Many thanks to Robert and Bruce.  I will try this as I have time over the
weekend, and report back how it goes.

On Thu, Aug 25, 2016 at 7:15 PM, Bruce Hohl <[email protected]> wrote:

> I can confirm as follows.
>
> With this in table1:
> ID time1 time2
> 1 01/01/16 11:30 PM 01/02/16 04:00 AM
> 2 01/02/16 01:30 AM 01/02/16 05:30 AM
> This query:
> SELECT "time1", "time2", DATEDIFF( 'minute', "time1", "time2" ) AS
> "diff_minutes", DATEDIFF( 'minute', "time1", "time2" ) / 60.00 AS
> "diff_hours" FROM "Table1"
>
> Returns:
> time1 time2 diff_minutes diff_hours
> 01/01/16 11:30 PM 01/02/16 04:00 AM 270 4.5
> 01/02/16 01:30 AM 01/02/16 05:30 AM 240 4
> This might be helpful:
> http://hsqldb.org/doc/guide/builtinfunctions-chapt.html#bfc_
> datetime_functions
>
> Be mindful of appropriate use of HSQLDB embedded:
> https://wiki.openoffice.org/wiki/FAQ_(Base)
>
>
> On Thu, Aug 25, 2016 at 3:29 PM, Robert Großkopf <
> [email protected]> wrote:
>
>> -----BEGIN PGP SIGNED MESSAGE-----
>> Hash: SHA1
>>
>> Hi Don,
>>
>> > It occurs to me my OP may not have been as clear as it should be.
>> > To clarify, I need to calculate the decimal value of the times
>> > involved: (([day_end]-[day_start])-([lunch_end]-[lunch_start])) =
>> > H.nn worked
>> >
>> > So, for example: ((17:00-08:00)-(13:00-12:00)) = 8.0
>>
>> This one works:
>> SELECT "ID", "DayBegin", "DayEnd", "LunchBegin", "LunchEnd",
>> DATEDIFF('hh',"DayBegin", "DayEnd")-DATEDIFF('hh',"LunchBegin",
>> "LunchEnd") AS "Worked" FROM "Times"
>>
>> Examples I have written down in Base-Handbook. But the newer versions
>> aren't translated yet from German to English.
>>
>> Be careful with the example. It does only work with the value for the
>> hour. So you get the same with
>> (17:00-08:00)-(13:00-12:00)
>> and for example
>> (17:10-08:00)-(13:00-12:00)
>>
>> So it would be better to write down the same code for minutes:
>> SELECT "ID", "DayBegin", "DayEnd", "LunchBegin", "LunchEnd",
>> DATEDIFF('mi',"DayBegin", "DayEnd")-DATEDIFF('mi',"LunchBegin",
>> "LunchEnd") AS "Worked" FROM "Times"
>>
>> So you get the workingtime in minutes. Devide it with 60.00 (.00 for
>> the decimal places) and you get a better diff for hours than in the
>> first query.
>>
>> Regards
>>
>> Robert
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>


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