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| How To Find The Day of Week for A Given Date? |
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| kirani |
 Posted: Mon Nov 08, 2004 11:22 pm Post subject: |
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Jedi Knight
Joined: 05 Sep 2001
Posts: 3779
Location: Torrance, CA, USA
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I guess he cannot use Extract function because he is working with CHAR STRING. EXTRACT function only works on date data types.
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Kiran
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| nathanw |
| Posted: Tue Nov 09, 2004 4:56 am Post subject: |
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Knight
Joined: 14 Jul 2004
Posts: 550
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beat me to it
but yes
trust me on this guys if yu put the appropriate date in it calculates the day
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| Nigelg |
| Posted: Wed Nov 10, 2004 6:57 am Post subject: |
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Grand Master
Joined: 02 Aug 2004
Posts: 1046
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Does the calculation get the day right for days in January and February of a leap year?
If this is based on the Doomsday algorithm there should be an adjustment for leap years, and I cannot see such an adjustment in the code. I rewrote this in C and it gets the days wrong before the leap day.
Here is the code. Feel free to find the bug...
| Code: |
/*
y is year number 2004
m is month number 1-12
d is day in month 1-31
*/
/* extract century, calc 1 */
/* RETURN (2 * (3 - Mod((Cast(Substring(DateInC From 7 for 2) as Int)), 4))) */
y1 = y / 100;
i1 = 2 * (3 - (y1 % 4));
/* extract year in century, calc 2 */
/* Mod((y / 12) + mod(y,12) + (mod(y,12)/ 4), 7) */
y2 = (y - 1900)%100;
i2 = ((y2/12) + y2 % 12 + ((y2%12)/4))%7;
/* transform month */
/* long mnums[] = {0,0,3,3,6,1,4,6,2,5,0,3,5}; */
m1 = mnums[m];
/* mod days */
d1 = d%7;
/* add it up.. */
res = i1 + i2 + m1 + d1;
/* final result is mod by 7 */
res %= 7;
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This works for days after the leap day in leap years, but not for before. Feb 29 is the same day as Mar 1, and all days up to Feb 29 are 1 day ahead. Here is the result of running it for this year
| Quote: |
$ dday 2004 3 1
Mon Mar 1 14:54:37 2004
Day is Mon
$ dday 2004 2 29
Sun Feb 29 14:54:42 2004
Day is Mon
$
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