CODEWITHSUNDEEP
node.jspromiseq
RaceBase
RaceBase

Reputation: 18848

Calling Q Promise on the basis of conditional logic

Below is my condition

if abc is true
    call async func1, func2
else
    call async func1

function test(): Q.Promise<boolean> {
    if(abc)
       Q.all([func1,func2])
    else
       Q.all([func1])
    //if failed throw reject reason all the way in the chain
}
  1. As shown, it can be done using if and else clause, is there a better way to call promise conditionally?
  2. How to throw back error from any one of the promises?

Upvotes: 1

Views: 149

Answers (2)

Johannes Merz
Johannes Merz

Reputation: 3342

You are actually pretty close already:

function test() {
    if(abc)
       return Q.all([func1(),func2()])
    else
       return func1();
}
test().then(() => {
    // do whatever
}).catch(err => console.log(err));

Make sure you always return the promises as they don't get chained otherwise.

Upvotes: 0

dfsq
dfsq

Reputation: 193261

I would put promises in array and append new one depending on condition:

function test(): Q.Promise<Boolean[]> {
    const promises = [func1()] 
    if (abc) promises.push(func2()) 
    return Q.all(promises)
}

I corrected type signature a little, because Q.all resolves with array of values (Boolean in your case) from each underlying promises. You also need to call func1 and func2. And finally, don't forget to return from test function.

Upvotes: 1

Related Questions