Reduce in TypeScript. How to correctly display the value of an object using the path to this value?

Question

typescript

Good afternoon. Tell me how to correctly implement the function in TypeScript:

extractValue(row: object, path: string) {
  return result = path.split('.')
      .reduce((accumulator: object, currentValue: string) => accumulator[currentValue], row);
}

How to correctly specify the return type of a function (string | number | undefined)?

Answer 1

I'll add my solution. Thanks Alexei Yarkov for the tip. I will draw attention to the fact that the any type is not used .

interface IObjectItem {
    [index: string]: IObjectItem | number | string;
}

extractValue(obj: IObjectItem, path: string): string | number {
    const value = path.split('.').reduce((accumulator: IObjectItem | string | number, currentValue: string) => {
        if (typeof accumulator === 'object' && accumulator[currentValue])
            return accumulator[currentValue];
        else
            throw new Error(`Unexpected parameter "${currentValue}"`);
    }, obj);

    // Проверка типа получившегося значения
    if (typeof value === 'string' || typeof value === 'number')
        return value;
    else
        throw new Error('The final value must be string or number');
}

The function itself consists of 2 parts: the first one gets the value based on the given path, the second one checks the type of the output value.

Answer 2

there is no normal built-in way. people have been asking for this for a long time, but to no avail: https://github.com/Microsoft/TypeScript/issues/12290
there are fierce perversions in the comments that people resort to to do something similar.
and in a simple way, to get one field, you can stupidly pass the desired generic type as a parameter and cast the received before returning. or pass a default value as a parameter in case there are no fields along the way, and deduce the return type from it.