Re: Hours available in a day
| From: | Shaun | Date: | Fri, 03 Oct 2003 21:43:36 +0000 |
| Subject: | Re: Hours available in a day | ||
| References: | 1 | Groups: | php.db |
| Request: | Send a blank email to php-db+get-30830@lists.php.net to get a copy of this message | ||
"Shaun" <shaun@mania.plus.com> wrote in message
news:20031003110254.34029.qmail@pb1.pair.com...
> Hi,
>
> I have a table called Bookings which has two important columns;
> Booking_Start_Time and Booking_End_Time. These columns are both of type
> DATETIME. Given any day how can I calculate how many hours are available
> between the hours of 09.00 and 17.30 so a user can see at a glance how
many
> hours they have unbooked on a particular day (i.e. 8.5 hours less the time
> of any bookings on that day), can this be done with a query
> or do I have to work it out in my code?
>
> Thanks for your help
Fixed:
$query = "SELECT
8.5 - (SUM(((DATE_FORMAT(B.Booking_End_Date, '%k') * 60 ) +
DATE_FORMAT(B.Booking_End_Date, '%i')) -
((DATE_FORMAT(B.Booking_Start_Date, '%k') * 60 ) +
DATE_FORMAT(B.Booking_Start_Date, '%i'))) / 60) AS
Available_Hours
FROM Bookings B
WHERE B.User_ID = '".$user_id."'
AND DATE_FORMAT(Booking_Start_Date, '%Y-%m-%d') =
'".date("Y-m-d",
strtotime("$year-$month-$day"))."'";