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"
8 Answers
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
1 Comment
AndreasHassing
Similarly, nullish coalescing operator also exists as of TS 3.7:
let x = foo ?? bar();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 :)
7 Comments
Da Yin
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?
Leonardo Chaia
As a side note typescript does have the
? operator but does not do that.
ProgrammingLlama
For anyone looking, here is the related javascript proposal.
santiago arizti
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
Franva
I'm using Typescript v3.7.2 but still not found this feature. could someone help please?
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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.
Comments
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
Comments
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
Comments
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.
Comments
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.
Comments
This will set item to null if there is no e-mail:
let item = user?.email;
null: (person!.getName())