Change the week period on differenceInBusinessDays?

[Stuart-wdas]

I am trying to make use of the differenceInBusinessDays and so far the function has been superb, but im wondering if there is a way to change the period of a business week i.e. the default is currently 5 - 2, 5 working days 2 weekend days, is there a way to change the options on the function so the period is 6 - 1, 6 working days 1 weekend day? const working_days = differenceInBusinessDays(current_date, previous_month, options: { business_days: 6 });

[MILLERMARRU]

There's no option for this, checked the current source of differenceInBusinessDays (pkgs/core/src/differenceInBusinessDays/index.ts):

export function differenceInBusinessDays(
  laterDate: DateArg<Date> & {},
  earlierDate: DateArg<Date> & {},
  options?: DifferenceInBusinessDaysOptions | undefined,
): number {
  ...
  let result = weeks * 5;
  ...
  while (!isSameDay(laterDate_, movingDate)) {
    result += isWeekend(movingDate, options) ? 0 : sign;
    movingDate = addDays(movingDate, sign);
  }
  ...
}

The weeks * 5 is a hardcoded literal, and DifferenceInBusinessDaysOptions only extends ContextOptions (the in option for time zone context), nothing weekmask-related. It leans on isWeekend, which is hardcoded too:

export function isWeekend(date, options) {
  const day = toDate(date, options?.in).getDay();
  return day === 0 || day === 6;
}

No configurability there either, Saturday/Sunday is baked in. This isn't just undocumented, there's genuinely no plumbing for a custom weekend definition anywhere in this function's option surface.

Given date-fns' general philosophy of small composable functions rather than parameterizing every variant into one function, the practical way to get a 6/1 week is to build it from eachDayOfInterval (which also has no weekday-aware options, confirmed the same way) plus your own day-of-week filter:

import { eachDayOfInterval } from "date-fns";

function differenceInSixDayWorkWeek(laterDate: Date, earlierDate: Date) {
  const [start, end] = laterDate < earlierDate
    ? [laterDate, earlierDate]
    : [earlierDate, laterDate];

  const sign = laterDate < earlierDate ? -1 : 1;

  const count = eachDayOfInterval({ start, end })
    .filter(d => d.getDay() !== 0) // only Sunday is off
    .length - 1; // exclude the start day itself, matching differenceInBusinessDays' semantics

  return count * sign;
}

Adjust the getDay() !== 0 predicate to whatever day(s) you want to exclude. That gives you full control over which day(s) count as the "weekend" without waiting on an API that doesn't exist today.