of 7500ms

.. still not what I'm trying to achieve.

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1 Year ago

answered 11 Years ago jarettajb · Accepted Answer

If you want to chain the callbacks, you have to return a new promise object from one of the callbacks. In your first example, you write

.then( delay( 7500 ) )

which means you are passing a promise object to .then, not a function. According to the Promise/A+ proposal (which Q follows), all non-function arguments must be ignored. So, basically it's the same as if you just write:

delay( 2500 )

  .then( function () { console.log( Step 1 done ) } )

  .then( function () { console.log( Step 2 done ) } );

Instead, pass function which calls delay and returns the promise object:

delay( 2500 )

  .then( function () { console.log( Step 1 done ); } ) 

  .then( function () { return delay( 7500 ); } )

  .then( function () { console.log( Step 2 done ); } );

Now the last callback will only be called once the promise object returned by delay in the second callback is resolved.