Re: Re: Display Mysql Result in drop down list

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Hi:

then now i'm trying to select a value from the drop down list, echo the value 
i have selected and lastly INSERT the value into DB with the following :
Below's a snippet of code:

<snip>
//retrieve all "tutor_name" from DB and display in drop down list

<SELECT class="textarea" name="tutor_name" />
<?


$sql = mysql_query("SELECT tutor_name FROM tutor ");
while ($row = mysql_fetch_array($sql))
{
 print "<OPTION VALUE=\"$tutor_name\" SELECTED>" .$row
["tutor_name"]. "</option>";
 }
 $result = $db->query($sql);

?>
</select>

</snip>

<snip>
***********************INSERT selected values into DB**************************
$tutor_name = $_POST["tutor_name"];

$sql = "INSERT INTO class (class_code, tutor_name, edu_level, timetable_day, 
timetable_time)
        VALUES 
('$class_code','$tutor_name','$edu_level','$timetable_day','$timetable_time')";

//execute query statement
$result = $db->query($sql);

</snip>


<snip>
*********************echo the value i have selected***************************
<?
if($selected_tutor_name == $tutor_name)
echo $_POST["tutor_name"];
?>

</snip>


Problem: I was unable to echo the value i selected from drop down as well as 
INSERT into DB...wonder where have i gone wrong???
Reali need some help here...all help are greatly appreciated =)
Thanks in advance.


Irin.

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