How to write omit function with proper types in typescript
What I ultimately want to do is to create a React HOC that would inject a property defined by a string and return a component without requiring the injected property. In essence though this boils down to a factory function returning an omit function.
As you can see in this example for some reason the type B ends up being 'never'.
const f = <A extends object, B extends keyof A>(arg: B) => (obj: A): Omit<A, B> => {
delete obj[arg]
return obj
}
// TS2345: Argument of type "test" is not assignable to parameter of type 'never'
const a = f('test')
const b = a({ test: 1})
const c = b.test
When I try keyof outside of generic parameter, it seem to be functioning better, but typescript does not infer the return type properly and I do not know how to type it as I do not know how to get the reference to the first string arg that could be used in Omit:
const f = <A extends object>(arg: keyof A) => (obj: A) => {
delete obj[arg]
return obj
}
const a = f('test')
const b = a({ test: 1 })
// Does not infer 'test' is no longer here
const c = b.test
For reference Omit is:
export type Omit<T, K> = Pick<T, Exclude<keyof T, K>>
reactjs typescript
asked Nov 25 '18 at 9:17
aocenasaocenas
5117
add a comment |
What I ultimately want to do is to create a React HOC that would inject a property defined by a string and return a component without requiring the injected property. In essence though this boils down to a factory function returning an omit function.
As you can see in this example for some reason the type B ends up being 'never'.
const f = <A extends object, B extends keyof A>(arg: B) => (obj: A): Omit<A, B> => {
delete obj[arg]
return obj
}
// TS2345: Argument of type "test" is not assignable to parameter of type 'never'
const a = f('test')
const b = a({ test: 1})
const c = b.test
When I try keyof outside of generic parameter, it seem to be functioning better, but typescript does not infer the return type properly and I do not know how to type it as I do not know how to get the reference to the first string arg that could be used in Omit:
const f = <A extends object>(arg: keyof A) => (obj: A) => {
delete obj[arg]
return obj
}
const a = f('test')
const b = a({ test: 1 })
// Does not infer 'test' is no longer here
const c = b.test
For reference Omit is:
export type Omit<T, K> = Pick<T, Exclude<keyof T, K>>
reactjs typescript
asked Nov 25 '18 at 9:17
aocenasaocenas
5117
add a comment |
What I ultimately want to do is to create a React HOC that would inject a property defined by a string and return a component without requiring the injected property. In essence though this boils down to a factory function returning an omit function.
As you can see in this example for some reason the type B ends up being 'never'.
const f = <A extends object, B extends keyof A>(arg: B) => (obj: A): Omit<A, B> => {
delete obj[arg]
return obj
}
// TS2345: Argument of type "test" is not assignable to parameter of type 'never'
const a = f('test')
const b = a({ test: 1})
const c = b.test
When I try keyof outside of generic parameter, it seem to be functioning better, but typescript does not infer the return type properly and I do not know how to type it as I do not know how to get the reference to the first string arg that could be used in Omit:
const f = <A extends object>(arg: keyof A) => (obj: A) => {
delete obj[arg]
return obj
}
const a = f('test')
const b = a({ test: 1 })
// Does not infer 'test' is no longer here
const c = b.test
For reference Omit is:
export type Omit<T, K> = Pick<T, Exclude<keyof T, K>>
reactjs typescript
asked Nov 25 '18 at 9:17
aocenasaocenas
5117
What I ultimately want to do is to create a React HOC that would inject a property defined by a string and return a component without requiring the injected property. In essence though this boils down to a factory function returning an omit function.
As you can see in this example for some reason the type B ends up being 'never'.
const f = <A extends object, B extends keyof A>(arg: B) => (obj: A): Omit<A, B> => {
delete obj[arg]
return obj
}
// TS2345: Argument of type "test" is not assignable to parameter of type 'never'
const a = f('test')
const b = a({ test: 1})
const c = b.test
When I try keyof outside of generic parameter, it seem to be functioning better, but typescript does not infer the return type properly and I do not know how to type it as I do not know how to get the reference to the first string arg that could be used in Omit:
const f = <A extends object>(arg: keyof A) => (obj: A) => {
delete obj[arg]
return obj
}
const a = f('test')
const b = a({ test: 1 })
// Does not infer 'test' is no longer here
const c = b.test
For reference Omit is:
export type Omit<T, K> = Pick<T, Exclude<keyof T, K>>
reactjs typescript
reactjs typescript
asked Nov 25 '18 at 9:17
aocenasaocenas
5117
asked Nov 25 '18 at 9:17
aocenasaocenas
5117
asked Nov 25 '18 at 9:17
aocenasaocenas
5117
asked Nov 25 '18 at 9:17
aocenasaocenas
5117
asked Nov 25 '18 at 9:17
aocenasaocenas
5117
5117
add a comment |
add a comment |
1 Answer
1
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Since the first function has both type arguments, typescript will try to infer both when that call happens and since there is no inference site for A it will probably infer {} for it, making B never. The way to fix this, is to make B and when the second call occurs infer A, with the constraint that A must have a B key:
export type Omit<T, K> = Pick<T, Exclude<keyof T, K>>
const f = <B extends keyof any>(arg: B) => <A extends Record<B, any>>(obj: A): Omit<A, B> => {
delete obj[arg]
return obj
}
const a = f('test')
const b = a({ test: 1, other: ""})
const c = b.test; // error as expected
const c2 = b.other; // ok
answered Nov 25 '18 at 9:57
Titian Cernicova-DragomirTitian Cernicova-Dragomir
67.6k34664
Thank you very much, this helps tremendously. I realised thought I have one more piece in the HOC, that I omitted in the question. Would it be possible to back propagate theAtype somehow to the first function call. What I want to do is<B extends keyof any>(arg: B, fn: (arg: A) => void) => .... So basically, I give you attribute name and a function that will take object I will give you later. If that does not make sense I can create new question, my bad I did not realise this will be important.
