I need a SQL...

Well, it's close... :)

But it looks like the case doesn't work..
If I run your sql, the timediff is negative.

But if I run this:
SELECT (stoptime-starttime+'24 hours') as timediff FROM mytable
the timediff has correct value..

Do you see any error in the case, cause I don't?

BTJ

On Thu, 2003-09-11 at 14:29, Mattias Kregert wrote:

Solution:

SELECT starttime, stoptime, (CASE WHEN stoptime-starttime >= 0 THEN
stoptime-starttime ELSE stoptime-starttime+'24 hours' END) as timediff
FROM mytable;

/Mattias

----- Original Message -----
From: Bjørn T Johansen
To: Andrew L. Gould
Cc: PostgreSQL general list
Sent: Thursday, September 11, 2003 2:12 PM
Subject: Re: [GENERAL] I need a SQL...

On Thu, 2003-09-11 at 14:07, Andrew L. Gould wrote:

On Thursday 11 September 2003 06:25 am, Bjørn T Johansen wrote:

I need to write a SQL that calculates the interval between a start time
and a stop time. This is the easy part. The problem is that I only have
the time part, i.e. no date, so how can I be sure to also calculate the
interval if the start time is before midnight and the stop time is after
midnight?

Regards,

BTJ

If the activity or period you are measuring can equal or exceed 12 hours, you
won't be able to calculate it reliably without a start date and a stop date.
If the periods are always less than 12 hours (and you assume all the data is
good), then stop times that are less than start times would indicate an
intervening midnight.

The dates do not have to be in the same fields as the times, since you can add
date and time data to create a timestamp for datetime calculations:

(stop_date + stop_time) - (start_date + start_time)

Best of luck,

Andrew Gould

Yes, the period can exceed 12 hours, so what you are saying is that this is not possible to solve
without the date part? I can write this logic in my business logic but I was hoping to
solve this in my database layer...

BTJ

---------------------------(end of broadcast)---------------------------
TIP 4: Don't 'kill -9' the postmaster

On Thu, 2003-09-11 at 14:29, Mattias Kregert wrote:

Solution:

SELECT starttime, stoptime, (CASE WHEN stoptime-starttime >= 0 THEN
stoptime-starttime ELSE stoptime-starttime+'24 hours' END) as timediff
FROM mytable;

/Mattias

----- Original Message -----
From: Bj�rn T Johansen
To: Andrew L. Gould
Cc: PostgreSQL general list
Sent: Thursday, September 11, 2003 2:12 PM
Subject: Re: [GENERAL] I need a SQL...

On Thu, 2003-09-11 at 14:07, Andrew L. Gould wrote:

On Thursday 11 September 2003 06:25 am, Bj�rn T Johansen wrote:

I need to write a SQL that calculates the interval between a start time
and a stop time. This is the easy part. The problem is that I only have
the time part, i.e. no date, so how can I be sure to also calculate the
interval if the start time is before midnight and the stop time is after
midnight?

Regards,

BTJ

If the activity or period you are measuring can equal or exceed 12 hours, you
won't be able to calculate it reliably without a start date and a stop date.
If the periods are always less than 12 hours (and you assume all the data is
good), then stop times that are less than start times would indicate an
intervening midnight.

The dates do not have to be in the same fields as the times, since you can add
date and time data to create a timestamp for datetime calculations:

(stop_date + stop_time) - (start_date + start_time)

Best of luck,

Andrew Gould

Yes, the period can exceed 12 hours, so what you are saying is that this is not possible to solve
without the date part? I can write this logic in my business logic but I was hoping to
solve this in my database layer...

BTJ

Yes, I am sure, I just use copy-and-paste and I have double checked....
I am running on 7.3.4 but that shouldn't make any difference?

BTJ

On Thu, 2003-09-11 at 14:56, Mattias Kregert wrote:

When i run it, it works as intended (on pg 7.3.3). Which version do you use?

Are you absolutely sure you copied it exactly? You typed in '>=' and not '=', right?

/Mattias

----- Original Message -----
From: "Bjørn T Johansen" <btj@havleik.no>
To: "Mattias Kregert" <mattias@kregert.se>
Cc: <pgsql-general@postgresql.org>
Sent: Thursday, September 11, 2003 3:02 PM
Subject: Re: [GENERAL] I need a SQL...

Well, it's close... :)

But it looks like the case doesn't work..
If I run your sql, the timediff is negative.

But if I run this:
SELECT (stoptime-starttime+'24 hours') as timediff FROM mytable
the timediff has correct value..

Do you see any error in the case, cause I don't?

