Assuming I have an object like this
foo = {
a: 1,
b: 2,
c: 3
};
I would destructure it this way
const {a, b, c} = foo;
However, what, if the keys a, b and c do not exist?
foo = null;
This won't work then
const {a, b, c} = foo;
I would have to do something like
let a, b, c;
if(foo) {
a = foo.a;
b = foo.b;
c = foo.c;
};
Is there a way to do this in one line? Something like
const {a, b, c} = foo ? foo : null; // this is not working
null is not an object. (evaluating '_ref2.a')
When foo is not an object, you are going to have to give it an object so the destructuring can happen. So set it to an empty object using an 'or'. Now if foo is defined it will use foo, if it is not defined it will use the default object.
const foo = undefined;
const { a, b, c } = foo || {};
console.log(a, b, c);If you want default values, set them in the object.
const foo = undefined;
const { a, b, c } = foo || {a : 'a', b: 'b', c: 'c' };
console.log(a, b, c);If you want default values if the object does not provide them all, you can set values in the destructuring.
const foo = { a: 1};
const { a = "a", b = "b", c = "c" } = foo || {};
console.log(a, b, c);What about destructuring default parameters?
If foo could be undefined set = foo || {} in destructuring to avoid exception
let foo = {};
const {a = null, b = null, c = null} = foo;
console.log('a:', a);
console.log('b:', b);
console.log('c:', c);It will work like you might expect, the variables a b and c will be created with the value of undefined. this is the fallback when also trying to access a property which doesn't exist on a certain object.
const foo = {};
console.log(foo.a); // undefined
const {a, b, c} = foo;
console.log(a); // undefined
console.log(b); // undefined
console.log(d); // Uncaught ReferenceError: d is not defined