How to exclude a key from an interface in TypeScript












10















In TypeScript, you can combine two interface types like this



interface Foo {
var1: string
}

interface Bar {
var2: string
}

type Combined = Foo & Bar


Instead of combining keys, I want to exclude keys from one interface to another. Is there anyway you can do it in TypeScript?



The reason is, I have an HOC, which manages a property value for other wrapped component like this



export default function valueHOC<P> (
Comp: React.ComponentClass<P> | React.StatelessComponent<P>
): React.ComponentClass<P> {
return class WrappedComponent extends React.Component<P, State> {
render () {
return (
<Comp
{...this.props}
value={this.state.value}
/>
)
}
}


With that I can write



const ValuedComponent = valueHOC(MyComponent)


then



<ValuedComponent />


but the problem is, the returned component type is also using the props type from given component, so TypeScript will complain and ask me to provide the value prop. As a result, I will have to write something like



<ValuedComponent value="foo" />


Which the value will not be used anyway. What I want here is to return an interface without specific keys, I want to have something like this



React.ComponentClass<P - {value: string}>


Then the value will not be needed in the returned component. Is it possible in TypeScript for now?










share|improve this question


















  • 1





    You may check out Pick<T, K keyof T>. But you can't pick things out dynamically.

    – unional
    Jul 9 '17 at 6:26
















10















In TypeScript, you can combine two interface types like this



interface Foo {
var1: string
}

interface Bar {
var2: string
}

type Combined = Foo & Bar


Instead of combining keys, I want to exclude keys from one interface to another. Is there anyway you can do it in TypeScript?



The reason is, I have an HOC, which manages a property value for other wrapped component like this



export default function valueHOC<P> (
Comp: React.ComponentClass<P> | React.StatelessComponent<P>
): React.ComponentClass<P> {
return class WrappedComponent extends React.Component<P, State> {
render () {
return (
<Comp
{...this.props}
value={this.state.value}
/>
)
}
}


With that I can write



const ValuedComponent = valueHOC(MyComponent)


then



<ValuedComponent />


but the problem is, the returned component type is also using the props type from given component, so TypeScript will complain and ask me to provide the value prop. As a result, I will have to write something like



<ValuedComponent value="foo" />


Which the value will not be used anyway. What I want here is to return an interface without specific keys, I want to have something like this



React.ComponentClass<P - {value: string}>


Then the value will not be needed in the returned component. Is it possible in TypeScript for now?










share|improve this question


















  • 1





    You may check out Pick<T, K keyof T>. But you can't pick things out dynamically.

    – unional
    Jul 9 '17 at 6:26














10












10








10


4






In TypeScript, you can combine two interface types like this



interface Foo {
var1: string
}

interface Bar {
var2: string
}

type Combined = Foo & Bar


Instead of combining keys, I want to exclude keys from one interface to another. Is there anyway you can do it in TypeScript?



The reason is, I have an HOC, which manages a property value for other wrapped component like this



export default function valueHOC<P> (
Comp: React.ComponentClass<P> | React.StatelessComponent<P>
): React.ComponentClass<P> {
return class WrappedComponent extends React.Component<P, State> {
render () {
return (
<Comp
{...this.props}
value={this.state.value}
/>
)
}
}


With that I can write



const ValuedComponent = valueHOC(MyComponent)


then



<ValuedComponent />


but the problem is, the returned component type is also using the props type from given component, so TypeScript will complain and ask me to provide the value prop. As a result, I will have to write something like



<ValuedComponent value="foo" />


Which the value will not be used anyway. What I want here is to return an interface without specific keys, I want to have something like this



React.ComponentClass<P - {value: string}>


Then the value will not be needed in the returned component. Is it possible in TypeScript for now?










share|improve this question














In TypeScript, you can combine two interface types like this



interface Foo {
var1: string
}

interface Bar {
var2: string
}

type Combined = Foo & Bar


Instead of combining keys, I want to exclude keys from one interface to another. Is there anyway you can do it in TypeScript?



