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Features added:

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* Wrote max(date) and min(date) aggregates
* Wrote operator "-" for date; date - date yields number of days
  difference
* Wrote operator+(date,int) and operator-(date,int); the int is the
  number of days.  Each operator returns a new date.


By: Tom Tromey <tromey@creche.cygnus.com>
This commit is contained in:
Marc G. Fournier
1996-11-14 21:39:14 +00:00
parent eedc75bfe1
commit 8abc490181
4 changed files with 112 additions and 5 deletions
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100
src/backend/utils/adt/datetimes.c
View File

@@ -7,7 +7,7 @@
*
*
* IDENTIFICATION
* $Header: /cvsroot/pgsql/src/backend/utils/adt/Attic/datetimes.c,v 1.6 1996/11/10 03:03:10 momjian Exp $
* $Header: /cvsroot/pgsql/src/backend/utils/adt/Attic/datetimes.c,v 1.7 1996/11/14 21:38:58 scrappy Exp $
*
*-------------------------------------------------------------------------
*/
@@ -224,6 +224,104 @@ date_cmp(int4 dateVal1, int4 dateVal2)
return 0;
}
int4
date_larger(int4 dateVal1, int4 dateVal2)
{
return (date_gt (dateVal1, dateVal2) ? dateVal1 : dateVal2);
}
int4
date_smaller(int4 dateVal1, int4 dateVal2)
{
return (date_lt (dateVal1, dateVal2) ? dateVal1 : dateVal2);
}
/* Compute difference between two dates in days. */
int32
date_mi(int4 dateVal1, int4 dateVal2)
{
DateADT *date1, *date2;
int32 days = 0;
int i;
date1 = (DateADT *) &dateVal1;
date2 = (DateADT *) &dateVal2;
/* Sum number of days in each full year between date1 and date2. */
for (i = date1->year + 1; i < date2->year; ++i)
days += isleap (i) ? 366 : 365;
/* Add in number of days in each full month from date1 to end of
year. */
for (i = date1->month + 1; i <= 12; ++i)
days += day_tab[isleap (date1->year)][i - 1];
/* Add in number of days in each full month from start of year to
date2. */
for (i = 1; i < date2->month; ++i)
days += day_tab[isleap (date2->year)][i - 1];
/* Add in number of days left in month for date1. */
days += day_tab[isleap (date1->year)][date1->month - 1] - date1->day;
/* Add in day of month of date2. */
days += date2->day;
return (days);
}
/* Add a number of days to a date, giving a new date.
Must handle both positive and negative numbers of days. */
int4
date_pli(int4 dateVal, int32 days)
{
DateADT *date1 = (DateADT *) &dateVal;
/* Use separate day variable because date1->day is a narrow type. */
int32 day = date1->day + days;
if (days > 0)
{
/* Loop as long as day has wrapped around end of month. */
while (day > day_tab[isleap (date1->year)][date1->month - 1])
{
day -= day_tab[isleap (date1->year)][date1->month - 1];
if (++date1->month > 12)
{
/* Month wrapped around. */
date1->month = 1;
++date1->year;
}
}
}
else
{
/* Loop as long as day has wrapped around beginning of month. */
while (day < 1)
{
/* Decrement month first, because a negative day number
should be held as relative to the previous month's end. */
if (--date1->month < 1)
{
/* Month wrapped around. */
date1->month = 12;
--date1->year;
}
day += day_tab[isleap (date1->year)][date1->month - 1];
}
}
date1->day = day;
return (dateVal);
}
/* Subtract a number of days from a date, giving a new date. */
int4
date_mii(int4 dateVal, int32 days)
{
return (date_pli (dateVal, -days));
}
/*****************************************************************************
* Time ADT
*****************************************************************************/