Way to define a type in typescript like [one of union of interfaces] & {//atrributes}
// The way types defined:
interface SudoA {
foo: string;
}
interface SudoB {
bar: string;
}
type Sudo = SudoA | SudoB;
type SuperSudo = Sudo & {
super: boolean;
}
const baz: SuperSudo = {
}
// typescript (3.1.6) says i have to define both `bar` and `foo` for the object
What I expect is that to put super attribute and other attributes (coming from Sudo type) should be optional. The question; is this a wrong expectation? Regardless if yes or no how is this achieved?
Edit
Corrected the type of baz, my mistake :/.
What I am expecting is for SuperSudo defining only super attribute should be enough and other attributes coming from union of interfaces should be optional.
I can define a new type liketype SuperSudo<T> = T & {} but this ends up using SuperSudo<T>which I find quite verbose.
Edit2
Changed the title
javascript typescript
asked Nov 21 '18 at 0:17
mdikici
4092617
add a comment |
// The way types defined:
interface SudoA {
foo: string;
}
interface SudoB {
bar: string;
}
type Sudo = SudoA | SudoB;
type SuperSudo = Sudo & {
super: boolean;
}
const baz: SuperSudo = {
}
// typescript (3.1.6) says i have to define both `bar` and `foo` for the object
What I expect is that to put super attribute and other attributes (coming from Sudo type) should be optional. The question; is this a wrong expectation? Regardless if yes or no how is this achieved?
Edit
Corrected the type of baz, my mistake :/.
What I am expecting is for SuperSudo defining only super attribute should be enough and other attributes coming from union of interfaces should be optional.
I can define a new type liketype SuperSudo<T> = T & {} but this ends up using SuperSudo<T>which I find quite verbose.
Edit2
Changed the title
javascript typescript
asked Nov 21 '18 at 0:17
mdikici
4092617
I don't quite understand what you're saying.bazis in error because it needs to have at least one of thefooorbarproperties in order to beSudoA | SudoB. And sincebazis of typeSudoI don't understand howSuperSudohas anything to do with it. Could you rephrase the question or add more examples of what you expect to work and what you expect to report an error?
– jcalz
Nov 21 '18 at 0:43
When i understand your question, then you find the answer in the documentation of the release notes 2.8 typescript. typescriptlang.org/docs/handbook/release-notes/… Here is everything covered what you need to know about typings and how to make attributes optional / readonly / modify datatypes in inheritance.
– Sly321
Nov 21 '18 at 0:59
1
You wantSuperSudoto only requiresuperand not require any of the properties fromSudoAorSudoB? What ifSudoAhas two or more properties (e.g.,{foo: string, qux: number})? Can aSuperSudocontain just one of those properties (e.g.,{super: true, foo: "a"})? It's still not clear to me what you're trying to do. Maybe you're looking for something liketype AllKeys<T> = T extends any ? keyof T : never; type SuperSudo = {[K in AllKeys<Sudo>]?: Extract<Sudo, Record<K, any>>[K]} & {super: boolean;}but it's hard to be sure.
– jcalz
Nov 21 '18 at 1:48
add a comment |
// The way types defined:
interface SudoA {
foo: string;
}
interface SudoB {
bar: string;
}
type Sudo = SudoA | SudoB;
type SuperSudo = Sudo & {
super: boolean;
}
const baz: SuperSudo = {
}
// typescript (3.1.6) says i have to define both `bar` and `foo` for the object
What I expect is that to put super attribute and other attributes (coming from Sudo type) should be optional. The question; is this a wrong expectation? Regardless if yes or no how is this achieved?
Edit
Corrected the type of baz, my mistake :/.
What I am expecting is for SuperSudo defining only super attribute should be enough and other attributes coming from union of interfaces should be optional.
I can define a new type liketype SuperSudo<T> = T & {} but this ends up using SuperSudo<T>which I find quite verbose.
Edit2
Changed the title
javascript typescript
asked Nov 21 '18 at 0:17
mdikici
4092617
// The way types defined:
interface SudoA {
foo: string;
}
interface SudoB {
bar: string;
}
type Sudo = SudoA | SudoB;
type SuperSudo = Sudo & {
super: boolean;
}
const baz: SuperSudo = {
}
// typescript (3.1.6) says i have to define both `bar` and `foo` for the object
What I expect is that to put super attribute and other attributes (coming from Sudo type) should be optional. The question; is this a wrong expectation? Regardless if yes or no how is this achieved?
Edit
Corrected the type of baz, my mistake :/.
What I am expecting is for SuperSudo defining only super attribute should be enough and other attributes coming from union of interfaces should be optional.
I can define a new type liketype SuperSudo<T> = T & {} but this ends up using SuperSudo<T>which I find quite verbose.
