Is it possible to create an Interface in TypeScript with optional function?
interface IElement {
name: string;
options: any;
type: string;
value?: string;
validation(any): boolean; // --> should be optional.
}
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2I found an answer, i think there is possible to implement optional function like this: validation?: (any) => boolean;Anton Selin– Anton Selin2014-12-17 10:16:58 +00:00Commented Dec 17, 2014 at 10:16
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2 Answers
There are currently three syntaxes that TypeScript allows for function declarations in interfaces:
Using your example of a validation function taking 1 parameter (of any type) and a boolean return value:
validation: {(flag: any): boolean};
or in the newer syntax:
validation(flag: any) : boolean;
or an alternative is:
validation: (flag: any) => boolean;
Solution:
so to make it optional with the old syntax is easy:
validation?: {(flag: any): boolean};
with the second syntax (recent addition - thanks to @toothbrush)
validation?(flag: any) : boolean;
or in the third syntax (as you found):
validation?: (flag: any) => boolean;
answered Dec 17, 2014 at 10:56
iCollect.it Ltd
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5 Comments
NYCdotNet
The way you've written it,
any is not a type, but is the name of a parameter that is implicitly of type any. Parameters must be named, even on interfaces. This code will fail to compile if --noImplicitAny is enabled. It should be something like this: validation?:(whatever:any) => boolean; where whatever is some reasonable parameter name.iCollect.it Ltd
@NYCdotNet: Yes, I should not have taken the original as-read. Corrected :)
Ilan Olkies
@gone-coding Hi! How do I know if the function is defined or not?
iCollect.it Ltd
@IlanOlkies: Same way you test for anything in JavaScript
if (object.validation) :) It needs to against the object context, so if (this.validation) may also be appropriate depending on your circumstance.Jamie Garcia
thank you! found this very helpful for optional function props and TS errors :)
Just like field properties, we can also declare optional methods in an interface, just by placing "?" after the method name
interface Vehicle{
run?(): void;
}
class Sedan implements Vehicle {}
class SUV implements Vehicle {
run() {
console.log("run run");
}
}
let suv = new SUV();
suv.run();