BTJ

On Thu, 2003-09-11 at 14:29, Mattias Kregert wrote:

Solution:

SELECT starttime, stoptime, (CASE WHEN stoptime-starttime >= 0 THEN
stoptime-starttime ELSE stoptime-starttime+'24 hours' END) as timediff
FROM mytable;

/Mattias

----- Original Message -----
From: Bjørn T Johansen
To: Andrew L. Gould
Cc: PostgreSQL general list
Sent: Thursday, September 11, 2003 2:12 PM
Subject: Re: [GENERAL] I need a SQL...

On Thu, 2003-09-11 at 14:07, Andrew L. Gould wrote:

On Thursday 11 September 2003 06:25 am, Bjørn T Johansen wrote:

I need to write a SQL that calculates the interval between a start time
and a stop time. This is the easy part. The problem is that I only have
the time part, i.e. no date, so how can I be sure to also calculate the
interval if the start time is before midnight and the stop time is after
midnight?

Regards,

BTJ

If the activity or period you are measuring can equal or exceed 12 hours, you
won't be able to calculate it reliably without a start date and a stop date.
If the periods are always less than 12 hours (and you assume all the data is
good), then stop times that are less than start times would indicate an
intervening midnight.

The dates do not have to be in the same fields as the times, since you can add
date and time data to create a timestamp for datetime calculations:

(stop_date + stop_time) - (start_date + start_time)

Best of luck,

Andrew Gould

Yes, the period can exceed 12 hours, so what you are saying is that this is not possible to solve
without the date part? I can write this logic in my business logic but I was hoping to
solve this in my database layer...

BTJ

---------------------------(end of broadcast)---------------------------
TIP 4: Don't 'kill -9' the postmaster

--
-----------------------------------------------------------------------------------------------
Bjørn T Johansen (BSc,MNIF)
Executive Manager
btj@havleik.no Havleik Consulting
Phone : +47 67 54 15 17 Conradisvei 4
Fax : +47 67 54 13 91 N-1338 Sandvika
Cellular : +47 926 93 298 http://www.havleik.no
-----------------------------------------------------------------------------------------------
"The stickers on the side of the box said "Supported Platforms: Windows
98, Windows NT 4.0,
Windows 2000 or better", so clearly Linux was a supported platform."
-----------------------------------------------------------------------------------------------

---------------------------(end of broadcast)---------------------------
TIP 7: don't forget to increase your free space map settings

On Thu, 2003-09-11 at 15:25, Mattias Kregert wrote:

Very strange indeed!

This is my output.
------------------------------------

Welcome to psql 7.3.3, the PostgreSQL interactive terminal.

Type: \copyright for distribution terms
\h for help with SQL commands
\? for help on internal slash commands
\g or terminate with semicolon to execute query
\q to quit

test=# create table mytable (starttime time, stoptime time);
CREATE TABLE
test=# insert into mytable values ('10:45', '22:30');
INSERT 103688 1
test=# insert into mytable values ('19:45', '04:30');
INSERT 103689 1
test=# insert into mytable values ('00:00', '00:00');
INSERT 103690 1
test=# insert into mytable values ('23:59', '00:01');
INSERT 103691 1
test=# insert into mytable values ('00:01', '23:59');
INSERT 103692 1
test=# select starttime,stoptime,(case when stoptime-starttime >= 0 then stoptime-starttime else stoptime-starttime+'24 hours' end) as timediff from mytable;
starttime | stoptime | timediff
-----------+----------+----------
10:45:00 | 22:30:00 | 11:45
19:45:00 | 04:30:00 | 08:45
00:00:00 | 00:00:00 | 00:00
23:59:00 | 00:01:00 | 00:02
00:01:00 | 23:59:00 | 23:58
(5 rows)

test=#
------------------------------------

As you see, it all works as it should. Can you do exactly the same and send me the complete output? If you get a different result, then it's time to send in a bug report...

/Mattias

----- Original Message -----
From: "Bj�rn T Johansen" <btj@havleik.no>

Yes, I am sure, I just use copy-and-paste and I have double checked....
I am running on 7.3.4 but that shouldn't make any difference?

BTJ

On Thu, 2003-09-11 at 14:56, Mattias Kregert wrote:

When i run it, it works as intended (on pg 7.3.3). Which version do you use?

Are you absolutely sure you copied it exactly? You typed in '>=' and not '=', right?

/Mattias

----- Original Message -----
From: "Bj�rn T Johansen" <btj@havleik.no>
To: "Mattias Kregert" <mattias@kregert.se>
Cc: <pgsql-general@postgresql.org>
Sent: Thursday, September 11, 2003 3:02 PM
Subject: Re: [GENERAL] I need a SQL...

Well, it's close... :)

But it looks like the case doesn't work..
If I run your sql, the timediff is negative.

But if I run this:
SELECT (stoptime-starttime+'24 hours') as timediff FROM mytable
the timediff has correct value..

Do you see any error in the case, cause I don't?

BTJ

On Thu, 2003-09-11 at 14:29, Mattias Kregert wrote:

Solution:

SELECT starttime, stoptime, (CASE WHEN stoptime-starttime >= 0 THEN
stoptime-starttime ELSE stoptime-starttime+'24 hours' END) as timediff
FROM mytable;

/Mattias

----- Original Message -----
From: Bj�rn T Johansen
To: Andrew L. Gould
Cc: PostgreSQL general list
Sent: Thursday, September 11, 2003 2:12 PM
Subject: Re: [GENERAL] I need a SQL...

On Thu, 2003-09-11 at 14:07, Andrew L. Gould wrote:

On Thursday 11 September 2003 06:25 am, Bj�rn T Johansen wrote:

I need to write a SQL that calculates the interval between a start time
and a stop time. This is the easy part. The problem is that I only have
the time part, i.e. no date, so how can I be sure to also calculate the
interval if the start time is before midnight and the stop time is after
midnight?

Regards,

BTJ

If the activity or period you are measuring can equal or exceed 12 hours, you
won't be able to calculate it reliably without a start date and a stop date.
If the periods are always less than 12 hours (and you assume all the data is
good), then stop times that are less than start times would indicate an
intervening midnight.

The dates do not have to be in the same fields as the times, since you can add
date and time data to create a timestamp for datetime calculations:

(stop_date + stop_time) - (start_date + start_time)

Best of luck,

Andrew Gould

Yes, the period can exceed 12 hours, so what you are saying is that this is not possible to solve
without the date part? I can write this logic in my business logic but I was hoping to
solve this in my database layer...

BTJ

---------------------------(end of broadcast)---------------------------
TIP 4: Don't 'kill -9' the postmaster

--
-----------------------------------------------------------------------------------------------
Bj�rn T Johansen (BSc,MNIF)
Executive Manager
btj@havleik.no Havleik Consulting
Phone : +47 67 54 15 17 Conradisvei 4
Fax : +47 67 54 13 91 N-1338 Sandvika
Cellular : +47 926 93 298 http://www.havleik.no
-----------------------------------------------------------------------------------------------
"The stickers on the side of the box said "Supported Platforms: Windows
98, Windows NT 4.0,
Windows 2000 or better", so clearly Linux was a supported platform."
-----------------------------------------------------------------------------------------------

---------------------------(end of broadcast)---------------------------
TIP 7: don't forget to increase your free space map settings

Well, here is my output..:

DT=# create table mytable (starttime time, stoptime time);
CREATE TABLE
DT=# insert into mytable values ('10:45', '22:30');
INSERT 20746 1
DT=# insert into mytable values ('19:45', '04:30');
INSERT 20747 1
DT=# insert into mytable values ('00:00', '00:00');
INSERT 20748 1
DT=# insert into mytable values ('23:59', '00:01');
INSERT 20749 1
DT=# insert into mytable values ('00:01', '23:59');
INSERT 20750 1
DT=# select starttime,stoptime,(case when stoptime-starttime >= 0 then
stoptime-starttime else stoptime-starttime+'24 hours' end) as timediff
from mytable;
starttime | stoptime | timediff
-----------+----------+----------
10:45:00 | 22:30:00 | 11:45
19:45:00 | 04:30:00 | -15:15
00:00:00 | 00:00:00 | 00:00
23:59:00 | 00:01:00 | -23:58
00:01:00 | 23:59:00 | 23:58
(5 rows)

DT=#

Strange....

On Thu, 2003-09-11 at 15:25, Mattias Kregert wrote:

Very strange indeed!

This is my output.
------------------------------------

Welcome to psql 7.3.3, the PostgreSQL interactive terminal.

Type: \copyright for distribution terms
\h for help with SQL commands
\? for help on internal slash commands
\g or terminate with semicolon to execute query
\q to quit

test=# create table mytable (starttime time, stoptime time);
CREATE TABLE
test=# insert into mytable values ('10:45', '22:30');
INSERT 103688 1
test=# insert into mytable values ('19:45', '04:30');
INSERT 103689 1
test=# insert into mytable values ('00:00', '00:00');
INSERT 103690 1
test=# insert into mytable values ('23:59', '00:01');
INSERT 103691 1
test=# insert into mytable values ('00:01', '23:59');
INSERT 103692 1
test=# select starttime,stoptime,(case when stoptime-starttime >= 0 then stoptime-starttime else stoptime-starttime+'24 hours' end) as timediff from mytable;
starttime | stoptime | timediff
-----------+----------+----------
10:45:00 | 22:30:00 | 11:45
19:45:00 | 04:30:00 | 08:45
00:00:00 | 00:00:00 | 00:00
23:59:00 | 00:01:00 | 00:02
00:01:00 | 23:59:00 | 23:58
(5 rows)

test=#
------------------------------------

As you see, it all works as it should. Can you do exactly the same and send me the complete output? If you get a different result, then it's time to send in a bug report...

