RE: [SQL] datediff function

At 17:37 +0300 on 17/08/1999, Pham, Thinh wrote:

select datediff(minute, timein, timeout) as totaltime from schedule

It would give me the number 60, that's it. I don't want any qualifier behind
the number since it blew up the stupid microsoft ADO driver like you
wouldn't believe.

If you don't want to write 'now'::datetime you can always write
datetime('now'). Same goes for '1 week'::timespan and timespan( '1 week' ).
I don't think this will blow up your Microsoft product, but then again,
anything can blow up a Microsoft product, being a Microsoft Product
included...

To make things clear, here is what Postgres can and cannot do:

It can give you the interval between two dates. The returned value is an
integer representing the number of days between them.

It can give you the interval between two datetimes. The returned value is a
timespan, expressing days, hours, minutes, etc. as needed.

Another method to get the same thing is using age( datetime1, datetime2 ).
This returns a timespan, but expressed in years, months, days, hours and
minutes. There is a subtle difference here, because a year is not always
365 days, and a month is 28-31 days, depending...

You can also truncate datetimes, dates, and other date related types, to
the part of your choice. Truncate it to the minute, and it drops the
seconds, and gives it back to you with 00 in the seconds. Truncate it to
days and it gives it back to you at 00:00:00. This is done with
date_trunc().

Another useful operation which can be done is taking one part of the
datetime (or related type). For example, the minutes, the seconds, the day,
the day of week, or the seconds since the epoch.

Now, I'm not sure these functions do exactly what you wanted. It depends on
what you expect from datediff(minute, timein, itmeout) when they are not on
the same day. For 13-oct-1999 14:00:00 and 14-oct-1999 14:00:05, do you
expect 5 or 24*60 + 5?

If only 5, then you can do it with

SELECT date_part( 'minute', datetime1 - datetime2 )

If not, you will have to do the 24*60 calculation in full.

Herouth

--
Herouth Maoz, Internet developer.
Open University of Israel - Telem project
http://telem.openu.ac.il/~herutma

I unfortunately do MS-SQL.
Datediff in MS-SQL gives you the number of boundaries between two dates.
DATEDIFF(day, '1/1/99 23:59:00', '1/2/99 00:01:00') gives 1
DATEDIFF(day, '1/2/99 00:01:00', '1/2/99 00:03:00') gives 0
The {PostgreSQL|postgres|pgsql|whatever} way of doing it is much nicer.

select datediff(minute, timein, timeout) as totaltime from schedule

It would give me the number 60, that's it. I don't want any

qualifier behind

the number since it blew up the stupid microsoft ADO driver like you
wouldn't believe.

If you don't want to write 'now'::datetime you can always write
datetime('now'). Same goes for '1 week'::timespan and

espan(

Here the SQL/92 expression supported also by PostgreSQL:
select extract(day from date '1999-01-02') - extract(day from date
'1999-01-01');

Jos�

Datediff in MS-SQL gives you the number of boundaries between two dates.
DATEDIFF(day, '1/1/99 23:59:00', '1/2/99 00:01:00') gives 1
DATEDIFF(day, '1/2/99 00:01:00', '1/2/99 00:03:00') gives 0
The {PostgreSQL|postgres|pgsql|whatever} way of doing it is much nicer.

select datediff(minute, timein, timeout) as totaltime from schedule

It would give me the number 60, that's it. I don't want any

qualifier behind

the number since it blew up the stupid microsoft ADO driver like you
wouldn't believe.

If you don't want to write 'now'::datetime you can always write
datetime('now'). Same goes for '1 week'::timespan and

espan( extract(day from date '1999-01-02') - extract(day from date '1999
-01-01');

John Ridout ha scritto: