Typescript extended types declaration
1
I have the following interfaces:
export interface x {
a: string;
}
export interface y extends x {
b: string;
}
export interface z extends x {
c: string;
}
What I want is to declare a property that accepts an Array of a mix of y and z Objects, or basically any Object that is an extended Version of x. Is that possible? Perhaps something like this?
export interface n {
d: <T extends x>;
}
typescript
asked Nov 27 '18 at 19:39
CocktailCocktail
82
add a comment |
1
I have the following interfaces:
export interface x {
a: string;
}
export interface y extends x {
b: string;
}
export interface z extends x {
c: string;
}
What I want is to declare a property that accepts an Array of a mix of y and z Objects, or basically any Object that is an extended Version of x. Is that possible? Perhaps something like this?
export interface n {
d: <T extends x>;
}
typescript
asked Nov 27 '18 at 19:39
CocktailCocktail
82
Only objects that extendx? Orxtoo?
– Frank Modica
Nov 27 '18 at 19:43
Either is fine. But out of interest... what would be the solution if I only wanted Objects that extend x?
– Cocktail
Nov 27 '18 at 20:33
I'm actually not sure, you might have to create a dummy interface likeinterface dummy extends x {}and then use that. Maybe someone else knows of a better way.
– Frank Modica
Nov 27 '18 at 20:38
add a comment |
1
1
1
I have the following interfaces:
export interface x {
a: string;
}
export interface y extends x {
b: string;
}
export interface z extends x {
c: string;
}
What I want is to declare a property that accepts an Array of a mix of y and z Objects, or basically any Object that is an extended Version of x. Is that possible? Perhaps something like this?
export interface n {
d: <T extends x>;
}
typescript
asked Nov 27 '18 at 19:39
CocktailCocktail
82
I have the following interfaces:
export interface x {
a: string;
}
export interface y extends x {
b: string;
}
export interface z extends x {
c: string;
}
What I want is to declare a property that accepts an Array of a mix of y and z Objects, or basically any Object that is an extended Version of x. Is that possible? Perhaps something like this?
export interface n {
d: <T extends x>;
}
typescript
typescript
asked Nov 27 '18 at 19:39
CocktailCocktail
82
asked Nov 27 '18 at 19:39
CocktailCocktail
82
asked Nov 27 '18 at 19:39
CocktailCocktail
82
asked Nov 27 '18 at 19:39
CocktailCocktail
82
asked Nov 27 '18 at 19:39
CocktailCocktail
82
82
Only objects that extendx? Orxtoo?
– Frank Modica
Nov 27 '18 at 19:43
Either is fine. But out of interest... what would be the solution if I only wanted Objects that extend x?
– Cocktail
Nov 27 '18 at 20:33
I'm actually not sure, you might have to create a dummy interface likeinterface dummy extends x {}and then use that. Maybe someone else knows of a better way.
– Frank Modica
Nov 27 '18 at 20:38
add a comment |
Only objects that extendx? Orxtoo?
– Frank Modica
Nov 27 '18 at 19:43
Either is fine. But out of interest... what would be the solution if I only wanted Objects that extend x?
– Cocktail
Nov 27 '18 at 20:33
I'm actually not sure, you might have to create a dummy interface likeinterface dummy extends x {}and then use that. Maybe someone else knows of a better way.
– Frank Modica
Nov 27 '18 at 20:38
Only objects that extend
x? Or x too?– Frank Modica
Nov 27 '18 at 19:43
Only objects that extend
x? Or x too?– Frank Modica
Nov 27 '18 at 19:43
Either is fine. But out of interest... what would be the solution if I only wanted Objects that extend x?
– Cocktail
Nov 27 '18 at 20:33
Either is fine. But out of interest... what would be the solution if I only wanted Objects that extend x?
– Cocktail
Nov 27 '18 at 20:33
I'm actually not sure, you might have to create a dummy interface like
interface dummy extends x {} and then use that. Maybe someone else knows of a better way.– Frank Modica
Nov 27 '18 at 20:38
I'm actually not sure, you might have to create a dummy interface like
interface dummy extends x {} and then use that. Maybe someone else knows of a better way.– Frank Modica
Nov 27 '18 at 20:38
add a comment |
1 Answer
1
active
oldest
votes
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export interface n {
d: Array<x>;
}
answered Nov 27 '18 at 19:43
Frank ModicaFrank Modica
6,5342728
add a comment |
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1 Answer
1
active
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votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
export interface n {
d: Array<x>;
}
answered Nov 27 '18 at 19:43
Frank ModicaFrank Modica
6,5342728
add a comment |
2
export interface n {
d: Array<x>;
}
answered Nov 27 '18 at 19:43
Frank ModicaFrank Modica
6,5342728
add a comment |
2
2
2
export interface n {
d: Array<x>;
}
answered Nov 27 '18 at 19:43
Frank ModicaFrank Modica
6,5342728
export interface n {
d: Array<x>;
}
answered Nov 27 '18 at 19:43
Frank ModicaFrank Modica
6,5342728
answered Nov 27 '18 at 19:43
Frank ModicaFrank Modica
6,5342728
answered Nov 27 '18 at 19:43
Frank ModicaFrank Modica
6,5342728
answered Nov 27 '18 at 19:43
Frank ModicaFrank Modica
6,5342728
6,5342728
add a comment |
add a comment |
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Only objects that extend
x? Orxtoo?– Frank Modica
Nov 27 '18 at 19:43
Either is fine. But out of interest... what would be the solution if I only wanted Objects that extend x?
– Cocktail
Nov 27 '18 at 20:33
I'm actually not sure, you might have to create a dummy interface like
interface dummy extends x {}and then use that. Maybe someone else knows of a better way.– Frank Modica
Nov 27 '18 at 20:38