mysqli_fetch_array iteration issue - php

Question

I have a page that is dynamically generated using PHP and a MySQL database. The page displays information about an event, and lists the dates that event will run.
My database has the following relevant tables:
events - containing the event name, cost etc
venue - containing venue information
instance - containing the event_id, venue_id, the date the event at that venue will run, and the capacity for that instance.
registration - containing the instance_id, and attendee_id.
To grab all the information to actually display the event information, I use the following code:
$eid = $_GET['event_id'];
$q = "SELECT e.event_name, e.event_description, e.event_byline, e.event_benefit, e.event_cost, e.event_exam, e.event_resi, i.venue_id, i.instance_id, i.instance_cap, v.venue_name, DATE_FORMAT( i.instance_date, '%M %D, %Y' ) AS DATE
FROM events AS e
INNER JOIN instance AS i ON e.event_id = i.event_id
INNER JOIN venue AS v ON i.venue_id = v.venue_id
WHERE e.event_id = $eid
ORDER BY i.venue_id, i.instance_date";
$cur_venue = 0;
$r = #mysqli_query ($dbc,$q) or die(mysqli_error($dbc));
$row = mysqli_fetch_array($r, MYSQLI_ASSOC);
Now, what I want to do is display a list, sorted by venue, of the instances for the relevant event, which I have done up to a point. However, what I also want to do is only display the instance if there is space left on that particular instance.
Since I know the capacity of each instance (from my instance_cap column), and I can COUNT the number of attendees registered to each instance, I figure I can do this thuswise:
do
{
$list_instance = $row['instance_id'];
$qRegs = "SELECT COUNT(delegate_id) AS regs FROM registration
WHERE registration.instance_id = $list_instance";
$rRegs = mysqli_query($dbc,$qRegs);
$registrations = mysqli_fetch_object($rRegs);
$capacity = $row['instance_cap'];
$availability = $capacity - $registrations->regs;
if ($availability > 0){ //if event has places available...
if ($cur_venue != $row['venue_id']) //and if the current venue is not the same as the venue id
{
echo '<li id="'.$row['venue_name'].'">'
$cur_venue = $row['venue_id'];
echo '<h4>'.$row['venue_name'].'</h4>';//display the venue name
}
echo ''.$row['DATE'].'' //display the date for current instance
echo '</li>';//close list tag
}
} while ($row = mysqli_fetch_array($r, MYSQLI_ASSOC));';
The problem I have is that this misses out the first instance and skips straight to the second one. I understand that this is probably due to the fact that I have called mysqli_fetch_array twice, so how I can work it so that this doesn't happen?

Answer 1

You should be able to run one query, and then just use the resultset. Try this:
SELECT e.event_name, e.event_description, e.event_byline,
e.event_benefit, e.event_cost, e.event_exam, e.event_resi,
i.venue_id, i.instance_id, i.instance_cap, v.venue_name,
DATE_FORMAT( i.instance_date, '%M %D, %Y' ) AS DATE
FROM events AS e
INNER JOIN instance AS i ON e.event_id = i.event_id
INNER JOIN venue AS v ON i.venue_id = v.venue_id
WHERE e.event_id = $eid AND i.instance_cap > (SELECT COUNT(r.delegate_id) FROM registration AS r WHERE r.instance_id = i.instance_id)
ORDER BY i.venue_id, i.instance_date
Granted, this is untested, so it might not work as-is.
EDIT: a sub-query is probably more correct. See the edited query above.

Answer 2

Put
mysqli_data_seek ( $r , 0 );
before the second mysqli_fetch_array to reset the internal pointer to the first instance.