Define a list of optional keys for Typescript Record
I want to type an object which can only have keys 'a', 'b' or 'c'.
So I can do it as follows:
Interface IList {
a?: string;
b?: string;
c?: string;
}
They are all optional!
Now I was wondering if this can be written with Record in just one line
type List = Record<'a' | 'b' | 'c', string>;
The only issue is that all keys need to be defined. So I ended up with
type List = Partial<Record<'a' | 'b' | 'c', string>>;
This works, but I can imagine there is a better way to do this without Partial. Is there an other way to make the keys optional inside Record ?
typescript typescript-typings
asked Nov 13 '18 at 8:30
Jeanluca ScaljeriJeanluca Scaljeri
6,96323104178
add a comment |
I want to type an object which can only have keys 'a', 'b' or 'c'.
So I can do it as follows:
Interface IList {
a?: string;
b?: string;
c?: string;
}
They are all optional!
Now I was wondering if this can be written with Record in just one line
type List = Record<'a' | 'b' | 'c', string>;
The only issue is that all keys need to be defined. So I ended up with
type List = Partial<Record<'a' | 'b' | 'c', string>>;
This works, but I can imagine there is a better way to do this without Partial. Is there an other way to make the keys optional inside Record ?
typescript typescript-typings
asked Nov 13 '18 at 8:30
Jeanluca ScaljeriJeanluca Scaljeri
6,96323104178
add a comment |
I want to type an object which can only have keys 'a', 'b' or 'c'.
So I can do it as follows:
Interface IList {
a?: string;
b?: string;
c?: string;
}
They are all optional!
Now I was wondering if this can be written with Record in just one line
type List = Record<'a' | 'b' | 'c', string>;
The only issue is that all keys need to be defined. So I ended up with
type List = Partial<Record<'a' | 'b' | 'c', string>>;
This works, but I can imagine there is a better way to do this without Partial. Is there an other way to make the keys optional inside Record ?
typescript typescript-typings
asked Nov 13 '18 at 8:30
Jeanluca ScaljeriJeanluca Scaljeri
6,96323104178
I want to type an object which can only have keys 'a', 'b' or 'c'.
So I can do it as follows:
Interface IList {
a?: string;
b?: string;
c?: string;
}
They are all optional!
Now I was wondering if this can be written with Record in just one line
type List = Record<'a' | 'b' | 'c', string>;
The only issue is that all keys need to be defined. So I ended up with
type List = Partial<Record<'a' | 'b' | 'c', string>>;
This works, but I can imagine there is a better way to do this without Partial. Is there an other way to make the keys optional inside Record ?
typescript typescript-typings
typescript typescript-typings
asked Nov 13 '18 at 8:30
Jeanluca ScaljeriJeanluca Scaljeri
6,96323104178
asked Nov 13 '18 at 8:30
Jeanluca ScaljeriJeanluca Scaljeri
6,96323104178
edited Nov 13 '18 at 9:18
Jeanluca Scaljeri
asked Nov 13 '18 at 8:30
Jeanluca ScaljeriJeanluca Scaljeri
6,96323104178
asked Nov 13 '18 at 8:30
Jeanluca ScaljeriJeanluca Scaljeri
6,96323104178
asked Nov 13 '18 at 8:30
Jeanluca ScaljeriJeanluca Scaljeri
6,96323104178
6,96323104178
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
There is no way to specify the optionality of members of Record. They are required by definition
type Record<K extends keyof any, T> = {
[P in K]: T; // Mapped properties are not optional, and it's not a homomorphic mapped type so it can't come from anywhere else.
};
You can define your own type if this is a common scenario for you:
type PartialRecord<K extends keyof any, T> = {
[P in K]?: T;
};
type List = PartialRecord<'a' | 'b' | 'c', string>
Or you can define PartialRecord using the predefined mapped types as well:
type PartialRecord<K extends keyof any, T> = Partial<Record<K, T>>
answered Nov 13 '18 at 8:36
Titian Cernicova-DragomirTitian Cernicova-Dragomir
58.4k33552
1
This is a good comprehensive answer.
– Fenton
Nov 13 '18 at 8:37
add a comment |
You can create the partial version of your List type:
type PartialList = Partial<List>;
And you could do it all on one line if you don't want the intermediate type:
type PartialList = Partial<Record<'a' | 'b' | 'c', string>>;
You might decide that, in the end, the most expressive for your future self is:
type List = {
a?: string;
b?: string;
c?: string;
}
answered Nov 13 '18 at 8:36
FentonFenton
152k42287311
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
There is no way to specify the optionality of members of Record. They are required by definition
type Record<K extends keyof any, T> = {
[P in K]: T; // Mapped properties are not optional, and it's not a homomorphic mapped type so it can't come from anywhere else.
};
You can define your own type if this is a common scenario for you:
type PartialRecord<K extends keyof any, T> = {
[P in K]?: T;
};
type List = PartialRecord<'a' | 'b' | 'c', string>
Or you can define PartialRecord using the predefined mapped types as well:
type PartialRecord<K extends keyof any, T> = Partial<Record<K, T>>
answered Nov 13 '18 at 8:36
Titian Cernicova-DragomirTitian Cernicova-Dragomir
58.4k33552
1
This is a good comprehensive answer.