– aocenas
Nov 25 '18 at 18:08
@aocenas could you post a new question with the full code ? It would make more sense to see it in full.. the comments is not a good place for a lot of code..
– Titian Cernicova-Dragomir
Nov 25 '18 at 18:10
Here is the new question, hope it makes sense stackoverflow.com/questions/53470559/…
– aocenas
Nov 25 '18 at 18:32
add a comment |
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Since the first function has both type arguments, typescript will try to infer both when that call happens and since there is no inference site for A it will probably infer {} for it, making B never. The way to fix this, is to make B and when the second call occurs infer A, with the constraint that A must have a B key:
export type Omit<T, K> = Pick<T, Exclude<keyof T, K>>
const f = <B extends keyof any>(arg: B) => <A extends Record<B, any>>(obj: A): Omit<A, B> => {
delete obj[arg]
return obj
}
const a = f('test')
const b = a({ test: 1, other: ""})
const c = b.test; // error as expected
const c2 = b.other; // ok
answered Nov 25 '18 at 9:57
Titian Cernicova-DragomirTitian Cernicova-Dragomir
67.6k34664
Thank you very much, this helps tremendously. I realised thought I have one more piece in the HOC, that I omitted in the question. Would it be possible to back propagate theAtype somehow to the first function call. What I want to do is<B extends keyof any>(arg: B, fn: (arg: A) => void) => .... So basically, I give you attribute name and a function that will take object I will give you later. If that does not make sense I can create new question, my bad I did not realise this will be important.
– aocenas
Nov 25 '18 at 18:08
@aocenas could you post a new question with the full code ? It would make more sense to see it in full.. the comments is not a good place for a lot of code..
– Titian Cernicova-Dragomir
Nov 25 '18 at 18:10
Here is the new question, hope it makes sense stackoverflow.com/questions/53470559/…
– aocenas
Nov 25 '18 at 18:32
add a comment |
Since the first function has both type arguments, typescript will try to infer both when that call happens and since there is no inference site for A it will probably infer {} for it, making B never. The way to fix this, is to make B and when the second call occurs infer A, with the constraint that A must have a B key:
export type Omit<T, K> = Pick<T, Exclude<keyof T, K>>
const f = <B extends keyof any>(arg: B) => <A extends Record<B, any>>(obj: A): Omit<A, B> => {
delete obj[arg]
return obj
}
const a = f('test')
const b = a({ test: 1, other: ""})
const c = b.test; // error as expected
const c2 = b.other; // ok
answered Nov 25 '18 at 9:57
Titian Cernicova-DragomirTitian Cernicova-Dragomir
67.6k34664
Thank you very much, this helps tremendously. I realised thought I have one more piece in the HOC, that I omitted in the question. Would it be possible to back propagate theAtype somehow to the first function call. What I want to do is<B extends keyof any>(arg: B, fn: (arg: A) => void) => .... So basically, I give you attribute name and a function that will take object I will give you later. If that does not make sense I can create new question, my bad I did not realise this will be important.
– aocenas
Nov 25 '18 at 18:08
@aocenas could you post a new question with the full code ? It would make more sense to see it in full.. the comments is not a good place for a lot of code..
