PHP Forms question

[Anon]

I have a section in my web site that allows people to update/change their profile and it uses alot of drop down menu's to ensure uniformity of answers such as:
<select name="Interests1"> <option selected value="none">
<option value="Acting">Acting / Drama
<option value="Aerobics">Aerobics
<option value="Backpacking">Backpacking
//SNIP
</select>

The problem is, I want people to be able to see what they had before from the drop down. Currently, the user has no idea what their previous selection was. How do I fix this?

[Anon]

okay, I had the same problem, and here is my solution.
the option which is selected automatically is the one with <option selected --> syntax. what you need to do is get the value of Interests1 from the database and compare it with the ones you want in the drop down box. If the value from the database matches any value then just add "selected" syntax in the code, and that particular option is selected.
let me try to show an example.
the user has "Aerobics" stored in the db.
<?
$result = @("select col_name from db");
$user_val = @mysql_fetch_array($result);
// now the options stuff
echo "<select name=Interests>";
if ($user_val == "Acting")
echo "<option selected value=Acting>Acting
else
echo "<option value=Acting>Acting

if ($user_val == "Aerobics")
echo "<option selected value=Aerobics>Aerobics
else
echo "<option value=Aerobics>Aerobics";

echo "</select>";

try to optimise the code, eg have all the <options> value in a table and you could write a loop..

cheers
Nilesh.

[Anon]

Here's my personal favourite solution to this:

$options is the array that contains the options (e.g. from database).
$useroption is a string that contains the user's pre-selection (from the user's record).

echo "<select name=interest size=1>\n";

foreach($options as $entry)
{
if($entry == $useroption)
$between = " selected";
else
$between = "";

echo "<option$between>$entry</option>\n";
}

echo "</select>\n\n";