Typescript define type for method argument

Question

I have a company object, that looks like this:

{ 
 tracking_hours: {
  open: '8:00',
  close: '20:00'
 }
}

I use it to set values this way:

set({
 openTime: setTime(company, 'open_time'),
 closeTime: setTime(company, 'close_time'),
})

I need somehow set a type for company in setTime function

export function setTime(
    company: {
        tracking_hours: null | any
    },
    type: string,
): number | null {
    if (company.tracking_hours) {
        if (company.tracking_hours[type]) {
            const time = Number(company.tracking_hours[type].split(':')[0])

            if (time) {
                return time
            } else {
                return null
            }
        }

        return null
    }

    return null
}

How can I replace any with its actual type?

@VLAZ TS7053: Element implicitly has an 'any' type because expression of type 'string' can't be used to index type '{ open: string; close: string; }'. No index signature with a parameter of type 'string' was found on type '{ open: string; close: string; }'. — Yana Trifonova
– Yana Trifonova, Commented Feb 25, 2022 at 12:41
So, what do you want to do, exactly? Allow any strings for keys of tracking_hours or restrict the values of type to be the kets of tracking_hours? Or something else? Note that right now you're passing in "open_hours" for type but the key found on the object is open. — VLAZ
– VLAZ, Commented Feb 25, 2022 at 12:47

Ben Wainwright · Accepted Answer · 2022-02-25 13:00:05Z

This can be done by

Creating a separate type for the Company
Using keyof to indicate the correct type for type

const company = {
  tracking_hours: {
    open: "8:00",
    close: "20:00",
  },
};

interface Company {
  tracking_hours: {
    open: string;
    close: string;
  };
}

function setTime(
  company: Company,
  type: keyof Company["tracking_hours"]
): number | null {
  const time = Number(company.tracking_hours[type].split(":")[0]);

  return time ?? null
}

setTime(company, "open");
setTime(company, "close");

Note, I've simplified some of your code because

You don't need to check for the presence of company.tracking_hours or company.tracking_hours[type] - TypeScript guarantees it will they will always be present because of the types that have been specified in the function signature
You only need to return null once, since all the other cases would have dropped out of the respective blocks and hit the final return statement anyway.

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