如何在ES6 Promise链中传播拒绝？

Question

Usually the two Promise objects are created across function calls. I have put the promises together showing what I would expect to happen:

new Promise((resolve, reject) => {

    //function call with returned promise...

    return new Promise((resolve, reject) => {

      reject("rejectCall_2")  //Prints nothing

    })

}).catch( e1 => {
  console.log('e1',e1) 
})

This should have propagated the reject into my parent promise. How can I capture the rejectCall_2 in the outer promise?

Answer 1

You don't return anything from inside new Promise . Whatever you return is just thrown away, What you do is resolve and reject. If you want to "return" something, including another promise, what you do is resolve with that something.

So what you want is

new Promise((resolve, reject) => {
    //function call with returned promise...
    resolve(new Promise((resolve, reject) => {
    ^^^^^^^
      reject("rejectCall_2")
    }));
}).catch(e1 => {
  console.log('e1', e1) 
})

Tested with babel-node.

FWIW, you might as well use the immediate return fat-arrow format and save yourself a few curly braces:

new Promise((resolve, reject) => 
    resolve(new Promise((resolve, reject) => 
      reject("rejectCall_2")
    ));
).catch( e1 => console.log('e1',e1));

I would take note of the comments which suggest you may be guilty of the "explicit promise constructor anti-pattern', where you construct a promise only to resolve or reject it with some other promise that you could have just used in the first place. I'm assuming your example is an artificial one designed to showcase your particular issue.