Subtract Business Days from a Date

[carmine1]

In Crystal Reports, how to subtract a number of business days (Monday through Friday) from an input date?

For example, October 29, 2024 - 16 business days = October 7, 2024

[leonardo0482]

To subtract business days in Crystal Reports (Monday to Friday), you can use a formula that iterates back through dates while skipping weekends. Since Crystal Reports doesn’t have a built-in function for business days, we’ll use date functions with a loop. Here’s a formula you can use: Define the starting date and the number of business days to subtract. In this case, the start date is #10/29/2024# and the business days to subtract are 16. Create a formula to perform the calculation:

[Ranjan__Panda]

Hi,

Use this function module to subtract a number of business days (Monday through Friday) from an input date.

DATA: lv_date TYPE sy-datum.  "Input Date
DATA: lv_business_days TYPE i VALUE 10. "Number of business days to subtract
DATA: lv_new_date TYPE sy-datum. "Output Date

CALL FUNCTION 'RP_CALC_DATE_IN_INTERVAL'
EXPORTING
date = lv_date     "October 29, 2024
days = lv_business_days    "16 business days
months = 0
years = 0
signum = '-'
IMPORTING
calc_date = lv_new_date.  "October 7, 2024

WRITE: / lv_date, lv_new_date.

Regards,
Ranjan Panda

[carmine1]

Thanks! I will give this a try.

[ido_millet]

See https://kenhamady.com/formulas/form17.shtml

[DonWilliams]

In ALL software if you hit the F1 key you will get the Help file attached.

Then search for "DateAdd (intervalType, nIntervals, startDateTime)"

Or something similar should work for you

[carmine1]

I know very well how to DateAdd. But that does not help with an elegant way to adjust for business days (weekdays) only.  For example, November 15, 2024 minus 127 weekdays = WhatDate?

[carmine1]

I found this solution elsewhere, however I am not certain that it always yields the correct result: DATEADD('d', -7*(no_of_days / 5) - no_of_days % 5, start_date)

[carmine1]

I found this solution elsewhere, however I am not certain that it always yields the correct result: DATEADD('d', -7*( no_of_days/ 5) - no_of_days mod 5, starting_date)