typescript remove optional property
I'm trying to build a dynamic type for a builder
export type Builder<T, K extends keyof T> = {
[P in K]: (value: T[P]) => Builder<T, K>;
} & {
build(): Readonly<T>;
};
If I have a class or an interface with optional properties, I get this kind of error:
$ tsc
test/OtherClass.ts:29:1 - error TS2722: Cannot invoke an object which is possibly 'undefined'.
29 OtherClass.builder().patate(Patate.AU_FOUR).build();
~~~~~~~~~~~~~~~~~~~~~~~~~~~
This is my class
export class OtherClass {
constructor(
public literal: string | undefined,
public patate?: Patate,
public hello: Readonly<Hello> = {},
) { }
static builder: () => Builder<OtherClass, keyof OtherClass> = builder.for(OtherClass);
}
I was expecting the type to create a builder with a non-optional method for each properties, but for some reason, the optionality of patate seem inherent to the key and not the type. I don't get this behavior with the property literal
It looks like an issue to me. I'm using typescript 3.1.4. Is there another way to remove the question mark dynamically?
I've tried to use the NonNullable helper to create first a copy of my type with nothing nullable, but patate remains optional.
This is the effective type that vscode gives me
(property) patate?: ((value: Patate | undefined) => Builder<OtherClass, "literal" | "patate" | "hello">) | undefined
typescript typescript3.0
asked Nov 15 '18 at 3:07
FrancisFrancis
2,7371243
add a comment |
I'm trying to build a dynamic type for a builder
export type Builder<T, K extends keyof T> = {
[P in K]: (value: T[P]) => Builder<T, K>;
} & {
build(): Readonly<T>;
};
If I have a class or an interface with optional properties, I get this kind of error:
$ tsc
test/OtherClass.ts:29:1 - error TS2722: Cannot invoke an object which is possibly 'undefined'.
29 OtherClass.builder().patate(Patate.AU_FOUR).build();
~~~~~~~~~~~~~~~~~~~~~~~~~~~
This is my class
export class OtherClass {
constructor(
public literal: string | undefined,
public patate?: Patate,
public hello: Readonly<Hello> = {},
) { }
static builder: () => Builder<OtherClass, keyof OtherClass> = builder.for(OtherClass);
}
I was expecting the type to create a builder with a non-optional method for each properties, but for some reason, the optionality of patate seem inherent to the key and not the type. I don't get this behavior with the property literal
It looks like an issue to me. I'm using typescript 3.1.4. Is there another way to remove the question mark dynamically?
I've tried to use the NonNullable helper to create first a copy of my type with nothing nullable, but patate remains optional.
This is the effective type that vscode gives me
(property) patate?: ((value: Patate | undefined) => Builder<OtherClass, "literal" | "patate" | "hello">) | undefined
typescript typescript3.0
asked Nov 15 '18 at 3:07
FrancisFrancis
2,7371243
add a comment |
I'm trying to build a dynamic type for a builder
export type Builder<T, K extends keyof T> = {
[P in K]: (value: T[P]) => Builder<T, K>;
} & {
build(): Readonly<T>;
};
If I have a class or an interface with optional properties, I get this kind of error:
$ tsc
test/OtherClass.ts:29:1 - error TS2722: Cannot invoke an object which is possibly 'undefined'.
29 OtherClass.builder().patate(Patate.AU_FOUR).build();
~~~~~~~~~~~~~~~~~~~~~~~~~~~
This is my class
export class OtherClass {
constructor(
public literal: string | undefined,
public patate?: Patate,
public hello: Readonly<Hello> = {},
) { }
static builder: () => Builder<OtherClass, keyof OtherClass> = builder.for(OtherClass);
}
I was expecting the type to create a builder with a non-optional method for each properties, but for some reason, the optionality of patate seem inherent to the key and not the type. I don't get this behavior with the property literal
It looks like an issue to me. I'm using typescript 3.1.4. Is there another way to remove the question mark dynamically?
I've tried to use the NonNullable helper to create first a copy of my type with nothing nullable, but patate remains optional.
This is the effective type that vscode gives me
(property) patate?: ((value: Patate | undefined) => Builder<OtherClass, "literal" | "patate" | "hello">) | undefined
typescript typescript3.0
asked Nov 15 '18 at 3:07
FrancisFrancis
2,7371243
I'm trying to build a dynamic type for a builder
export type Builder<T, K extends keyof T> = {
[P in K]: (value: T[P]) => Builder<T, K>;
} & {
build(): Readonly<T>;
};
If I have a class or an interface with optional properties, I get this kind of error:
$ tsc
test/OtherClass.ts:29:1 - error TS2722: Cannot invoke an object which is possibly 'undefined'.
29 OtherClass.builder().patate(Patate.AU_FOUR).build();
~~~~~~~~~~~~~~~~~~~~~~~~~~~
This is my class
export class OtherClass {
constructor(
public literal: string | undefined,
public patate?: Patate,
public hello: Readonly<Hello> = {},
) { }
static builder: () => Builder<OtherClass, keyof OtherClass> = builder.for(OtherClass);
}
I was expecting the type to create a builder with a non-optional method for each properties, but for some reason, the optionality of patate seem inherent to the key and not the type. I don't get this behavior with the property literal
It looks like an issue to me. I'm using typescript 3.1.4. Is there another way to remove the question mark dynamically?
I've tried to use the NonNullable helper to create first a copy of my type with nothing nullable, but patate remains optional.