The reason is, I have an HOC, which manages a property value for other wrapped component like this



export default function valueHOC<P> (
Comp: React.ComponentClass<P> | React.StatelessComponent<P>
): React.ComponentClass<P> {
return class WrappedComponent extends React.Component<P, State> {
render () {
return (
<Comp
{...this.props}
value={this.state.value}
/>
)
}
}


With that I can write



const ValuedComponent = valueHOC(MyComponent)


then



<ValuedComponent />


but the problem is, the returned component type is also using the props type from given component, so TypeScript will complain and ask me to provide the value prop. As a result, I will have to write something like



<ValuedComponent value="foo" />


Which the value will not be used anyway. What I want here is to return an interface without specific keys, I want to have something like this



React.ComponentClass<P - {value: string}>


Then the value will not be needed in the returned component. Is it possible in TypeScript for now?







typescript






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Jul 8 '17 at 7:13









Fang-Pen LinFang-Pen Lin

5,50484882




5,50484882








  • 1





    You may check out Pick<T, K keyof T>. But you can't pick things out dynamically.

    – unional
    Jul 9 '17 at 6:26














  • 1





    You may check out Pick<T, K keyof T>. But you can't pick things out dynamically.

    – unional
    Jul 9 '17 at 6:26








1




1





You may check out Pick<T, K keyof T>. But you can't pick things out dynamically.

– unional
Jul 9 '17 at 6:26





You may check out Pick<T, K keyof T>. But you can't pick things out dynamically.

– unional
Jul 9 '17 at 6:26












3 Answers
3






active

oldest

votes


















11














In TypeScript 2.8 you can now do the following:



interface Foo {
attribute1: string;
optional2?: string;
excludePlease: string;
}

// Define Omit. Can be defined in a utilities package
type Omit<T, K extends keyof T> = Pick<T, Exclude<keyof T, K>>

// Use Omit to exclude one or more fields (use "excludePlease"|"field2"|"field3" etc to exclude multiple)
type Bar = Omit<Foo, "excludePlease">
const b: Bar = {
attribute1: ''
};


So in relation to your question the following might be what you want:



export default function valueHOC<P> (
Comp: React.ComponentClass<P> | React.StatelessComponent<P>
): React.ComponentClass<Omit<P, "value">> {
return class WrappedComponent extends React.Component<Omit<P, "value">, State> {
render () {
return (
<Comp
{...this.props}
value={this.state.value}
/>
)
}
}





share|improve this answer





















  • 1





    Whilst this works, it will make optional parameters required. You can do this to prevent that: stackoverflow.com/a/48216010/616589

    – Simon
    Aug 14 '18 at 17:07











  • I don't dispute the accuracy of the answer, but that's an absurdly round the houses way of achieving a simple end - it will take anyone reading it some non-trivial amount of time to work out just what the hell is going in. Is there any logic at all as to why the typescript team went this way?

    – havlock
    Sep 26 '18 at 17:49













  • @havlock sorry I recently edited it to simplify and improve the answer but I forget to also use the simplified Omit code in the simplify example. Is that better now?

    – AJP
    Sep 27 '18 at 13:37











  • @AJP I was moaning about typescript, not your answer, but thanks - yes, that is slightly more straightforward. Still absolutely bonkers that they make us jump through such ridiculous syntax to achieve something simple - I fear for the language if they really think this sort of thing is OK.

    – havlock
    Sep 28 '18 at 6:56











  • @havlock :) Would be interesting to see examples of other typed languages where this is dealt with better :)

    – AJP
    Sep 28 '18 at 16:41



















0














You can't remove properties from already existing interfaces. Even trying to extend existing interface with the interface having value property set as optional will return an error.



To avoid this issue, modify the typings of the target component so value property is optional.



e.g.



// ...
class MyComponent extends React.Component<{value?: string}, State> {
// ...
}


and then component produced when using High Order Function



const valuedComponent = valueHOC(MyComponent);


won't ask for value prop.






share|improve this answer































    0














    There is utility-types library that has Substract mapped type:



    import { Subtract } from 'utility-types';

    type Props = { name: string; age: number; visible: boolean };
    type DefaultProps = { age: number };

    type RequiredProps = Subtract<Props, DefaultProps>;
    // Expect: { name: string; visible: boolean; }





    share|improve this answer

























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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      11














      In TypeScript 2.8 you can now do the following:



      interface Foo {
      attribute1: string;
      optional2?: string;
      excludePlease: string;
      }

      // Define Omit. Can be defined in a utilities package
      type Omit<T, K extends keyof T> = Pick<T, Exclude<keyof T, K>>

      // Use Omit to exclude one or more fields (use "excludePlease"|"field2"|"field3" etc to exclude multiple)
      type Bar = Omit<Foo, "excludePlease">
      const b: Bar = {
      attribute1: ''
      };


