[RESOLVED] mysql_fetch_array Issues

[scriderpsd]

I have a form creating an absence form for teachers at a high school. I have 9 rows (for each of the 9 possible periods) and each row has a drop down box with a list of subs. Therefore, I need to create the same dropdown box 9 times.

I pull the subs from a database table as follows....

$sql_subbed="SELECT * FROM personnel where loc_id='".$row_name[loc_id]."' order by last_name";
$conn = db_absence_connect();
$result_subbed = mysql_query($sql_subbed,$conn);

Next, I do something like this to create the dropdown...

echo "<select name='subbed_0' onchange=\"reload3(this.form)\"><option value=''>Select one</option>";
    while($row_subbed_0 = mysql_fetch_array($result_subbed)) { 
        echo  "<option value='$row_subbed_0[personnel_id]'>$row_subbed_0[last_name], $row_subbed_0[first_name]</option>";

}
echo "</SELECT>";

Now, my question is....can I re-use $result_subbed for each of the nine rows or do I need to need to do 9 separate database queries and run mysql_fetch_array on each of those?

I tried doing something like this for the second row but it doesn't work.

echo "<select name='subbed_1' onchange=\"reload3(this.form)\"><option value=''>Select one</option>";
    while($row_subbed_1 = mysql_fetch_array($result_subbed)) { 
        echo  "<option value='$row_subbed_1[personnel_id]'>$row_subbed_1[last_name], $row_subbed_1[first_name]</option>";

}
echo "</SELECT>";

It seems like there should be an efficient way to do this.

As always, any help is very much appreciated.

Thanks,

Scott

[reddrum]

Hi Scott,
If you save your result set in an array you can use it to build your other selects.

$subID = array();
$subLastName = array();
$subFirstName = array();
$i=0;
echo "<select name='subbed_0' onchange=\"reload3(this.form)\"><option value=''>Select one</option>";
while($row_subbed_0 = mysql_fetch_array($result_subbed)) {
   $subID[$i] = $row_subbed_0['personnel_id'];
   $subLastName[$i] = $row_subbed_0['last_name'];
   $subFirstName[$i] = $row_subbed_0['first_name'];
   echo "<option value='$subID[$i]'>$subLastName[$i],  $subFirstName</option>";      

   ++$i;
}
echo "</SELECT>";

[Weedpacket]

Or, even cleaner:

$sql_subbed="SELECT * FROM personnel where loc_id='".$row_name[loc_id]."' order by last_name";
$conn = db_absence_connect();
$result_subbed = mysql_query($sql_subbed,$conn); 

$subbed_personnel = array();
while($result = mysql_fetch_array($result_subbed))
{
	$subbed_personnel[] = $result;
}

$subbed_options = "";
foreach($subbed_personnel as $person)
{
	$subbed_options .= "<option value='$person[personnel_id]'>$person[last_name], $person[first_name]</option>";
}



echo '<select name="subbed_0" onchange="reload3(this.form)"><option value="">Select one</option>';
echo $subbed_options;
echo '</select>';

echo '<select name="subbed_1" onchange="reload3(this.form)"><option value="">Select one</option>';
echo $subbed_options;
echo '</select>';