displaying names in a drop down from a database

timmybuck

i have a two column table called ' places' in there is a countryID and eventCountry column.

The idea is that a user selects a country from the drop down menu and when they choose go they get taken to another page.

this is how far i have got


mysql_connect('localhost', 'poo', 'poo123')or die ("Database connection error");

mysql_select_db('booking');

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

print("<option value=\"b\">Select");
print("<option value=\"b\">-----");
while($row = mysql_fetch_array ($places))
{
print("<option value=\"$row[countryID]\">$row[eventCountry] \n");
}
print("</select>\n");
?>

kellog

Seems OK to me, except you might want to consider chainging:

print("<option value=\"$row[countryID]\">$row[eventCountry] \n");

to:

print("<option value=\"$row['countryID']\">$row['eventCountry'] \n");

Not having the quote there will cause problems in later versions of php.

Originally posted by timmybuck
Hi

i have a two column table called ' places' in there is a countryID and eventCountry column.

The idea is that a user selects a country from the drop down menu and when they choose go they get taken to another page.

this is how far i have got
mysql_connect('localhost', 'poo', 'poo123')or die ("Database connection error");

mysql_select_db('booking');

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

print("<option value=\"b\">Select");
print("<option value=\"b\">-----");
while($row = mysql_fetch_array ($places))
{
print("<option value=\"$row[countryID]\">$row[eventCountry] \n");
}
print("</select>\n");
?>
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