Creé un menú desplegable de PHP que se rellena desde una base de datos MySql y funciona bien, el problema ocurre cuando quiero publicar el seleccionado en otro script. La pregunta es ¿cómo publicar los datos en el otro script? Este es el código fuente del script que implementa los menús desplegables. ¡¡¡¡Por favor ayuda!!!!

<?php
$conn = mysql_connect("localhost", "admin", "admin");

if (!$conn) {
    echo "Unable to connect to DB: " . mysql_error();
    exit;
}

if (!mysql_select_db("ekupuvac")) {
    echo "Unable to select EKupuvac: " . mysql_error();
    exit;
}

$query = "SELECT ImeK, KupuvacID FROM kupuvac ORDER BY Saldo DESC";
$result = mysql_query($query) or die(mysql_error());

if (!$result) {
    echo "Could not successfully run query ($query) from DB: " . mysql_error();
    exit;
}

if (mysql_num_rows($result) == 0) {
    echo "No rows found, nothing to print so I am exiting";
    exit;
}

$dropdown = "<select name='ImeK'>";
while($row = mysql_fetch_assoc($result)) {
$dropdown.= "\r\n<option value='{$row['KupuvacID']}'>{$row['ImeK']}</option>";
}
$dropdown .= "\r\n</select>";

echo"Izberi Kupuvac:";
echo $dropdown;

// Second Combo

$conn = mysql_connect("localhost", "admin", "admin");

if (!$conn) {
    echo "Unable to connect to DB: " . mysql_error();
    exit;
}

if (!mysql_select_db("ekupuvac")) {
    echo "Unable to select EKupuvac: " . mysql_error();
    exit;
}

$query2 = "SELECT ImeP, ProzivodID FROM proizvod ORDER BY ImeP";
$result2 = mysql_query($query2) or die(mysql_error());

if (!$result2) {
    echo "Could not successfully run query ($query2) from DB: " . mysql_error();
    exit;
}

if (mysql_num_rows($result2) == 0) {
    echo "No rows found, nothing to print so I am exiting";
    exit;
}

$dropdown2 = "<select name='ImeP'>";
while($row = mysql_fetch_assoc($result2)) {
$dropdown2.= "\r\n<option value='{$row['ProzivodID']}'>{$row['ImeP']}</option>";
}
$dropdown2.= "\r\n</select>";

echo"<br> Izberi Proizvod:";
echo $dropdown2;
echo"<br>";

mysql_free_result($result);
?>