Help Please (Select Problems)

[DEWD]

Hello,

I've been looking at Sitepoint for a few days now. Very nice site. Lots of good information. Keep up the good work!!!

I am trying my hand at PHP/mySQL for the first time. I followed Kevin Yank's tutorial and learned a great deal. I've got everything working but one item. I have a form on my site with a <SELECT> populated by PHP from a mySQL database. For some reason, the VALUE always ends up being the last record in the array, not the value that is selected. As far as I can tell I coded everything correctly. Can someone please look at my code below and tell me what I'm doing wrong. I greatly appreciate it. Thank you.

<?php

$connect = @mysql_connect("localhost", "username", "password");

if (! $connect) {
echo( "<p>Unable to connect</p>");
exit();
}

mysql_select_db("mydatabase");

$artists= mysql_query("SELECT ID_Artist, Name from Artist");

?>


<h2 align="center">CD Entry Screen</h2>

<p>Enter the title of the CD here and press SUBMIT to continue.</p>

<FORM ACTION="cd_entry_screen.php" METHOD=POST>

<P>Artist<br>
<select size="1" name="artistdd" tabindex="1">
<OPTION SELECTED VALUE ="">Select One
<OPTION VALUE="">----------
<?php
while ($artist = mysql_fetch_array($artists)) {
$artistdd = $artist[ID_Artist];
$aname = $artist[Name];
echo( "<OPTION VALUE='$artistdd'>$aname\n");
}
?>
</select></P>

<P>CD TitleBR>
<INPUT NAME="Title" size="35" tabindex="2">

<P>Yearbr>
<input type="text" name="year" size="5" tabindex="3">

<P>Genrebr>
<input type="text" name="genre" size="10" tabindex="4">

<P>Orginalbr>
<select name="original" size="1" tabindex="5">
<OPTION SELECTED VALUE="N">N
<OPTION VALUE="Y">Y
</select>
<P>

<INPUT TYPE=submit NAME="submitcd" VALUE="SUBMIT" tabindex="6">
</select></FORM>

<?php

if ("SUBMIT" == $submitcd) {

$sql = "INSERT INTO CD_Info SET " .
"Artist_ID*='$artistdd', Title='$Title', Year='$year', Genre='$genre',original='$original', Date_Added=CURDATE()";

if (mysql_query($sql)) {
echo( "<P>CD has been added.</P>");
} else {
echo( "<P>Error adding CD: " .
mysql_error() . "</P>");
}
}


?>

BTW - This is basically a CD Collection database. I am trying to get the correct artist in with the correct CD title by inserting the Artist_ID in the CD_Info table.

[DEWD]

DUH! I figured it out. I made the variable $artistdd = artist.id_artist. $artistdd is also the name of my control on my form. Of course this will make it not work. I change the variable to $artistid and it works. Thanks to anyone who looked at this. The correct and working version is copied below.

<select size="1" name="artistdd" tabindex="1">
<OPTION SELECTED VALUE ="">Select One
<OPTION VALUE="">----------
<?php
while ($artist = mysql_fetch_array($artists)) {
$artistid = $artist[ID_Artist];
$aname = $artist[Name];
echo( "<OPTION VALUE='$artistid'>$aname\n");
}