Edit2
Changed the title
javascript typescript
javascript typescript
asked Nov 21 '18 at 0:17
mdikici
4092617
asked Nov 21 '18 at 0:17
mdikici
4092617
edited Nov 21 '18 at 1:32
asked Nov 21 '18 at 0:17
mdikici
4092617
asked Nov 21 '18 at 0:17
mdikici
4092617
asked Nov 21 '18 at 0:17
mdikici
4092617
4092617
I don't quite understand what you're saying.bazis in error because it needs to have at least one of thefooorbarproperties in order to beSudoA | SudoB. And sincebazis of typeSudoI don't understand howSuperSudohas anything to do with it. Could you rephrase the question or add more examples of what you expect to work and what you expect to report an error?
– jcalz
Nov 21 '18 at 0:43
When i understand your question, then you find the answer in the documentation of the release notes 2.8 typescript. typescriptlang.org/docs/handbook/release-notes/… Here is everything covered what you need to know about typings and how to make attributes optional / readonly / modify datatypes in inheritance.
– Sly321
Nov 21 '18 at 0:59
1
You wantSuperSudoto only requiresuperand not require any of the properties fromSudoAorSudoB? What ifSudoAhas two or more properties (e.g.,{foo: string, qux: number})? Can aSuperSudocontain just one of those properties (e.g.,{super: true, foo: "a"})? It's still not clear to me what you're trying to do. Maybe you're looking for something liketype AllKeys<T> = T extends any ? keyof T : never; type SuperSudo = {[K in AllKeys<Sudo>]?: Extract<Sudo, Record<K, any>>[K]} & {super: boolean;}but it's hard to be sure.
– jcalz
Nov 21 '18 at 1:48
add a comment |
I don't quite understand what you're saying.bazis in error because it needs to have at least one of thefooorbarproperties in order to beSudoA | SudoB. And sincebazis of typeSudoI don't understand howSuperSudohas anything to do with it. Could you rephrase the question or add more examples of what you expect to work and what you expect to report an error?
– jcalz
Nov 21 '18 at 0:43
When i understand your question, then you find the answer in the documentation of the release notes 2.8 typescript. typescriptlang.org/docs/handbook/release-notes/… Here is everything covered what you need to know about typings and how to make attributes optional / readonly / modify datatypes in inheritance.
– Sly321
Nov 21 '18 at 0:59
1
You wantSuperSudoto only requiresuperand not require any of the properties fromSudoAorSudoB? What ifSudoAhas two or more properties (e.g.,{foo: string, qux: number})? Can aSuperSudocontain just one of those properties (e.g.,{super: true, foo: "a"})? It's still not clear to me what you're trying to do. Maybe you're looking for something liketype AllKeys<T> = T extends any ? keyof T : never; type SuperSudo = {[K in AllKeys<Sudo>]?: Extract<Sudo, Record<K, any>>[K]} & {super: boolean;}but it's hard to be sure.
– jcalz
Nov 21 '18 at 1:48
I don't quite understand what you're saying.
baz is in error because it needs to have at least one of the foo or bar properties in order to be SudoA | SudoB. And since baz is of type Sudo I don't understand how SuperSudo has anything to do with it. Could you rephrase the question or add more examples of what you expect to work and what you expect to report an error?– jcalz
Nov 21 '18 at 0:43
I don't quite understand what you're saying.
baz is in error because it needs to have at least one of the foo or bar properties in order to be SudoA | SudoB. And since baz is of type Sudo I don't understand how SuperSudo has anything to do with it. Could you rephrase the question or add more examples of what you expect to work and what you expect to report an error?– jcalz
Nov 21 '18 at 0:43
When i understand your question, then you find the answer in the documentation of the release notes 2.8 typescript. typescriptlang.org/docs/handbook/release-notes/… Here is everything covered what you need to know about typings and how to make attributes optional / readonly / modify datatypes in inheritance.
– Sly321
Nov 21 '18 at 0:59
When i understand your question, then you find the answer in the documentation of the release notes 2.8 typescript. typescriptlang.org/docs/handbook/release-notes/… Here is everything covered what you need to know about typings and how to make attributes optional / readonly / modify datatypes in inheritance.
– Sly321
Nov 21 '18 at 0:59
1
1
You want
SuperSudo to only require super and not require any of the properties from SudoA or SudoB? What if SudoA has two or more properties (e.g., {foo: string, qux: number})? Can a SuperSudo contain just one of those properties (e.g., {super: true, foo: "a"})? It's still not clear to me what you're trying to do. Maybe you're looking for something like type AllKeys<T> = T extends any ? keyof T : never; type SuperSudo = {[K in AllKeys<Sudo>]?: Extract<Sudo, Record<K, any>>[K]} & {super: boolean;} but it's hard to be sure.– jcalz
Nov 21 '18 at 1:48
You want
SuperSudo to only require super and not require any of the properties from SudoA or SudoB? What if SudoA has two or more properties (e.g., {foo: string, qux: number})? Can a SuperSudo contain just one of those properties (e.g., {super: true, foo: "a"})? It's still not clear to me what you're trying to do. Maybe you're looking for something like type AllKeys<T> = T extends any ? keyof T : never; type SuperSudo = {[K in AllKeys<Sudo>]?: Extract<Sudo, Record<K, any>>[K]} & {super: boolean;} but it's hard to be sure.– jcalz
Nov 21 '18 at 1:48
add a comment |
1 Answer
1
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Have you considered using a type discriminator?
