ES6 change a value while destructing

问题

Lets say I have an object

const obj = { width: 100, height: 200 }

I wish to pass that object to a method

myFunc( obj );

Inside that method I wish to pull out the height, but at the same time subtract a value. I only wish to do this once and after that it will never change.

Doing the following will get me the correct height I want of say 150.

let {height: localHeight} = obj;
localHeight = localHeight - 50;

How do I do the above in a single line ? Something like this - but I know this doesn't work.

const { height: (localHeight = localHeight - 50) } = obj

回答1:

It's possible, although it's a hack that hurts readability, might produce an error (if the object will contain the fake property in the future), and shouldn't be used in a real product.

You can destructure a non existing property, and use the default value to do the subtraction.

const obj = { width: 100, height: 200 }

const { height, localHeight = height - 50 } = obj

console.log(localHeight)

回答2:

No, this is not possible. Destructuring does just that, it assigns properties to target expressions. Assignment syntax does not have any modifiers for altering the assigned value (default initialisers are already a stretch).

Just don't use destructuring and you get to do it in a single statement:

const localHeight = obj.height - 50;

Much shorter and less complicated than anything you could achieve with destructuring.

来源：https://stackoverflow.com/questions/54345307/es6-change-a-value-while-destructing

标签

ecmascript-6

destructuring