      So in relation to your question the following might be what you want:



      export default function valueHOC<P> (
      Comp: React.ComponentClass<P> | React.StatelessComponent<P>
      ): React.ComponentClass<Omit<P, "value">> {
      return class WrappedComponent extends React.Component<Omit<P, "value">, State> {
      render () {
      return (
      <Comp
      {...this.props}
      value={this.state.value}
      />
      )
      }
      }





      share|improve this answer





















      • 1





        Whilst this works, it will make optional parameters required. You can do this to prevent that: stackoverflow.com/a/48216010/616589

        – Simon
        Aug 14 '18 at 17:07











      • I don't dispute the accuracy of the answer, but that's an absurdly round the houses way of achieving a simple end - it will take anyone reading it some non-trivial amount of time to work out just what the hell is going in. Is there any logic at all as to why the typescript team went this way?

        – havlock
        Sep 26 '18 at 17:49













      • @havlock sorry I recently edited it to simplify and improve the answer but I forget to also use the simplified Omit code in the simplify example. Is that better now?

        – AJP
        Sep 27 '18 at 13:37











      • @AJP I was moaning about typescript, not your answer, but thanks - yes, that is slightly more straightforward. Still absolutely bonkers that they make us jump through such ridiculous syntax to achieve something simple - I fear for the language if they really think this sort of thing is OK.

        – havlock
        Sep 28 '18 at 6:56











      • @havlock :) Would be interesting to see examples of other typed languages where this is dealt with better :)

        – AJP
        Sep 28 '18 at 16:41
















      11














      In TypeScript 2.8 you can now do the following:



      interface Foo {
      attribute1: string;
      optional2?: string;
      excludePlease: string;
      }

      // Define Omit. Can be defined in a utilities package
      type Omit<T, K extends keyof T> = Pick<T, Exclude<keyof T, K>>

      // Use Omit to exclude one or more fields (use "excludePlease"|"field2"|"field3" etc to exclude multiple)
      type Bar = Omit<Foo, "excludePlease">
      const b: Bar = {
      attribute1: ''
      };


      So in relation to your question the following might be what you want:



      export default function valueHOC<P> (
      Comp: React.ComponentClass<P> | React.StatelessComponent<P>
      ): React.ComponentClass<Omit<P, "value">> {
      return class WrappedComponent extends React.Component<Omit<P, "value">, State> {
      render () {
      return (
      <Comp
      {...this.props}
      value={this.state.value}
      />
      )
      }
      }





      share|improve this answer





















      • 1





        Whilst this works, it will make optional parameters required. You can do this to prevent that: stackoverflow.com/a/48216010/616589

        – Simon
        Aug 14 '18 at 17:07











      • I don't dispute the accuracy of the answer, but that's an absurdly round the houses way of achieving a simple end - it will take anyone reading it some non-trivial amount of time to work out just what the hell is going in. Is there any logic at all as to why the typescript team went this way?

        – havlock
        Sep 26 '18 at 17:49













      • @havlock sorry I recently edited it to simplify and improve the answer but I forget to also use the simplified Omit code in the simplify example. Is that better now?

        – AJP
        Sep 27 '18 at 13:37











      • @AJP I was moaning about typescript, not your answer, but thanks - yes, that is slightly more straightforward. Still absolutely bonkers that they make us jump through such ridiculous syntax to achieve something simple - I fear for the language if they really think this sort of thing is OK.

        – havlock
        Sep 28 '18 at 6:56











      • @havlock :) Would be interesting to see examples of other typed languages where this is dealt with better :)

        – AJP
        Sep 28 '18 at 16:41














      11












      11








      11







      In TypeScript 2.8 you can now do the following:



      interface Foo {
      attribute1: string;
      optional2?: string;
      excludePlease: string;
      }

      // Define Omit. Can be defined in a utilities package
      type Omit<T, K extends keyof T> = Pick<T, Exclude<keyof T, K>>

      // Use Omit to exclude one or more fields (use "excludePlease"|"field2"|"field3" etc to exclude multiple)
      type Bar = Omit<Foo, "excludePlease">
      const b: Bar = {
      attribute1: ''
      };


      So in relation to your question the following might be what you want:



      export default function valueHOC<P> (
      Comp: React.ComponentClass<P> | React.StatelessComponent<P>
      ): React.ComponentClass<Omit<P, "value">> {
      return class WrappedComponent extends React.Component<Omit<P, "value">, State> {
      render () {
      return (
      <Comp
      {...this.props}
      value={this.state.value}
      />
      )
      }
      }





      share|improve this answer















      In TypeScript 2.8 you can now do the following:



      interface Foo {
      attribute1: string;
      optional2?: string;
      excludePlease: string;
      }

