Destucture object in typescript

Question

One question for some reason I can't figure out how to do it. I have 2 objects and need to use optional operator to get value in typescript. I can break them into 2 line but I am wondering how can I do in one line code?

// in python
car = {
  "brand": "Ford",
}
name = {
  "brand": "Ford",
}

x, y = car.get("price", 150), name.get('print', 200)
print(x, y)

// Something like this?
const {price: carPrice  = 150, price: namePrice =200} = {car?.price, name?.price}

//It works but I would like to be one line code
const carPrice = car?.price || 150
const namePrice =  name?.price || 200

I feel like you're abusing syntax to achieve superficial brevity. @Psidom shows you how you can write it but I don't think you should. — Aluan Haddad
– Aluan Haddad, Commented Jul 31, 2021 at 22:53

akuiper · Accepted Answer · 2021-07-31 18:51:30Z

2

Tuple destructure would be something similar:

const [ carPrice, namePrice ] = [ car1?.price || 150, name?.price || 200 ]

Or to be more precise, use ?? instead of ||:

const [ carPrice, namePrice ] = [ car1?.price ?? 150, name?.price ?? 200 ]

edited Jul 31, 2021 at 18:51

answered Jul 31, 2021 at 18:46

akuiper

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2 Comments

VLAZ Over a year ago

There is also the newer ?? operator that can be used. As well as const [ carPrice = 150, namePrice = 200 ] = [ car1?.price, name?.price ]

akuiper Over a year ago

@VLAZ Wow, good to know that. ?? looks awesome.

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