Does TypeScript have a Null-conditional operator

Question

Is there any operator like ?. in TypeScript that can check if the variable is null or not defined like Kotlin? Like

person?.getName()?.firstName ?: "None"

Not to my knowledge, there's just a operator which tells the compiler that a value won't be `null`: (`person!.getName()`) — cyr_x, Aug 01 '17 at 14:52
Possible duplicate of [Does Typescript support the ?. operator? (And, what's it called?)](https://stackoverflow.com/questions/15260732/does-typescript-support-the-operator-and-whats-it-called) — Daria Pydorenko, Aug 01 '17 at 14:53
Yes. But I think this operator is just for compiler but does nothing to check. If the person is null in runtime, I suppose it will cause an error. Is is right? — Da Yin, Aug 01 '17 at 15:01
Yes, you are right. When I search ?. in typescript, stackoverflow did not give that. Sorry for the duplication. — Da Yin, Aug 01 '17 at 15:02
This is now available as of TS 3.7, see https://stackoverflow.com/a/60594782/618441, https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Optional_chaining and https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Nullish_coalescing_operator — AndreasHassing, Jun 03 '20 at 17:07

score 25 · Answer 1 · answered Mar 09 '20 at 04:36

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

Similarly, [nullish coalescing operator](https://www.typescriptlang.org/docs/handbook/release-notes/typescript-3-7.html#nullish-coalescing) also exists as of TS 3.7: `let x = foo ?? bar();` — AndreasHassing, Jun 03 '20 at 17:09

Loonquawl · Accepted Answer · 2017-08-01T15:00:36.210

18

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

edited Aug 01 '17 at 15:00

answered Aug 01 '17 at 14:52

Loonquawl

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Thank you. I have looked through the handbook but did not find anything. I think this operator is useful and beautiful that eliminate a lot of shit code. Do you have any idea whether this operator is intended to be included in ES standards in the future? – Da Yin Aug 01 '17 at 14:58
As a side note typescript does have the ``?`` operator but does not do that. – Leonardo Chaia Aug 01 '17 at 14:58
1

For anyone looking, here is the [related javascript proposal](https://github.com/tc39/proposal-optional-chaining). – ProgrammingLlama Sep 04 '18 at 02:00
1

_so maybe v3_ v3 is out, maybe this answer needs to be updated (even if it is just to say it was not accepted if that is the case – santiago arizti Dec 06 '18 at 19:06
I'm using Typescript v3.7.2 but still not found this feature. could someone help please? – Franva Nov 15 '19 at 04:32
it is now possible see [my answer](https://stackoverflow.com/a/60594782/224410) – wal Jun 04 '20 at 00:37
been possible since 3.7.x version somewhere, and as of ES2020, also just available in javascript. – bvdb Jan 28 '21 at 21:06

score 7 · Answer 3 · answered Apr 06 '18 at 13:21

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.

score 3 · Answer 4 · answered Aug 08 '17 at 09:11

3

Actually this is related to javascript null safety discussion that is mentioned in this answer. I guess they want it to be supported on javascript before they will get to typescript.

answered Aug 08 '17 at 09:11

ApriOri

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Ebuka · Answer 5 · 2018-11-30T05:42:48.883

3

I used this in my code to check for null

this.firstname = (this.firstname != null) ? this.firstname : this.firstname;

This could do till v3 is out

edited Nov 30 '18 at 05:42

answered Sep 14 '18 at 12:11

Ebuka

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score 3 · Answer 6 · answered Sep 10 '20 at 12:55

The actual good answer is in Andreas' comment. Your example would be implemented as below from Typescript 3.7 onwards :

person?.getName()?.firstName ?? "None"

See https://www.typescriptlang.org/docs/handbook/release-notes/typescript-3-7.html#nullish-coalescing

bvdb · Answer 7 · 2021-02-03T10:57:18.300

1

Not only is it supported in typescript (since version 3.7.2). It is also supported in javascript now (since ES2020).

In practice that means you can use it in frameworks like Angular (since v9 which supports typescript 3.7).
But especially now that it is in the ES2020 spec, we will be seeing it everywhere.

edited Feb 03 '21 at 10:57

answered Jan 28 '21 at 21:15

bvdb

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score -1 · Answer 8 · edited Feb 03 '21 at 11:25

-1

This will set item to null if there is no e-mail:

let item = user?.email;

edited Feb 03 '21 at 11:25

sɐunıɔןɐqɐp

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answered Feb 03 '21 at 11:01

armel sauvy

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Does TypeScript have a Null-conditional operator

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