ecpg-timestamp- interval problem V8.2.x
От | paul tilles |
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Тема | ecpg-timestamp- interval problem V8.2.x |
Дата | |
Msg-id | 47FB75C1.8080209@noaa.gov обсуждение исходный текст |
Ответы |
Re: ecpg-timestamp- interval problem V8.2.x
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Список | pgsql-general |
A colleague of mine is working with the following code (postgres Version 8.2.4): #include <stdio.h> #include <stdlib.h> #include "pgtypes_timestamp.h" #include "pgtypes_interval.h" EXEC SQL INCLUDE sqlca; main(argc, argv) int argc; char *argv[]; { timestamp dt1st, dt2nd; interval *int_day, *int_pt; char timebeg[40]; char timeend[40]; int rc; strcpy(timebeg,"2008-04-01"); strcpy(timeend,"2008-04-05"); printf("timebeg: %s\n", timebeg); printf("timeend: %s\n", timeend); dt1st=PGTYPEStimestamp_from_asc(timebeg,NULL); dt2nd=PGTYPEStimestamp_from_asc(timeend,NULL); int_day=PGTYPESinterval_from_asc("02 01:02:03", NULL); printf("int_day : %s\n", PGTYPESinterval_to_asc(int_day)); rc=PGTYPEStimestamp_sub(&dt2nd, &dt1st, int_pt); printf("rc, date2 - date1 = %d, %s\n", rc, PGTYPESinterval_to_asc(int_pt)); exit(0); } -------------------------------------------------------------------------------------------- The output is timebeg: 2008-04-01 timeend: 2008-04-05 int_day : @ 2 days 1 hour 2 mins 3 secs rc, date2 - date1 = 0, @ 1 sec ago -------------------------------------------------------------------------------------------- The result of the subtraction of the the two timestamps is not correct. What's going on? TIA. Paul Tilles
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