/Mattias

----- Original Message -----
From: "Bjørn T Johansen" <btj@havleik.no>

Yes, I am sure, I just use copy-and-paste and I have double checked....
I am running on 7.3.4 but that shouldn't make any difference?

BTJ

On Thu, 2003-09-11 at 14:56, Mattias Kregert wrote:

When i run it, it works as intended (on pg 7.3.3). Which version do you use?

Are you absolutely sure you copied it exactly? You typed in '>=' and not '=', right?

/Mattias

----- Original Message -----
From: "Bjørn T Johansen" <btj@havleik.no>
To: "Mattias Kregert" <mattias@kregert.se>
Cc: <pgsql-general@postgresql.org>
Sent: Thursday, September 11, 2003 3:02 PM
Subject: Re: [GENERAL] I need a SQL...

Well, it's close... :)

But it looks like the case doesn't work..
If I run your sql, the timediff is negative.

But if I run this:
SELECT (stoptime-starttime+'24 hours') as timediff FROM mytable
the timediff has correct value..

Do you see any error in the case, cause I don't?

BTJ

On Thu, 2003-09-11 at 14:29, Mattias Kregert wrote:

Solution:

SELECT starttime, stoptime, (CASE WHEN stoptime-starttime >= 0 THEN
stoptime-starttime ELSE stoptime-starttime+'24 hours' END) as timediff
FROM mytable;

/Mattias

----- Original Message -----
From: Bjørn T Johansen
To: Andrew L. Gould
Cc: PostgreSQL general list
Sent: Thursday, September 11, 2003 2:12 PM
Subject: Re: [GENERAL] I need a SQL...

On Thu, 2003-09-11 at 14:07, Andrew L. Gould wrote:

On Thursday 11 September 2003 06:25 am, Bjørn T Johansen wrote:

I need to write a SQL that calculates the interval between a start time
and a stop time. This is the easy part. The problem is that I only have
the time part, i.e. no date, so how can I be sure to also calculate the
interval if the start time is before midnight and the stop time is after
midnight?

Regards,

BTJ

If the activity or period you are measuring can equal or exceed 12 hours, you
won't be able to calculate it reliably without a start date and a stop date.
If the periods are always less than 12 hours (and you assume all the data is
good), then stop times that are less than start times would indicate an
intervening midnight.

The dates do not have to be in the same fields as the times, since you can add
date and time data to create a timestamp for datetime calculations:

(stop_date + stop_time) - (start_date + start_time)

Best of luck,

Andrew Gould

Yes, the period can exceed 12 hours, so what you are saying is that this is not possible to solve
without the date part? I can write this logic in my business logic but I was hoping to
solve this in my database layer...

BTJ

---------------------------(end of broadcast)---------------------------
TIP 4: Don't 'kill -9' the postmaster

--
-----------------------------------------------------------------------------------------------
Bjørn T Johansen (BSc,MNIF)
Executive Manager
btj@havleik.no Havleik Consulting
Phone : +47 67 54 15 17 Conradisvei 4
Fax : +47 67 54 13 91 N-1338 Sandvika
Cellular : +47 926 93 298 http://www.havleik.no
-----------------------------------------------------------------------------------------------
"The stickers on the side of the box said "Supported Platforms: Windows
98, Windows NT 4.0,
Windows 2000 or better", so clearly Linux was a supported platform."
-----------------------------------------------------------------------------------------------

---------------------------(end of broadcast)---------------------------
TIP 7: don't forget to increase your free space map settings

On Thu, 2003-09-11 at 16:26, Tom Lane wrote:

=?ISO-8859-1?Q?Bj=F8rn?= T Johansen <btj@havleik.no> writes:

Yes, I am sure, I just use copy-and-paste and I have double checked....
I am running on 7.3.4 but that shouldn't make any difference?

It'll probably work better if you quote the zero. Unquoted, you get
some weird textual comparison. Compare:

regression=# explain select * from time_tbl where f1-f1 >= 0;
QUERY PLAN
--------------------------------------------------------
Seq Scan on time_tbl (cost=0.00..1.14 rows=3 width=8)
Filter: (((f1 - f1))::text >= '0'::text)
(2 rows)

regression=# explain select * from time_tbl where f1-f1 >= '0';
QUERY PLAN
--------------------------------------------------------
Seq Scan on time_tbl (cost=0.00..1.12 rows=3 width=8)
Filter: ((f1 - f1) >= '00:00'::interval)
(2 rows)

In "C" locale, the textual comparison accidentally manages to give
the desired answers, but in other locales it would not.

(Just another example of why implicit coercions to text are evil.)

regards, tom lane