On Sun, October 1, 2006 3:48 pm, Deckard wrote:
> I have this code:
> -------------------------------------------------------
> // use mysql database
> mysql_select_db('mysql') or die('Cannot select MySQL database.' .
> mysql_error());
>
> // insert new user in mysql user table
> $sqlInsertUser = "INSERT INTO user (Host, User , Password ,
> Select_priv, Insert_priv, Update_priv, Delete_priv) VALUES
> ('$hostname',
> '$mysql_username', '$mysql_password', 'N', 'N', 'N', 'N')";
>
> if (mysql_query($sqlInsertUser, $conn))
> $messagesqlInsertUser = 'Success...';
> else
> $messagesqlInsertUser = 'Error: ' . mysql_error();
> $result = mysql_query($sqlInsertUser);
> -------------------------------------------------------
>
> that gives me the error:
> ---------------------------------------------------
> Error: Duplicate entry 'localhost-gamito' for key 1
> ---------------------------------------------------
>
> although the pair 'localhost-gamito' is not duplicated and i can be
> assured os this, because it works perfectly well in the MySQL prompt.
Once you have put the user 'gamito' in to that table, with the host
'localhost', you cannot add a SECOND user 'gamito' for 'localhost'
because there is a UNIQUE KEY constraint on that 2-field combination.
You could add '%', 'gamito' or you could add 'localhost', 'gamito2'
but not another 'localhost', 'gamito'
I'm also going to wonder aloud why you are doing this when there are a
plethora of tools available to manipulate the MySQL user table...
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