Does TypeScript have a Null-conditional operator

Yes, as of Typescript 3.7 you can now do this via optional-chaining

person?.getName()?.firstName

gets transpiled to

let firstName = person === null || person === void 0 ? void 0 : (_person$getName = person.getName()) === null || _person$getName === void 0 ? void 0 : _person$getName.firstName;

Note the check for null. This will work as expected if for example person is defined as

let person:any = null; //no runtime TypeError when calling person?.getName()

However if person is defined as

let person:any = {};//Uncaught TypeError: person.getName is not a function

See also this similar stackoverflow question

I've encountered the same situation, and the following worked for me.

First, I defined a separate class with a property that can be null:

export class Foo {
    id: number; //this can be null
}

Then in my main class I set a property foo to this new type:

foo: Foo

After that, I was able to use it like this:

var fooId = (this.foo || ({} as Foo)).id;

This is the closest I could get to have a null propagation in a current version of TypeScript.

No, as of now safe navigation operatior is still not implemented in Typescript: https://github.com/Microsoft/TypeScript/issues/16

However according to the latest standardization meeting notes it has been proposed, so maybe v3 :)

https://github.com/tc39/agendas/blob/master/2017/07.md