keithl — Sun, 17 Dec 2000 04:08:39 +0000

Thanks for taking the time to answer my question Rob!
That did it...

Thanks,
Keith

Rob Pengelly — Fri, 15 Dec 2000 02:28:37 +0000

Sorry I did not have time to test it, but the following should work

<form action="<?=$PHP_SELF?>">
<select name="song">
<?
//Define variables
$mysql_host = 'localhost';
$mysql_username = 'username';
$mysql_password = 'password';
$table = 'song';

//Connect to database and choose table
//connect to mysql
$cnx = @mysql_connect($mysql_host,$mysql_username,$mysql_password) or die ("Could not connect:".mysql_error());
mysql_select_db($mysql_database) or die ("Could not choose database:".$mysql_error());

//Select the SONG from the table
$query = mysql_query("SELECT Song FROM $table");
while($row = mysql_fetch_array($query))
{
     echo "<option value=\"".$row["Song"]."\">".$row["Song"]."</option>";
}
mysql_close($cnx);
?>
</select>
<input type="submit" name="submitsong" value="Submit Song">
</form>

I am not sure what you mean by "it will drop down to that particular song".

With the code above, you could use:

if($submitsong)
{
     //A song has been submitted, do something here
}

Hope that helps.