Converting value of interface in Typescript

Question

given an interface of Person i would like to create a new interface HasmapOfPersonsOverTime based on that. im gonna push updates of Person objects (i.e if a person has a birthday i will push the new age number, and the time the person celebrated the birthday as a object into the array on the age property).

... symbolizes the rest of the properties

interface Person {
    age:number;
    name:string;
    ...
}

i would like an interface looking like :

interface HasmapOfPersonsOverTime {
  [id:string]:{
    age:[
          {
            age:number
            time:string
          }
        ]
    },
    name:[
          {
            name:string
            time:string
          }
        ]
    },
    ...
}

and to give an example on how i would like to use it:

const list:HasmapOfPersonsOverTime = {
  "1515":{ 
    age:[
      {
        age:24,
        time:"2020-03-05T21:53:37"
      },
      {
        age:25,
        time:"2021-03-05T18:22:05"
      }
    ],
    name:[
      {
        name:"bar",
        time:"1996-03-05T21:53:37"
      }
    ],
    ...
  }
}

i have come this far but i cant get type safety on properties. i create "value":string where i want the correct person property.

interface Person {
  age:number;
  name:string;
}
type HasmapOfPersonsOverTime <T extends Object> = Record<string,Record<keyof T, {'value':string,'time':string}[]>>

const list: HasmapOfPersonsOverTime<Person> = {
  "1515":{ 
    age:[
      {
        value:"24",
        time:"2020-03-05T21:53:37"
      },
      {
        value:"25",
        time:"2021-03-05T18:22:05"
      }
    ],
    name:[
      {
        value:"name",
        time:"1996-03-05T21:53:37"
      }
    ]
  }
}

Tal Ohana · Accepted Answer · 2021-07-05 18:19:39Z

2

You can use the following mapped type, also used with other builtin types such as Pick

type Person = {
  age: number;
  name: string;
};

type OverTime<T> = {
  [K in keyof T]: (Pick<T, K> & { time: string })[];
};

type HashmapOfPersonOverTime = Record<string, OverTime<Person>>;

edited Jul 5, 2021 at 18:19

answered Jul 5, 2021 at 18:14

Tal Ohana

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