Dynamically determine second argument when the first argument is a key from an interface The Next CEO of Stack OverflowTypescript: How to define interface for dynamical object like thisConvert a list of key-value-pair interfaces into a single mapped typeUsing generic keyof for a set action, for both the key and its typeTypeScript. How to use not exported type definitions?Post FormData (key, value): controller not receiving value when stringGeneric interface for conversion [keys].option.value.number to option.[Keys].numberTypescript clarification on index types and mapped typesTypeScript Interface With Known and Unknown Keys Not Using `any`Require child object property value to be same type as parent keyIs it possible to generate a string-enum in typescript with random or same strings?
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Dynamically determine second argument when the first argument is a key from an interface
The Next CEO of Stack OverflowTypescript: How to define interface for dynamical object like thisConvert a list of key-value-pair interfaces into a single mapped typeUsing generic keyof for a set action, for both the key and its typeTypeScript. How to use not exported type definitions?Post FormData (key, value): controller not receiving value when stringGeneric interface for conversion [keys].option.value.number to option.[Keys].numberTypescript clarification on index types and mapped typesTypeScript Interface With Known and Unknown Keys Not Using `any`Require child object property value to be same type as parent keyIs it possible to generate a string-enum in typescript with random or same strings?
I have an interface like so that's passed as options to a module factory function:
interface FactoryOpts
name?: string
radius?: number
// many more entries...
For example:
factory( radius: 10 )
Then I have an interface that defines the returned value from the factory. It has a few functions, but the one I'm stuck on is called "option" and can't see how I can avoid using "any":
interface ReturnedFactoryInstance
// I don't want to use "any" for the value, but I also don't
// want to write a method for each options
option(name: keyof FactoryOpts, value: any) => void
interface Factory String, options: FactoryOpts): ReturnedFactoryInstance
Is it possible to dynamically determine the type of value? For example:
const instance = factory( radius: 10 )
// Correct - this will compile
factory.option('radius', 10)
// Error - this will not compile
factory.option('radius', 'some string')
Is the only way to achieve this by making a function for each entry in FactoryOpts?
Solution using Karol's advice:
interface FactoryOpts
name?: string
radius?: number
// many more entries...
interface Factory
someProps: string
(selector: HTMLElement
interface ReturnedFactoryInstance<ValidOptions>
option: <Key extends keyof ValidOptions>(property: Key, value: ValidOptions[Key]) => void
You can then export a Factory e.g declare const f: Factory etc.
typescript typescript-typings
asked Mar 21 at 17:01
Evan ShortissEvan Shortiss
1,093715
add a comment |
I have an interface like so that's passed as options to a module factory function:
interface FactoryOpts
name?: string
radius?: number
// many more entries...
For example:
factory( radius: 10 )
Then I have an interface that defines the returned value from the factory. It has a few functions, but the one I'm stuck on is called "option" and can't see how I can avoid using "any":
interface ReturnedFactoryInstance
// I don't want to use "any" for the value, but I also don't
// want to write a method for each options
option(name: keyof FactoryOpts, value: any) => void
interface Factory String, options: FactoryOpts): ReturnedFactoryInstance
Is it possible to dynamically determine the type of value? For example:
const instance = factory( radius: 10 )
// Correct - this will compile
factory.option('radius', 10)
// Error - this will not compile
factory.option('radius', 'some string')
Is the only way to achieve this by making a function for each entry in FactoryOpts?
Solution using Karol's advice:
interface FactoryOpts
name?: string
radius?: number
// many more entries...
interface Factory
someProps: string
(selector: HTMLElement
interface ReturnedFactoryInstance<ValidOptions>
option: <Key extends keyof ValidOptions>(property: Key, value: ValidOptions[Key]) => void
You can then export a Factory e.g declare const f: Factory etc.
typescript typescript-typings
asked Mar 21 at 17:01
Evan ShortissEvan Shortiss
1,093715
add a comment |
I have an interface like so that's passed as options to a module factory function:
interface FactoryOpts
name?: string
radius?: number
// many more entries...
For example:
factory( radius: 10 )
Then I have an interface that defines the returned value from the factory. It has a few functions, but the one I'm stuck on is called "option" and can't see how I can avoid using "any":
interface ReturnedFactoryInstance
// I don't want to use "any" for the value, but I also don't
// want to write a method for each options
option(name: keyof FactoryOpts, value: any) => void
interface Factory String, options: FactoryOpts): ReturnedFactoryInstance
Is it possible to dynamically determine the type of value? For example:
const instance = factory( radius: 10 )
// Correct - this will compile
factory.option('radius', 10)
// Error - this will not compile
factory.option('radius', 'some string')
Is the only way to achieve this by making a function for each entry in FactoryOpts?
Solution using Karol's advice:
interface FactoryOpts
name?: string
radius?: number
// many more entries...
interface Factory
someProps: string
(selector: HTMLElement
interface ReturnedFactoryInstance<ValidOptions>
option: <Key extends keyof ValidOptions>(property: Key, value: ValidOptions[Key]) => void
You can then export a Factory e.g declare const f: Factory etc.
typescript typescript-typings
asked Mar 21 at 17:01
Evan ShortissEvan Shortiss
1,093715
I have an interface like so that's passed as options to a module factory function:
interface FactoryOpts
name?: string
radius?: number
// many more entries...
