Okay I have the following code so far.
$sql = "SELECT stass_warrior FROM staff_assignments WHERE
stass_pos = $pos_id AND stass_weekend = $initiation";
if (!$query = mysql_query($sql)) error(mysql_error());
$query = mysql_fetch_row($query);
$sql = "SELECT war_id, CONCAT(war_lname, ', ', war_fname) FROM
warrior, staff_roster WHERE war_id != 1 AND staff_weekend = $initiation AND
staff_warrior=war_id ORDER BY war_lname";
if (!$res = mysql_query($sql)) error(mysql_error());
tr();
echo TTT . '<th class="format" width="25%">' . $row[1] . '</th>' . B;
echo TTT . '<td class="format">' . B;
echo TTT . '<select name="position_' . $n . '[]" size="5"
multiple>' . B;
while($rec = mysql_fetch_row($res)) {
($query[0] == $rec[0] ? $selected =" selected" : $selected
= "");
printf("\t\t\t\t<option value=\"%s_%s\"%s>%s</option>\n",
$pos_id, $rec[0], $selected, $rec[1]);
}
echo TTT . '</select>' . B;
echo TTT . '</td>' . B;
trc();
Now my problem is with the multiple feature of the drop down menu.
The first query to the db resulting in the $query variable may contain more
then one result.
However only the last result will appear with the selected option.
I need some code that is is going to display the selected option next to
all the correct results in the drop down.
Understand what I am needing?
Thanks for your help
Phillip
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