      // Define Omit. Can be defined in a utilities package
      type Omit<T, K extends keyof T> = Pick<T, Exclude<keyof T, K>>

      // Use Omit to exclude one or more fields (use "excludePlease"|"field2"|"field3" etc to exclude multiple)
      type Bar = Omit<Foo, "excludePlease">
      const b: Bar = {
      attribute1: ''
      };


      So in relation to your question the following might be what you want:



      export default function valueHOC<P> (
      Comp: React.ComponentClass<P> | React.StatelessComponent<P>
      ): React.ComponentClass<Omit<P, "value">> {
      return class WrappedComponent extends React.Component<Omit<P, "value">, State> {
      render () {
      return (
      <Comp
      {...this.props}
      value={this.state.value}
      />
      )
      }
      }






      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited Sep 27 '18 at 13:35

























      answered Jun 4 '18 at 21:37









      AJPAJP

      9,153155589




      9,153155589








      • 1





        Whilst this works, it will make optional parameters required. You can do this to prevent that: stackoverflow.com/a/48216010/616589

        – Simon
        Aug 14 '18 at 17:07











      • I don't dispute the accuracy of the answer, but that's an absurdly round the houses way of achieving a simple end - it will take anyone reading it some non-trivial amount of time to work out just what the hell is going in. Is there any logic at all as to why the typescript team went this way?

        – havlock
        Sep 26 '18 at 17:49













      • @havlock sorry I recently edited it to simplify and improve the answer but I forget to also use the simplified Omit code in the simplify example. Is that better now?

        – AJP
        Sep 27 '18 at 13:37











      • @AJP I was moaning about typescript, not your answer, but thanks - yes, that is slightly more straightforward. Still absolutely bonkers that they make us jump through such ridiculous syntax to achieve something simple - I fear for the language if they really think this sort of thing is OK.

        – havlock
        Sep 28 '18 at 6:56











      • @havlock :) Would be interesting to see examples of other typed languages where this is dealt with better :)

        – AJP
        Sep 28 '18 at 16:41














      • 1





        Whilst this works, it will make optional parameters required. You can do this to prevent that: stackoverflow.com/a/48216010/616589

        – Simon
        Aug 14 '18 at 17:07











      • I don't dispute the accuracy of the answer, but that's an absurdly round the houses way of achieving a simple end - it will take anyone reading it some non-trivial amount of time to work out just what the hell is going in. Is there any logic at all as to why the typescript team went this way?

        – havlock
        Sep 26 '18 at 17:49













      • @havlock sorry I recently edited it to simplify and improve the answer but I forget to also use the simplified Omit code in the simplify example. Is that better now?

        – AJP
        Sep 27 '18 at 13:37











      • @AJP I was moaning about typescript, not your answer, but thanks - yes, that is slightly more straightforward. Still absolutely bonkers that they make us jump through such ridiculous syntax to achieve something simple - I fear for the language if they really think this sort of thing is OK.

        – havlock
        Sep 28 '18 at 6:56











      • @havlock :) Would be interesting to see examples of other typed languages where this is dealt with better :)

        – AJP
        Sep 28 '18 at 16:41








      1




      1





      Whilst this works, it will make optional parameters required. You can do this to prevent that: stackoverflow.com/a/48216010/616589

      – Simon
      Aug 14 '18 at 17:07





      Whilst this works, it will make optional parameters required. You can do this to prevent that: stackoverflow.com/a/48216010/616589

      – Simon
      Aug 14 '18 at 17:07













      I don't dispute the accuracy of the answer, but that's an absurdly round the houses way of achieving a simple end - it will take anyone reading it some non-trivial amount of time to work out just what the hell is going in. Is there any logic at all as to why the typescript team went this way?

      – havlock
      Sep 26 '18 at 17:49







      I don't dispute the accuracy of the answer, but that's an absurdly round the houses way of achieving a simple end - it will take anyone reading it some non-trivial amount of time to work out just what the hell is going in. Is there any logic at all as to why the typescript team went this way?

      – havlock
      Sep 26 '18 at 17:49















      @havlock sorry I recently edited it to simplify and improve the answer but I forget to also use the simplified Omit code in the simplify example. Is that better now?

