Re: Week of the year routine

How about something like this?

1-Convert the target date to Julian Day.
2-Divide the Julian Day by 7. That will give you a week number.
3-Save the remainder from your division. It will range from 1 -7.
4-If the remainder is greater than zero, add one to the week number.

Then the fun starts. You have to find the offset vs the remainder to see if you need to add one more to your week number because of the day of the week that the year started.

My bet is that what I am about to write won't really work, but it should show the algorithm, and you can finesse it from there.

a-Find the day of the week for January 1st, UYEAR, as a number 0-6.
b-add 1 to the day-of-week so that the range is 1-7.
b-make a comparison with the remainder. Some combination of that comparison will tell you whether or not to add 1 to the week number.


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(A day of the week calculation is here:)
http://www.martinvt.com/Code_Samples/Date_Math/date_math.html

note: This initializing date is easiest if you set it for the first date of your week, so that day that is the first day of the week become day 0. It might even be useful to set it to the beginning of the week on or before January 1.

D wTestDate s d inz(d'2010-01-04')

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Anyway, this may give you some thoughts to play with, even if V5R3 does not support true date fields.




On 4/15/2011 11:23 AM, M. Lazarus wrote:

Before I reinvent the "calendar file" wheel, does anyone have an
*RPG* routine that will calculate the week of the year.

A few parameters:

- This machine is on an older OS version (5.3) and does not have the
SQL feature installed.

- Week 1 of the year starts on January 1. So, for example, in 2010,
Dec. 26-31 would be the last week of the year (53?). Saturday, Jan
1, 2001, is week 1. Monday, Jan 2-8, is week 2, etc.


Thank you.

-mark