Without one, then specifying either foo, bar or both foo and bar will always satisfy the union type.
With one might look like:
interface SudoA {
_type: "a";
foo: string;
}
interface SudoB {
_type: "b";
bar: string;
}
type Sudo = SudoA | SudoB;
type SuperSudo = Sudo & {
super: boolean;
}
const baz1_valid: SuperSudo = {
_type: "a",
super: true,
foo: "bar",
}
const baz2_valid: SuperSudo = {
_type: "b",
super: true,
bar: "foo",
}
The field name does not have to be _type, it simply must be the same field name with different type values on each variant.
answered Nov 21 '18 at 4:49
Alex
1,9841415
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Have you considered using a type discriminator?
Without one, then specifying either foo, bar or both foo and bar will always satisfy the union type.
With one might look like:
interface SudoA {
_type: "a";
foo: string;
}
interface SudoB {
_type: "b";
bar: string;
}
type Sudo = SudoA | SudoB;
type SuperSudo = Sudo & {
super: boolean;
}
const baz1_valid: SuperSudo = {
_type: "a",
super: true,
foo: "bar",
}
const baz2_valid: SuperSudo = {
_type: "b",
super: true,
bar: "foo",
}
The field name does not have to be _type, it simply must be the same field name with different type values on each variant.
answered Nov 21 '18 at 4:49
Alex
1,9841415
add a comment |
Have you considered using a type discriminator?
Without one, then specifying either foo, bar or both foo and bar will always satisfy the union type.
With one might look like:
interface SudoA {
_type: "a";
foo: string;
}
interface SudoB {
_type: "b";
bar: string;
}
type Sudo = SudoA | SudoB;
type SuperSudo = Sudo & {
super: boolean;
}
const baz1_valid: SuperSudo = {
_type: "a",
super: true,
foo: "bar",
}
const baz2_valid: SuperSudo = {
_type: "b",
super: true,
bar: "foo",
}
The field name does not have to be _type, it simply must be the same field name with different type values on each variant.
answered Nov 21 '18 at 4:49
Alex
1,9841415
add a comment |
Have you considered using a type discriminator?
Without one, then specifying either foo, bar or both foo and bar will always satisfy the union type.
With one might look like:
interface SudoA {
_type: "a";
foo: string;
}
interface SudoB {
_type: "b";
bar: string;
}
type Sudo = SudoA | SudoB;
type SuperSudo = Sudo & {
super: boolean;
}
const baz1_valid: SuperSudo = {
_type: "a",
super: true,
foo: "bar",
}
const baz2_valid: SuperSudo = {
_type: "b",
super: true,
bar: "foo",
}
The field name does not have to be _type, it simply must be the same field name with different type values on each variant.
answered Nov 21 '18 at 4:49
Alex
1,9841415
Have you considered using a type discriminator?
Without one, then specifying either foo, bar or both foo and bar will always satisfy the union type.
With one might look like:
interface SudoA {
_type: "a";
foo: string;
}
interface SudoB {
_type: "b";
bar: string;
}
type Sudo = SudoA | SudoB;
type SuperSudo = Sudo & {
super: boolean;
}
const baz1_valid: SuperSudo = {
_type: "a",
super: true,
foo: "bar",
}
const baz2_valid: SuperSudo = {
_type: "b",
super: true,
bar: "foo",
}
The field name does not have to be _type, it simply must be the same field name with different type values on each variant.
answered Nov 21 '18 at 4:49
Alex
1,9841415
answered Nov 21 '18 at 4:49
Alex
1,9841415
answered Nov 21 '18 at 4:49
Alex
1,9841415
answered Nov 21 '18 at 4:49
Alex
1,9841415
1,9841415
add a comment |
add a comment |
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I don't quite understand what you're saying.
bazis in error because it needs to have at least one of thefooorbarproperties in order to beSudoA | SudoB. And sincebazis of typeSudoI don't understand howSuperSudohas anything to do with it. Could you rephrase the question or add more examples of what you expect to work and what you expect to report an error?– jcalz
Nov 21 '18 at 0:43
When i understand your question, then you find the answer in the documentation of the release notes 2.8 typescript. typescriptlang.org/docs/handbook/release-notes/… Here is everything covered what you need to know about typings and how to make attributes optional / readonly / modify datatypes in inheritance.
– Sly321
Nov 21 '18 at 0:59
1
You want
SuperSudoto only requiresuperand not require any of the properties fromSudoAorSudoB? What ifSudoAhas two or more properties (e.g.,{foo: string, qux: number})? Can aSuperSudocontain just one of those properties (e.g.,{super: true, foo: "a"})? It's still not clear to me what you're trying to do. Maybe you're looking for something liketype AllKeys<T> = T extends any ? keyof T : never; type SuperSudo = {[K in AllKeys<Sudo>]?: Extract<Sudo, Record<K, any>>[K]} & {super: boolean;}but it's hard to be sure.– jcalz
Nov 21 '18 at 1:48