– Fenton
Nov 13 '18 at 8:37
add a comment |
There is no way to specify the optionality of members of Record. They are required by definition
type Record<K extends keyof any, T> = {
[P in K]: T; // Mapped properties are not optional, and it's not a homomorphic mapped type so it can't come from anywhere else.
};
You can define your own type if this is a common scenario for you:
type PartialRecord<K extends keyof any, T> = {
[P in K]?: T;
};
type List = PartialRecord<'a' | 'b' | 'c', string>
Or you can define PartialRecord using the predefined mapped types as well:
type PartialRecord<K extends keyof any, T> = Partial<Record<K, T>>
answered Nov 13 '18 at 8:36
Titian Cernicova-DragomirTitian Cernicova-Dragomir
58.4k33552
1
This is a good comprehensive answer.
– Fenton
Nov 13 '18 at 8:37
add a comment |
There is no way to specify the optionality of members of Record. They are required by definition
type Record<K extends keyof any, T> = {
[P in K]: T; // Mapped properties are not optional, and it's not a homomorphic mapped type so it can't come from anywhere else.
};
You can define your own type if this is a common scenario for you:
type PartialRecord<K extends keyof any, T> = {
[P in K]?: T;
};
type List = PartialRecord<'a' | 'b' | 'c', string>
Or you can define PartialRecord using the predefined mapped types as well:
type PartialRecord<K extends keyof any, T> = Partial<Record<K, T>>
answered Nov 13 '18 at 8:36
Titian Cernicova-DragomirTitian Cernicova-Dragomir
58.4k33552
There is no way to specify the optionality of members of Record. They are required by definition
type Record<K extends keyof any, T> = {
[P in K]: T; // Mapped properties are not optional, and it's not a homomorphic mapped type so it can't come from anywhere else.
};
You can define your own type if this is a common scenario for you:
type PartialRecord<K extends keyof any, T> = {
[P in K]?: T;
};
type List = PartialRecord<'a' | 'b' | 'c', string>
Or you can define PartialRecord using the predefined mapped types as well:
type PartialRecord<K extends keyof any, T> = Partial<Record<K, T>>
answered Nov 13 '18 at 8:36
Titian Cernicova-DragomirTitian Cernicova-Dragomir
58.4k33552
answered Nov 13 '18 at 8:36
Titian Cernicova-DragomirTitian Cernicova-Dragomir
58.4k33552
answered Nov 13 '18 at 8:36
Titian Cernicova-DragomirTitian Cernicova-Dragomir
58.4k33552
answered Nov 13 '18 at 8:36
Titian Cernicova-DragomirTitian Cernicova-Dragomir
58.4k33552
58.4k33552
1
This is a good comprehensive answer.
– Fenton
Nov 13 '18 at 8:37
add a comment |
1
This is a good comprehensive answer.
– Fenton
Nov 13 '18 at 8:37
1
1
This is a good comprehensive answer.
– Fenton
Nov 13 '18 at 8:37
This is a good comprehensive answer.
– Fenton
Nov 13 '18 at 8:37
add a comment |
You can create the partial version of your List type:
type PartialList = Partial<List>;
And you could do it all on one line if you don't want the intermediate type:
type PartialList = Partial<Record<'a' | 'b' | 'c', string>>;
You might decide that, in the end, the most expressive for your future self is:
type List = {
a?: string;
b?: string;
c?: string;
}
answered Nov 13 '18 at 8:36
FentonFenton
152k42287311
add a comment |
You can create the partial version of your List type:
type PartialList = Partial<List>;
And you could do it all on one line if you don't want the intermediate type:
type PartialList = Partial<Record<'a' | 'b' | 'c', string>>;
You might decide that, in the end, the most expressive for your future self is:
type List = {
a?: string;
b?: string;
c?: string;
}
answered Nov 13 '18 at 8:36
FentonFenton
152k42287311
add a comment |
You can create the partial version of your List type:
type PartialList = Partial<List>;
And you could do it all on one line if you don't want the intermediate type:
type PartialList = Partial<Record<'a' | 'b' | 'c', string>>;
You might decide that, in the end, the most expressive for your future self is:
type List = {
a?: string;
b?: string;
c?: string;
}
answered Nov 13 '18 at 8:36
FentonFenton
152k42287311
You can create the partial version of your List type:
type PartialList = Partial<List>;
And you could do it all on one line if you don't want the intermediate type:
type PartialList = Partial<Record<'a' | 'b' | 'c', string>>;
You might decide that, in the end, the most expressive for your future self is:
type List = {
a?: string;
b?: string;
c?: string;
}
answered Nov 13 '18 at 8:36
FentonFenton
152k42287311
answered Nov 13 '18 at 8:36
FentonFenton
152k42287311
answered Nov 13 '18 at 8:36
FentonFenton
152k42287311
answered Nov 13 '18 at 8:36
FentonFenton
152k42287311
152k42287311
add a comment |
add a comment |
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