Drop down list nr 1 to Drop down list nr 2

[keo]

Hi,

I have 2 drop down lists, and I want the 2nd drop down list to be generated off the first drop down list selection . But I'm missing something on how to pass the variable of the first drop down list that the user selects to the second drop down list.

Any help appreciated

Here is the code.

if (!isset($HTTP_POST_VARS[submit])) {
/ User has not submitted form data /

echo "<table width=\"100%\" border=\"1\" cellspacing=\"2\" cellpadding=\"2\"><tr><td>";

// Creates the first dropdown list

echo "<form action=\"$PHP_SELF\" method=\"POST\"><select name=\'city_id[]\'><option value=\"\" \"selected\">Choose City</option>";

while ($row = mysql_fetch_array($query))
{
echo "<option value=\"{$row['city_id']}\">{$row['cities']}</option>";
}

echo "</select>";

echo "<input type=\"submit\" name=\"submit\" value=\"Go\"></form>";

} else {

// This could be completly wrong I'm a novice
$sql = "select * from provinces where city_id={$_POST['city_id']} order by provincename";
$query2 = mysql_query($sql);

// Creates second dropdown list with the provinces

echo "<form action=\"$PHP_SELF\" method=\"POST\"><select name=\"province name\"><option value=\"\" \"selected\">Chosse Province</option>";

while ($row = mysql_fetch_array($query2))
{
echo "<option value=\"{$row['pro_id']}\">{$row['provincename']}</option>";
}

echo "</select>";

echo "<input type=\"submit\" name=\"submit\" value=\"Go\"></form>";

}

echo "</table></tr></td>";
?>

[bpat1434]

echo "<form action=\"$PHP_SELF\" method=\"POST\"><select name=\'city_id[]\'><option value=\"\" \"selected\">Choose City</option>";

should be

echo "<form action=\"$PHP_SELF\" method=\"POST\"><select name=\'city_id\'><option value=\"\" \"selected\">Choose City</option>";

$sql = "select * from provinces where city_id={$_POST['city_id']} order by provincename";

should probably be

$sql = "SELECT * FROM `provinces` WHERE city_id='".$_POST['city_id']."' ORDER BY provincename";

~Brett

[keo]

Hi Brett thanx for the reply!

Have now tried your answer but it didn't work, I just get an empty drop down nr 2. Still some problem with the $_POST and cathing the selected variable. What could it be I'm going nuts!.

[bpat1434]

change

action=\"$PHP_SELF\"

to

action=\"$_SERVER['PHP_SELF']\"

~Brett

[keo]

No that doesn't work. If I do a test and put it like this skipping the $_POST, then I get the drop down list nr2 populated, so the action=\"$PHP_SELF\" is not wrong.

$query2 = mysql_query("SELECT * FROM provinces ORDER BY provincename");

[bpat1434]

OKay fine. But this:

<select name=\'city_id[]\'>

should not be an array. It should be just "city_id".

~Brett