– Titian Cernicova-Dragomir
Nov 25 '18 at 18:10
Here is the new question, hope it makes sense stackoverflow.com/questions/53470559/…
– aocenas
Nov 25 '18 at 18:32
add a comment |
Since the first function has both type arguments, typescript will try to infer both when that call happens and since there is no inference site for A it will probably infer {} for it, making B never. The way to fix this, is to make B and when the second call occurs infer A, with the constraint that A must have a B key:
export type Omit<T, K> = Pick<T, Exclude<keyof T, K>>
const f = <B extends keyof any>(arg: B) => <A extends Record<B, any>>(obj: A): Omit<A, B> => {
delete obj[arg]
return obj
}
const a = f('test')
const b = a({ test: 1, other: ""})
const c = b.test; // error as expected
const c2 = b.other; // ok
answered Nov 25 '18 at 9:57
Titian Cernicova-DragomirTitian Cernicova-Dragomir
67.6k34664
Since the first function has both type arguments, typescript will try to infer both when that call happens and since there is no inference site for A it will probably infer {} for it, making B never. The way to fix this, is to make B and when the second call occurs infer A, with the constraint that A must have a B key:
export type Omit<T, K> = Pick<T, Exclude<keyof T, K>>
const f = <B extends keyof any>(arg: B) => <A extends Record<B, any>>(obj: A): Omit<A, B> => {
delete obj[arg]
return obj
}
const a = f('test')
const b = a({ test: 1, other: ""})
const c = b.test; // error as expected
const c2 = b.other; // ok
answered Nov 25 '18 at 9:57
Titian Cernicova-DragomirTitian Cernicova-Dragomir
67.6k34664
edited Nov 25 '18 at 10:04
answered Nov 25 '18 at 9:57
Titian Cernicova-DragomirTitian Cernicova-Dragomir
67.6k34664
answered Nov 25 '18 at 9:57
Titian Cernicova-DragomirTitian Cernicova-Dragomir
67.6k34664
answered Nov 25 '18 at 9:57
Titian Cernicova-DragomirTitian Cernicova-Dragomir
67.6k34664
67.6k34664
Thank you very much, this helps tremendously. I realised thought I have one more piece in the HOC, that I omitted in the question. Would it be possible to back propagate theAtype somehow to the first function call. What I want to do is<B extends keyof any>(arg: B, fn: (arg: A) => void) => .... So basically, I give you attribute name and a function that will take object I will give you later. If that does not make sense I can create new question, my bad I did not realise this will be important.
– aocenas
Nov 25 '18 at 18:08
@aocenas could you post a new question with the full code ? It would make more sense to see it in full.. the comments is not a good place for a lot of code..
– Titian Cernicova-Dragomir
Nov 25 '18 at 18:10
Here is the new question, hope it makes sense stackoverflow.com/questions/53470559/…
– aocenas
Nov 25 '18 at 18:32
add a comment |
Thank you very much, this helps tremendously. I realised thought I have one more piece in the HOC, that I omitted in the question. Would it be possible to back propagate theAtype somehow to the first function call. What I want to do is<B extends keyof any>(arg: B, fn: (arg: A) => void) => .... So basically, I give you attribute name and a function that will take object I will give you later. If that does not make sense I can create new question, my bad I did not realise this will be important.
– aocenas
Nov 25 '18 at 18:08
@aocenas could you post a new question with the full code ? It would make more sense to see it in full.. the comments is not a good place for a lot of code..
– Titian Cernicova-Dragomir
Nov 25 '18 at 18:10
Here is the new question, hope it makes sense stackoverflow.com/questions/53470559/…
– aocenas
Nov 25 '18 at 18:32
Thank you very much, this helps tremendously. I realised thought I have one more piece in the HOC, that I omitted in the question. Would it be possible to back propagate the
A type somehow to the first function call. What I want to do is <B extends keyof any>(arg: B, fn: (arg: A) => void) => .... So basically, I give you attribute name and a function that will take object I will give you later. If that does not make sense I can create new question, my bad I did not realise this will be important.– aocenas
Nov 25 '18 at 18:08
Thank you very much, this helps tremendously. I realised thought I have one more piece in the HOC, that I omitted in the question. Would it be possible to back propagate the
A type somehow to the first function call. What I want to do is <B extends keyof any>(arg: B, fn: (arg: A) => void) => .... So basically, I give you attribute name and a function that will take object I will give you later. If that does not make sense I can create new question, my bad I did not realise this will be important.– aocenas
Nov 25 '18 at 18:08
@aocenas could you post a new question with the full code ? It would make more sense to see it in full.. the comments is not a good place for a lot of code..
– Titian Cernicova-Dragomir
Nov 25 '18 at 18:10
@aocenas could you post a new question with the full code ? It would make more sense to see it in full.. the comments is not a good place for a lot of code..
– Titian Cernicova-Dragomir
Nov 25 '18 at 18:10
Here is the new question, hope it makes sense stackoverflow.com/questions/53470559/…
– aocenas
Nov 25 '18 at 18:32
Here is the new question, hope it makes sense stackoverflow.com/questions/53470559/…
– aocenas
Nov 25 '18 at 18:32
add a comment |
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