For example:
factory( radius: 10 )
Then I have an interface that defines the returned value from the factory. It has a few functions, but the one I'm stuck on is called "option" and can't see how I can avoid using "any":
interface ReturnedFactoryInstance
// I don't want to use "any" for the value, but I also don't
// want to write a method for each options
option(name: keyof FactoryOpts, value: any) => void
interface Factory String, options: FactoryOpts): ReturnedFactoryInstance
Is it possible to dynamically determine the type of value? For example:
const instance = factory( radius: 10 )
// Correct - this will compile
factory.option('radius', 10)
// Error - this will not compile
factory.option('radius', 'some string')
Is the only way to achieve this by making a function for each entry in FactoryOpts?
Solution using Karol's advice:
interface FactoryOpts
name?: string
radius?: number
// many more entries...
interface Factory
someProps: string
(selector: HTMLElement
interface ReturnedFactoryInstance<ValidOptions>
option: <Key extends keyof ValidOptions>(property: Key, value: ValidOptions[Key]) => void
You can then export a Factory e.g declare const f: Factory etc.
typescript typescript-typings
typescript typescript-typings
asked Mar 21 at 17:01
Evan ShortissEvan Shortiss
1,093715
asked Mar 21 at 17:01
Evan ShortissEvan Shortiss
1,093715
edited Mar 21 at 20:04
Evan Shortiss
asked Mar 21 at 17:01
Evan ShortissEvan Shortiss
1,093715
asked Mar 21 at 17:01
Evan ShortissEvan Shortiss
1,093715
asked Mar 21 at 17:01
Evan ShortissEvan Shortiss
1,093715
1,093715
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
interface Factory<T>
option<K extends keyof T>(this: Factory<T>, name: K, value: T[K]): void;
type FactoryConstructor = <T>(arg: T) => Factory<T>;
Usage:
declare const factory: FactoryConstructor;
factory( radius: 10 ).option('radius', 1);
See TypeScript Playground.
answered Mar 21 at 17:40
Karol MajewskiKarol Majewski
4,3191216
1
Thank you! This answer is really close! In my case "this" isn't applicable, but the rest is. Updated the question with the solution for this specific case.
– Evan Shortiss
Mar 21 at 20:01
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
interface Factory<T>
option<K extends keyof T>(this: Factory<T>, name: K, value: T[K]): void;
type FactoryConstructor = <T>(arg: T) => Factory<T>;
Usage:
declare const factory: FactoryConstructor;
factory( radius: 10 ).option('radius', 1);
See TypeScript Playground.
answered Mar 21 at 17:40
Karol MajewskiKarol Majewski
4,3191216
1
Thank you! This answer is really close! In my case "this" isn't applicable, but the rest is. Updated the question with the solution for this specific case.
– Evan Shortiss
Mar 21 at 20:01
add a comment |
interface Factory<T>
option<K extends keyof T>(this: Factory<T>, name: K, value: T[K]): void;
type FactoryConstructor = <T>(arg: T) => Factory<T>;
Usage:
declare const factory: FactoryConstructor;
factory( radius: 10 ).option('radius', 1);
See TypeScript Playground.
answered Mar 21 at 17:40
Karol MajewskiKarol Majewski
4,3191216
1
Thank you! This answer is really close! In my case "this" isn't applicable, but the rest is. Updated the question with the solution for this specific case.
– Evan Shortiss
Mar 21 at 20:01
add a comment |
interface Factory<T>
option<K extends keyof T>(this: Factory<T>, name: K, value: T[K]): void;
type FactoryConstructor = <T>(arg: T) => Factory<T>;
Usage:
declare const factory: FactoryConstructor;
factory( radius: 10 ).option('radius', 1);
See TypeScript Playground.
answered Mar 21 at 17:40
Karol MajewskiKarol Majewski
4,3191216
interface Factory<T>
option<K extends keyof T>(this: Factory<T>, name: K, value: T[K]): void;
type FactoryConstructor = <T>(arg: T) => Factory<T>;
Usage:
declare const factory: FactoryConstructor;
factory( radius: 10 ).option('radius', 1);
See TypeScript Playground.
answered Mar 21 at 17:40
Karol MajewskiKarol Majewski
4,3191216
answered Mar 21 at 17:40
Karol MajewskiKarol Majewski
4,3191216
answered Mar 21 at 17:40
Karol MajewskiKarol Majewski
4,3191216
answered Mar 21 at 17:40
Karol MajewskiKarol Majewski
4,3191216
4,3191216
1
Thank you! This answer is really close! In my case "this" isn't applicable, but the rest is. Updated the question with the solution for this specific case.
– Evan Shortiss
Mar 21 at 20:01
add a comment |
1
Thank you! This answer is really close! In my case "this" isn't applicable, but the rest is. Updated the question with the solution for this specific case.
– Evan Shortiss
Mar 21 at 20:01
1
1
Thank you! This answer is really close! In my case "this" isn't applicable, but the rest is. Updated the question with the solution for this specific case.
– Evan Shortiss
Mar 21 at 20:01
Thank you! This answer is really close! In my case "this" isn't applicable, but the rest is. Updated the question with the solution for this specific case.
– Evan Shortiss
Mar 21 at 20:01
add a comment |
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