Paul,
You're awesome...thank you. One more question pertaining to this...
I'm accustomed to selecting a variable and using something like this:
$firstin = mysql_result($result,$row,firstin);
to set my variable.
What do I want to do here to get the right result into my variable?
$query2 = "SELECT SEC_TO_TIME(TIME_TO_SEC('$lastout') - TIME_TO_SEC('$firstin'))";
$result2 = mysql_db_query($database, $query, $connection) or die ("Error in query:
$query. " . mysql_error());
This isn't working, I get 23:
$total = mysql_result($result2,0);
Thanks again!
Nathan Cowles
On Mon, 4 Mar 2002, Paul DuBois wrote:
> >Hello,
> >
> >I am trying to get an accurate (nearest 15 minutes would be
> >fine) difference between two times that I have in my database. They are
> >both of type time, and are in the format 00:00:00.
> >
> >For example, if I want the difference between 09:20:00 and 10:43:00, I
> >would like it to give me 01:23:00 as the answer, but all I have been able
> >to get it to do is give me one. It is rounding to the nearest hour.
> >
> >I found this query which seems like it would work, but gives me a NULL
> >result set:
> >
> >SELECT DATE_SUB('11:48:02', INTERVAL '10:37:37' HOUR_SECOND);
>
> For two times t1 and t2, compute the difference in seconds like this:
>
> SELECT TIME_TO_SEC(t2) - TIME_TO_SEC(t1)
>
> If you want the result as a time value, pass it to SEC_TO_TIME():
>
> SELECT SEC_TO_TIME(TIME_TO_SEC(t2) - TIME_TO_SEC(t1))
>
>
> >
> >I am trying to do all of this from PHP, so if you happen to know of a way
> >to make it do the subtraction in PHP without losing all accuracy, please
> >let me know.
> >
> >Thanks for any help you can offer.
> >
> >Nathan Cowles
> >StormNet Communications
> >530.897.4069
>
Nathan Cowles
StormNet Communications
530.897.4069
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