Hongmei,
Though you said I completely understood your question, I myself feel confused. Let me try one more time.
For static (or semi-dynamic) page:
<?
if (isset($submit)){
echo "Your data have been successfully saved. Here is what"
." you submitted: Your name = $_POST[yourname] and "
."Test = $_POST[test]";
}
echo "<form method=post action=$PHP_SELF>"
."<input type=text name=yourname value=\"$_POST[yourname]\">"
."<select name=test>";
if ($_POST[test]=="1"){$sel1="selected";}
elseif ($_POST[test]=="2"){$sel2="selected";}
elseif ($_POST[test]=="3"){$sel3="selected";}
echo"<option value=1 $sel1>X</opition>"
."<option value=2 $sel2>Y</option>"
."<option value=3 $sel3>Z</option>"
."</select>"
."<input type=submit name=submit value=\"Save your data\">"
."</form>";
?>
Save this script as test1.php and run it.
We need to modify the form part for database-driven dropdown menu. Assume you have a database called "menu" and table called "menu" with two columns, one called menu_name (X, Y, Z), the other called menu_value (1, 2, 3). Let's do it again:
<?
if (isset($submit)){
echo "Your data have been successfully saved. Here is what "
."you submitted: Your name = $_POST[yourname] and "
."Test = $_POST[test]";
}
echo "<form method=post action=\"$PHP_SELF\">"
."<input type=text name=yourname value=\"$_POST[yourname]\">"
."<select name=test>";
mysql_connect ("localhost", "your_usernamet", "your_password");
mysql_select_db ("menu");
$query="select * from menu";
$result=mysql_query($query);
while ($row=mysql_fetch_array($result))
{
if ($row[menu_value] ==$_POST[test])
{$sel="selected";}
echo "<option value=$row[menu_value] $sel>$row[menu_name]</opition>";
}
echo "</select>"
."<input type=submit name=submit value=\"Save your data\">"
."</form>";
?>
Save this script as test2.php, and run it.
These two examples should help you understand form applications.
Happy coding,
Paul
