Passing Variables

cueball2000uk

This is my code so far

<?php
echo "<script language=Javascript>";
echo "</script>";
include "../includes/ali.inc";
include "../includes/select_statments.inc";

$country = "$_GET[country]";
$county = "$_GET[county]";
$town = "$_GET[town]";

echo "$country<BR>";
echo convert_country_id_to_name_get();
echo " $county<BR> $town";
?>
<?php
$dropquery1 = mysql_query("SELECT DISTINCT country_ID,country_name FROM cs_country ORDER BY country_name ASC");
$dropquery2 = mysql_query("SELECT DISTINCT county_ID,country_ID,county_name FROM cs_county WHERE country_ID = '$_GET[country]' ORDER BY country_ID,county_name ASC");
$dropquery3 = mysql_query("SELECT DISTINCT town_ID,country_ID,county_ID,town_name FROM cs_towns WHERE county_ID = '$_GET[county]' ORDER BY country_ID,county_ID,town_name ASC");

$country1 .= "<form name=country>";
$country1 .= "<select name=country onChange=submit(this.form)>";
while($array = mysql_fetch_array($dropquery1)) {
	$country1 .= "<option value=$array[0]>$array[1]</option>";
}
$country1 .= "</select>";
$country1 .= "</form>";
echo "$country1";

$county1 .= "<form name=county1>";
$county1 .= "<select name=county onChange=submit(this.form)>";
while($array = mysql_fetch_array($dropquery2)) {
	$county1 .= "<option value=$array[0]>$array[2]</option>";
}
$county1 .= "</select>";
$county1 .= "</form>";
echo "$county1";

$town1 .= "<form name=town1>";
$town1 .= "<select name=town>";
while($array = mysql_fetch_array($dropquery3)) {
	$town1 .= "<option value=$array[0]>$array[3]</option>";
}
$town1 .= "</select>";
$town1 .= "</form>";
echo "$town1";
?>

Problem is that it forgets the information in the menus.

how do i use three dynamic menus in one page?

GBahle

Hm, I don't know what exactly your code is doing at the moment. Some things that spring to attention though:

You are using three forms instead of one. This means that submitting one form will actually lose the information of the other two. Second, your form tags don't have any action directive. Third, you print a <Script> tag without any script in it.

Please clarify your question.

cueball2000uk

i need to generate some dynamic menus.

country
county
town.

when the country is picked, the county menu should populate with the countys fot that county. when the county is picked the town menu needs to be populated with the towns for that county.

When the user has picked all three choices, there should be a submit button so they can begin.

the search results are dealt with on a different page.

i need the script to remember 3 values

country_ID, county_ID, town_ID

IE:- 26 22 1

hope this clears my question up a little

GBahle

ok, that shouldnt be too hard:
You got three possibilities

a) First, use a database to store the intermediate values (not really interesting here)

b) Use a session. This is entirely feasible (but still somewhat too elaborate here, imo)

c) Pass the values by using a query string attached to your URL. The simplest solution here, imo.

First, change your onChange event to a javascript function that creates the query string, then redirects to the same page (the string should look sth like ?Country=....&County=....&......)
with the query string attached. Then, use either $GET["Country"] or else HTTP_GET_VARS[....] depending on your PHP version to retrieve the values.
Be sure to define default values in case that the page is called without a valid query string.
This would look sth like that: $countryVal = isset($GET["Country"]) ? $_GET["Country"] : $yourDefault;