Hi,
I have some problems with making this code work! I'm trying to upload an image to my server with a link in my database. Everything but the image work!

Please help!

Database:

ID, Name, EMail, description, picture_name

Code:

<!-- newauthor.php -->
<HTML>
<HEAD>
<TITLE> Add New Author </TITLE>
</HEAD>
<BODY>
<?php

if ($submit): // A new author has been entered
// using the form below.

$dbcnx = @mysql_connect("localhost", "****", "****");
mysql_select_db("****");

$sql = "INSERT INTO Authors SET " .

"Name='$name', " .
"description='$description', ".
"picture_name='$picture_name', ".
"EMail='$email'";
exec("cp $picture /public_html/joke/images/$picture_name");

if (mysql_query($sql)) {
echo("<P>New author added</P>");

echo "Name: $name<br>\n";
echo "EMail: $email<br>\n";
echo "description: $description<br>\n";
echo "temp file: $picture<br>\n";
echo "file name: $picture_name<br>\n";
echo "file size: $picture_size<br>\n";
echo "file type: $picture_type<br>\n";
echo "<br>\n";
echo "<img src=images/$picture_name><br>\n";

} else {
echo("<P>Error adding new author: " .
mysql_error() . "</P>");
}

?>

<P><A HREF="<?php echo($PHP_SELF); ?>">Add another Author</A></P>
<P><A HREF="authors.php">Return to Authors list</A></P>

<?php
else: // Allow the user to enter a new author
?>

<FORM ACTION="<?php echo($PHP_SELF); ?>" METHOD=POST>
<P>Enter the new authorBR>
Name: <INPUT TYPE="TEXT" NAME="name" SIZE=20 MAXLENGTH=100><BR>
eMail: <INPUT TYPE="TEXT" NAME="email" SIZE=20 MAXLENGTH=100><BR>
Description: <INPUT TYPE="TEXT" NAME="description" SIZE=20 MAXLENGTH=100><BR>
Picture: <INPUT TYPE="FILE" NAME="picture" SIZE=20 MAXLENGTH=100><BR>
<INPUT TYPE="SUBMIT" NAME="submit" VALUE="SUBMIT"></P>
</FORM>

<?php endif; ?>

</BODY>
</HTML>