> Torsten
>
> I managed to work through it and this is the code:
>
> $NewUserID= $_POST['TXT_UserID'];
> $NewUserPassword= $_POST['TXT_UserPassword'];
> $NewUserComments= $_POST['TXT_Comments'];
> $NewUserFurtherComments=$_POST['TXT_FurtherComments'];
> $sql = "INSERT INTO RegisteredMembers
> (UserID,UserPassword,Comments,FurtherComments)
>
> VALUES('".$NewUserID."','".$NewUserPassword."','".$NewUserComments
> ."','".$Ne
> wUserFurtherComments."')";
> $query = mysql_query($sql);
> $mysql_result = mysql_query ($sql, $Connection) or die ("Invalid
> Query - " . mysql_error());
>
> echo "<pre>Username:\t\t$NewUserID</pre><br>";
> echo "<pre>Password:\t\t$NewUserPassword</pre><br>";
> echo "<pre>Comments:\t\t$NewUserComments</pre><br>";
> echo "<pre>Further Comments:\t$NewUserFurtherComments</pre>";
>
> mysql_close ($Connection);
>
> However. I still get an error although the input is actually
> commited to the
> database:
>
> Invalid Query - Duplicate entry 'SherylJo' for key 1
>
> I don't follow this because although I have set that particular
> field up as
> a primary key this value is not duplicated... Strange.
Hi Michael,
please always anser to the list (Answer All). Could you post your table
structure?
Regards, Torsten
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