Javascript return same value in then and catch-CodePudding

In JS, let's say I call a promise in a 'fire and forget' way. I don't care if it was successful or not, and I don't care what it returns as I want to return the same value anyways.

I can call

return myPromise().then(() => foo).catch(() => foo);

Is there a way to not repeat then then and catch parts?

What I've tried

I thought about using finally, thinking that it will take what finally returns (similar to what Java does by taking the return value of the finally block) : return myPromise().finally(() => foo); But it turns out it didn't do what I expected. According to mdn, this return value will be used to reject in case the original promise rejected.

CodePudding user response：

You can declare your result returning function once and then re-use it twice for the .then() and .catch() handlers

const myReturn = () => foo;

return myPromise()
    .then(myReturn)
    .catch(myReturn);

You can further shorten this by using both parameters of .then() which handle both the success and failure path:

const myReturn = () => foo;

return myPromise()
    .then(myReturn, myReturn);

If you find needing this a lot, you might make a small library function that takes a value and returns two functions that both just evaluate to it:

const doubleThunk = val =>
  [() => val, () => val];

Which would be used like this:

return myPromise()
    .then(...doubleThunk(foo));

CodePudding user response：

You could have a look at Promise.allSettled if you're targeting newer browsers or Node.js versions.

This function will return a Promise that resolves whether or not the passed Promise resolves or rejects.

  
function foo(results) { 
    console.log('foo():', ...results)
}

// Success path
Promise.allSettled([Promise.resolve('Yay')]).then(foo)

// Failure path
Promise.allSettled([Promise.reject('Doh')]).then(foo)

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