      – AJP
      Sep 27 '18 at 13:37





      @havlock sorry I recently edited it to simplify and improve the answer but I forget to also use the simplified Omit code in the simplify example. Is that better now?

      – AJP
      Sep 27 '18 at 13:37













      @AJP I was moaning about typescript, not your answer, but thanks - yes, that is slightly more straightforward. Still absolutely bonkers that they make us jump through such ridiculous syntax to achieve something simple - I fear for the language if they really think this sort of thing is OK.

      – havlock
      Sep 28 '18 at 6:56





      @AJP I was moaning about typescript, not your answer, but thanks - yes, that is slightly more straightforward. Still absolutely bonkers that they make us jump through such ridiculous syntax to achieve something simple - I fear for the language if they really think this sort of thing is OK.

      – havlock
      Sep 28 '18 at 6:56













      @havlock :) Would be interesting to see examples of other typed languages where this is dealt with better :)

      – AJP
      Sep 28 '18 at 16:41





      @havlock :) Would be interesting to see examples of other typed languages where this is dealt with better :)

      – AJP
      Sep 28 '18 at 16:41













      0














      You can't remove properties from already existing interfaces. Even trying to extend existing interface with the interface having value property set as optional will return an error.



      To avoid this issue, modify the typings of the target component so value property is optional.



      e.g.



      // ...
      class MyComponent extends React.Component<{value?: string}, State> {
      // ...
      }


      and then component produced when using High Order Function



      const valuedComponent = valueHOC(MyComponent);


      won't ask for value prop.






      share|improve this answer




























        0














        You can't remove properties from already existing interfaces. Even trying to extend existing interface with the interface having value property set as optional will return an error.



        To avoid this issue, modify the typings of the target component so value property is optional.



        e.g.



        // ...
        class MyComponent extends React.Component<{value?: string}, State> {
        // ...
        }


        and then component produced when using High Order Function



        const valuedComponent = valueHOC(MyComponent);


        won't ask for value prop.






        share|improve this answer


























          0












          0








          0







          You can't remove properties from already existing interfaces. Even trying to extend existing interface with the interface having value property set as optional will return an error.



          To avoid this issue, modify the typings of the target component so value property is optional.



          e.g.



          // ...
          class MyComponent extends React.Component<{value?: string}, State> {
          // ...
          }


          and then component produced when using High Order Function



          const valuedComponent = valueHOC(MyComponent);


          won't ask for value prop.






          share|improve this answer













          You can't remove properties from already existing interfaces. Even trying to extend existing interface with the interface having value property set as optional will return an error.



          To avoid this issue, modify the typings of the target component so value property is optional.



          e.g.



          // ...
          class MyComponent extends React.Component<{value?: string}, State> {
          // ...
          }


          and then component produced when using High Order Function



          const valuedComponent = valueHOC(MyComponent);


          won't ask for value prop.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Jul 8 '17 at 11:45









          Michal PietraszkoMichal Pietraszko

          2,39321120




          2,39321120























              0














              There is utility-types library that has Substract mapped type:



              import { Subtract } from 'utility-types';

              type Props = { name: string; age: number; visible: boolean };
              type DefaultProps = { age: number };

              type RequiredProps = Subtract<Props, DefaultProps>;
              // Expect: { name: string; visible: boolean; }





              share|improve this answer






























                0














                There is utility-types library that has Substract mapped type:



                import { Subtract } from 'utility-types';

                type Props = { name: string; age: number; visible: boolean };
                type DefaultProps = { age: number };

                type RequiredProps = Subtract<Props, DefaultProps>;
                // Expect: { name: string; visible: boolean; }





                share|improve this answer




























                  0












                  0








                  0







                  There is utility-types library that has Substract mapped type:



                  import { Subtract } from 'utility-types';

                  type Props = { name: string; age: number; visible: boolean };
                  type DefaultProps = { age: number };

                  type RequiredProps = Subtract<Props, DefaultProps>;
                  // Expect: { name: string; visible: boolean; }





                  share|improve this answer















                  There is utility-types library that has Substract mapped type:



                  import { Subtract } from 'utility-types';

                  type Props = { name: string; age: number; visible: boolean };
                  type DefaultProps = { age: number };

                  type RequiredProps = Subtract<Props, DefaultProps>;
                  // Expect: { name: string; visible: boolean; }






                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited Dec 29 '18 at 8:31

























                  answered Dec 29 '18 at 8:09









                  ChintsuChintsu

                  316510




                  316510






























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