display problem

[timmybuck]

hi im having a problem displaying some fields from my table.

I want to display all the eventtypes from the event table in a drop down menu. but all i get is a list of numbers from the eventid column

any ideas guys

here is my code

$event=mysql_query("SELECT * FROM event ORDER BY Date ASC")

OR DIE("Unable to find events. Error: ".mysql_error());

print("<select name=\"Eventtype\" size=1>\n");

print("<option value=\"b\">Select");
print("<option value=\"b\">-----");
while($row = mysql_fetch_array ($event))
{
print("<option value=\"$row[1]\">$row[0]\n");
}
print("</select>\n");

[laserlight]

that probably implies that the eventid column is the first column, and the eventtype column the second.

A solution would be to use:
print("<option value=\"{$row[0]}\">{$row[1]}\n");

Better would be to take advantage of the associative array:

echo "<option value=\"{$row[eventid]}\">{$row[eventtype]}\n";

[timmybuck]

hi that's great it works. thank u but now i have a diff problem. In my event types column i have the same types opf events eg sports occurs more than once. so in my drop down menu i have sports more than once but really i need just one. I need one of each really, eg sports, education, music etc. The idea is the user chooses and event type and then another page is displayed showing all the available types of events eg for sport it could be football, hockey etc

[timmybuck]

i have solved the problem. i needed another table