This is the effective type that vscode gives me
(property) patate?: ((value: Patate | undefined) => Builder<OtherClass, "literal" | "patate" | "hello">) | undefined
typescript typescript3.0
typescript typescript3.0
asked Nov 15 '18 at 3:07
FrancisFrancis
2,7371243
asked Nov 15 '18 at 3:07
FrancisFrancis
2,7371243
asked Nov 15 '18 at 3:07
FrancisFrancis
2,7371243
asked Nov 15 '18 at 3:07
FrancisFrancis
2,7371243
asked Nov 15 '18 at 3:07
FrancisFrancis
2,7371243
2,7371243
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Does this help?
export type Builder<T, K extends keyof T> = {
[P in K]-?: (value: T[P]) => Builder<T, K>;
} & {
build(): Readonly<T>;
};
Note the -? which does what you want - removes optionality.
answered Nov 15 '18 at 3:24
Nurbol AlpysbayevNurbol Alpysbayev
3,9621227
Yes thanks, but do you know why it inherits the optionality from the key name?
– Francis
Nov 15 '18 at 14:00
3
When the "constraint" of a mapped type (the type afterin, which determines the set of properties) is of the formkeyof Tor is a type parameter constrained bykeyof Tfor some typeT, the mapped type is "homomorphic" and copies modifiers (optional and readonly) fromT. There's a bit about this here.
– Matt McCutchen
Nov 15 '18 at 14:09
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53311845%2ftypescript-remove-optional-property%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Does this help?
export type Builder<T, K extends keyof T> = {
[P in K]-?: (value: T[P]) => Builder<T, K>;
} & {
build(): Readonly<T>;
};
Note the -? which does what you want - removes optionality.
answered Nov 15 '18 at 3:24
Nurbol AlpysbayevNurbol Alpysbayev
3,9621227
Yes thanks, but do you know why it inherits the optionality from the key name?
– Francis
Nov 15 '18 at 14:00
3
When the "constraint" of a mapped type (the type afterin, which determines the set of properties) is of the formkeyof Tor is a type parameter constrained bykeyof Tfor some typeT, the mapped type is "homomorphic" and copies modifiers (optional and readonly) fromT. There's a bit about this here.
– Matt McCutchen
Nov 15 '18 at 14:09
add a comment |
Does this help?
export type Builder<T, K extends keyof T> = {
[P in K]-?: (value: T[P]) => Builder<T, K>;
} & {
build(): Readonly<T>;
};
Note the -? which does what you want - removes optionality.
answered Nov 15 '18 at 3:24
Nurbol AlpysbayevNurbol Alpysbayev
3,9621227
Yes thanks, but do you know why it inherits the optionality from the key name?
– Francis
Nov 15 '18 at 14:00
3
When the "constraint" of a mapped type (the type afterin, which determines the set of properties) is of the formkeyof Tor is a type parameter constrained bykeyof Tfor some typeT, the mapped type is "homomorphic" and copies modifiers (optional and readonly) fromT. There's a bit about this here.
– Matt McCutchen
Nov 15 '18 at 14:09
add a comment |
Does this help?
export type Builder<T, K extends keyof T> = {
[P in K]-?: (value: T[P]) => Builder<T, K>;
} & {
build(): Readonly<T>;
};
Note the -? which does what you want - removes optionality.
answered Nov 15 '18 at 3:24
Nurbol AlpysbayevNurbol Alpysbayev
3,9621227
Does this help?
export type Builder<T, K extends keyof T> = {
[P in K]-?: (value: T[P]) => Builder<T, K>;
} & {
build(): Readonly<T>;
};
Note the -? which does what you want - removes optionality.
answered Nov 15 '18 at 3:24
Nurbol AlpysbayevNurbol Alpysbayev
3,9621227
answered Nov 15 '18 at 3:24
Nurbol AlpysbayevNurbol Alpysbayev
3,9621227
answered Nov 15 '18 at 3:24
Nurbol AlpysbayevNurbol Alpysbayev
3,9621227
answered Nov 15 '18 at 3:24
Nurbol AlpysbayevNurbol Alpysbayev
3,9621227
3,9621227
Yes thanks, but do you know why it inherits the optionality from the key name?
– Francis
Nov 15 '18 at 14:00
3
When the "constraint" of a mapped type (the type afterin, which determines the set of properties) is of the formkeyof Tor is a type parameter constrained bykeyof Tfor some typeT, the mapped type is "homomorphic" and copies modifiers (optional and readonly) fromT. There's a bit about this here.
– Matt McCutchen
Nov 15 '18 at 14:09
add a comment |
Yes thanks, but do you know why it inherits the optionality from the key name?
– Francis
Nov 15 '18 at 14:00
3
When the "constraint" of a mapped type (the type afterin, which determines the set of properties) is of the formkeyof Tor is a type parameter constrained bykeyof Tfor some typeT, the mapped type is "homomorphic" and copies modifiers (optional and readonly) fromT. There's a bit about this here.
– Matt McCutchen
Nov 15 '18 at 14:09
Yes thanks, but do you know why it inherits the optionality from the key name?
– Francis
Nov 15 '18 at 14:00
Yes thanks, but do you know why it inherits the optionality from the key name?
– Francis
Nov 15 '18 at 14:00
3
3
When the "constraint" of a mapped type (the type after
in, which determines the set of properties) is of the form keyof T or is a type parameter constrained by keyof T for some type T, the mapped type is "homomorphic" and copies modifiers (optional and readonly) from T. There's a bit about this here.– Matt McCutchen
Nov 15 '18 at 14:09
When the "constraint" of a mapped type (the type after
in, which determines the set of properties) is of the form keyof T or is a type parameter constrained by keyof T for some type T, the mapped type is "homomorphic" and copies modifiers (optional and readonly) from T. There's a bit about this here.– Matt McCutchen
Nov 15 '18 at 14:09
add a comment |
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53311845%2ftypescript-remove-optional-property%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
