PLC3 Day of Week

Oceansoul said:

S1:2 = Wednesday and S1:3 = > or =12 then set your Bit true.

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ControlJoe said:

OkiePC: Thank you for the correction. S1:2 is the day of month. I believe I will be stuck writing code that will determine the Wednesdays for the foreseeable future, but I am having trouble visualizing this. I have no idea how to build a lookup table.

Oceansoul: The Problem with the PLC3 is each of the following words: Year, month, day of month, hours, minutes, seconds and milliseconds are represented by an integer and not a string. So the day of the month, for example, is a value from 1-31 for January.

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' **************************************************
' * *
' * WEEKDAY.BAS *
' * *
' * Calculates Day of Week from Date *
' * Using Zeller's Congruence *
' * *
' * Returns 0=Sun, 1=Mon, 2=Tue, ... 6=Sat *
' * *
' * Compile using QuickBasic 4.5 *
' * *
' * Judson D McClendon *
' * Sun Valley Systems *
' * 4522 Shadow Ridge Pkwy *
' * Pinson, AL 35126 *
' ' * *
' **************************************************
'
DECLARE FUNCTION WeekDay (Month AS INTEGER, Day AS INTEGER, Year AS INTEGER)
DIM Month AS INTEGER, Day AS INTEGER, Year AS INTEGER, DOW AS INTEGER

INPUT "Enter Date: MM,DD,YYYY: ", Month, Day, Year
DOW = WeekDay(Month, Day, Year)
PRINT MID$("SunMonTueWedThuFriSat", DOW * 3 + 1, 3)
END

FUNCTION WeekDay (Month AS INTEGER, Day AS INTEGER, Year AS INTEGER)
DIM M AS INTEGER, D AS INTEGER, Y AS INTEGER
IF (Month < 3) THEN
Y = Year - 1
M = Month + 12
D = Day
ELSE
Y = Year
M = Month
D = Day
END IF
WeekDay = ((13 * M + 3) \ 5 + D + Y + Y \ 4 - Y \ 100 + Y \ 400 + 1) MOD 7
END FUNCTION

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daba said:

Fairly straightforward bit of maths (except the MOD function!!).



Notes:
I just tried this in Excel, and it works....
"\" is Integer Divide
MOD can be done by subtracting 7 until the result is less than 7 (JMP->LBL)

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OkiePC said:

That's simple enough, I would definitely do this in favor of a lookup table...

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