select option problem

Anon

I have two drop down menus populated with values from a database using the following code for both (changed names and variables...):

<xmp>
<TR>
<TD><B>State:</B><BR>
<SELECT NAME="statename[]">
<?php

$prod = mysql_query("SELECT * FROM state", $db);

while ($data2 = mysql_fetch_array($prod)) {

printf("<OPTION VALUE=\"%s\">%s</OPTION>\n", $data2[stateID], $data2[statename])
;
}
?>
</SELECT></TD></TR>
</xmp>

now, on this same page I check for the presence of $submit from the form at the top, and if it exists passes the info to my database. the problem I am having is the selection from the two dropdown menus are not being passed to the database.

other drop down menus where I have actually written the HTML into the page work just fine.

How do I need to alter the script to allow the selection from my two menus to be passed to the dbase with the rest of the information?

Anon

you could view the source of the first page and see if there are the value in the option.
if ok,then
try in the next page print the variables of the select from the last page before querying in the databse.

snoopy kun

Anon

this is the code i always use and it has always worked:

<?php
$result = mysql_query("SELECT * FROM state",$db);

echo "<select name=Member>";

if ($myrow = mysql_fetch_array($result)) {

do {
echo "<option value=$myrow[stateID]>$myrow[statename]</option>\n";
} while ($myrow = mysql_fetch_array($result));

echo "</select>";
}
?>

Anon

yeah, I looked at the source code and the value attribute for both of the drop down columns is correct. it is beeing filled in during the population of the menu. I also posted the submit values to a new screen and it worked as well. just, when I click submit and go to the dbase the values for the two tables are not inserted, everything else is.