Is this possible? Opposite of Partial Type #12578

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1. mhegazy commented on Nov 30, 2016

   

   Mohamed Hegazy (mhegazy)

   on Nov 30, 2016

   Contributor

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   You can use inference from homomorphic mapped types in #12528 to get a type that removes the undefined

   for instance:

   interface Test {
       optional?: string;
       required: number;
   }
   
   type OptionalMap<T> = {
       [P in keyof T]: T[P] | undefined;
   };
   
   function getProperty<T, K extends keyof T>(t: T, key: K): T[K] {
       return t[key];
   }
   
   function getPropertyRemoveUndefined<T, K extends keyof T>(t: OptionalMap<T>, key: K): T[K] {
       // Put it in a value, since `t[p]` is not narrowed only t.p which is not possible here
       var value = t[key];
       if (value) {
           return value;
       }
       throw new Error('Value does not exist at key');
   }
   
   var t: Test = {required: 0};
   
   
   var o1 = getProperty(t, 'optional'); // string | undefined
   var r1 = getProperty(t, 'required'); // number
   
   var o2 = getPropertyRemoveUndefined(t, 'optional'); // string
   var r2 = getPropertyRemoveUndefined(t, 'required'); // number

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2. ahejlsberg commented on Nov 30, 2016

   

   Anders Hejlsberg (ahejlsberg)

   on Nov 30, 2016

   Member

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   I just merged #12589 and now you can simply use the predefined Partial<T> type:

   interface Test {
       optional?: string;
       required: number;
   }
   
   function getProperty<T, K extends keyof T>(t: T, key: K): T[K] {
       return t[key];
   }
   
   function getPropertyRemoveUndefined<T, K extends keyof T>(t: Partial<T>, key: K): T[K] {
       // Put it in a value, since `t[p]` is not narrowed only t.p which is not possible here
       var value = t[key];
       if (value) {
           return value;
       }
       throw new Error('Value does not exist at key');
   }
   
   var t: Test = {required: 0};
   
   
   var o1 = getProperty(t, 'optional'); // string | undefined
   var r1 = getProperty(t, 'required'); // number
   
   var o2 = getPropertyRemoveUndefined(t, 'optional'); // string
   var r2 = getPropertyRemoveUndefined(t, 'required'); // number

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3. ahejlsberg commented on Nov 30, 2016

   

   Anders Hejlsberg (ahejlsberg)

   on Nov 30, 2016

   Member

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   opensrcken BTW, I noticed that you used a union Test[K] | never. Just wanted to point out that the never type is redundant here as it is always removed from union types (it effectively represents an empty union type).

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4. opensrcken commented on Nov 30, 2016

   

   opensrcken

   on Nov 30, 2016

   Author

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   It works. And point taken on never in unions. Thanks guys. 👍
5. opensrcken

   closed this as completedon Nov 30, 2016
6. Ryan Cavanaugh (RyanCavanaugh)

   added

   QuestionAn issue which isn't directly actionable in codeAn issue which isn't directly actionable in code

   on Nov 30, 2016
7. k8w commented on Jun 23, 2017

   

   Zhongyang Wang (k8w)

   on Jun 23, 2017

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   So I must write a function to get these opposite type ?
   Can I just remove undefined or null from a type just like use ! ?

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8. mjomble commented on Jun 1, 2018

   

   Andres Kalle (mjomble)

   on Jun 1, 2018

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   TypeScript 2.8 introduces Required<T> that may give you what you need: https://www.typescriptlang.org/docs/handbook/release-notes/typescript-2-8.html#example-6
9. microsoft

   locked and limited conversation to collaborators on Jul 31, 2018