### TECHNOLOGY

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- Miscellaneous

technology:datetools

This is a set of utility routines to provide guidance to visual calendar building tools. Written in 2001, this class provides mathematical information to calendar systems.

```
/**
* Insert the type's description here.
* Creation date: (9/26/01 11:57:50 AM)
* @author: cfreyer
*/
import java.util.Date;
import java.util.Calendar;
public class DateTools {
public static String days[] =
{
"Sunday",
"Monday",
"Tuesday",
"Wednesday",
"Thursday",
"Friday",
"Saturday" };
public static String months[] =
{
"January",
"February",
"March",
"April",
"May",
"June",
"July",
"August",
"September",
"October",
"November",
"December" };
public static int daysPerMonth[] =
{ 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
/**
* Insert the method's description here.
* Creation date: (9/26/01 2:19:17 PM)
* @param userYear int
* @param userMonth int
*/
public static int CalcEndOfMonth(int userYear, int userMonth) {
int daysPerMonth[] =
{ 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
return daysPerMonth[userMonth]
+ ((IsLeapYear(userYear) && (userMonth == Calendar.FEBRUARY)) ? 1 : 0);
}
public static int CalcFirstOfMonth(int year, int month)
/*
USE: Calculates day of the week the first day of the month falls on.
IN: year = given year after 1582 (start of the Gregorian calendar).
month = 0 for January, 1 for February, etc.
OUT: First day of month: 0 = Sunday, 1 = Monday, etc.
*/ {
int daysAMonth[] =
{ 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
int firstDay; /* day of week for Jan 1, then first day of month */
int i; /* to traverse months before given month */
/* Start at 1582, when modern calendar starts. */
if (year < 1582)
return (-1);
/* Catch month out of range. */
if ((month < 0) || (month > 11))
return (-1);
/* Get day of week for Jan 1 of given year. */
firstDay = CalcJanuaryFirst(year);
/* Increase firstDay by days in year before given month to get first day
* of month.
*/
for (i = 0; i < month; i++)
firstDay += daysAMonth[i];
/* Increase by one if month after February and leap year. */
if ((month > Calendar.FEBRUARY) && IsLeapYear(year))
firstDay++;
/* Convert to day of the week and return. */
return (firstDay % 7);
} // CalcFirstOfMonth
public static int CalcJanuaryFirst(int year)
/*
USE: Calculate day of the week on which January 1 falls for given year.
IN: year = given year after 1582 (start of the Gregorian calendar).
OUT: Day of week for January 1: 0 = Sunday, 1 = Monday, etc.
NOTE: Formula starts with a 5, since January 1, 1582 was a Friday; then
advances the day of the week by one for every year, adding the
number of leap years intervening, because those years Jan 1
advanced by two days. Calculate mod 7 to get the day of the week.
*/ {
/* Start at 1582, when modern calendar starts. */
if (year < 1582)
return (-1);
/* Start Fri 01-01-1582; advance a day for each year, 2 for leap yrs. */
return ((5 + (year - 1582) + CalcLeapYears(year)) % 7);
} // CalcJanuaryFirst
public static int CalcLeapYears(int year)
/*
USE: Calculate number of leap years since 1582.
IN: year = given year after 1582 (start of the Gregorian calendar).
OUT: number of leap years since the given year, -1 if year < 1582
NOTE: Count doesn't include the given year if it is a leap year.
In the Gregorian calendar, used since 1582, every fourth year
is a leap year, except for years that are a multiple of a
hundred, but not a multiple of 400, which are no longer leap
years. Years that are a multiple of 400 are still leap years:
1700, 1800, 1990 were not leap years, but 2000 will be.
*/ {
int leapYears; /* number of leap years to return */
int hundreds; /* number of years multiple of a hundred */
int fourHundreds; /* number of years multiple of four hundred */
/* Start at 1582, when modern calendar starts. */
if (year < 1582)
return (-1);
/* Calculate number of years in interval that are a multiple of 4. */
leapYears = (year - 1581) / 4;
/* Calculate number of years in interval that are a multiple of 100;
* subtract, since they are not leap years.
*/
hundreds = (year - 1501) / 100;
leapYears -= hundreds;
/* Calculate number of years in interval that are a multiple of 400;
* add back in, since they are still leap years.
*/
fourHundreds = (year - 1201) / 400;
leapYears += fourHundreds;
return (leapYears);
} // CalcLeapYears
/**
* Insert the method's description here.
* Creation date: (9/26/01 2:19:17 PM)
* @param userYear int
* @param userMonth int
*/
public static void DaysInMonth(int userYear, int userMonth) {
int numDays =
daysPerMonth[userMonth]
+ ((IsLeapYear(userYear) && (userMonth == Calendar.FEBRUARY)) ? 1 : 0);
}
public static boolean IsLeapYear(int year)
/*
USE: Determines if given year is a leap year.
IN: year = given year after 1582 (start of the Gregorian calendar).
OUT: TRUE if given year is leap year, FALSE if not.
NOTE: Formulas capture definition of leap years; cf CalcLeapYears().
*/ {
/* If multiple of 100, leap year iff multiple of 400. */
if ((year % 100) == 0)
return ((year % 400) == 0);
/* Otherwise leap year iff multiple of 4. */
return ((year % 4) == 0);
} // IsLeapYear
public static int NumberRowsNeeded(int year, int month)
/*
USE: Calculates number of rows needed for calendar.
IN: year = given year after 1582 (start of the Gregorian calendar).
month = 0 for January, 1 for February, etc.
OUT: Number of rows: 5 or 6, except for a 28 day February with
the first of the month on Sunday, requiring only four rows.
*/ {
int firstDay; /* day of week for first day of month */
int numCells; /* number of cells needed by the month */
int daysPerMonth[] =
{ 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
/* Start at 1582, when modern calendar starts. */
if (year < 1582)
return (-1);
/* Catch month out of range. */
if ((month < 0) || (month > 11))
return (-1);
/* Get first day of month. */
firstDay = CalcFirstOfMonth(year, month);
/* Non leap year February with 1st on Sunday: 4 rows. */
if ((month == Calendar.FEBRUARY) && (firstDay == 0) && !IsLeapYear(year))
return (4);
/* Number of cells needed = blanks on 1st row + days in month. */
numCells = firstDay + daysPerMonth[month];
/* One more cell needed for the Feb 29th in leap year. */
if ((month == Calendar.FEBRUARY) && (IsLeapYear(year)))
numCells++;
/* 35 cells or less is 5 rows; more is 6. */
return ((numCells <= 35) ? 5 : 6);
} // NumberRowsNeeded
}
```

/home/cfreyer/public_html/data/pages/technology/datetools.txt · Last modified: 2008/03/17 01:06 by Chris Freyer