You probably have an anonymous user ''@'localhost' or ''@'127.0.0.1'.
As per the manual:
When multiple matches are possible, the server must determine which of
them to use. It resolves this issue as follows: (...)
- When a client attempts to connect, the server looks through the rows [of table mysql.user] in sorted order.
- The server uses the first row that matches the client host name and user name.
(...)
The server uses sorting rules that order rows with the most-specific Host values first.
Literal host names [such as 'localhost'] and IP addresses are the most specific.
Hence, such an anonymous user would "mask" any other user like '[any_username]'@'%' when connecting from localhost.
'bill'@'localhost' does match 'bill'@'%', but would match (e.g.) ''@'localhost' beforehands.
The recommended solution is to drop this anonymous user (this is usually a good thing to do anyways).
Below edits are mostly irrelevant to the main question. These are only meant to answer some questions raised in other comments within this thread.
Edit 1
Authenticating as 'bill'@'%' through a socket.
root@myhost:/home/mysql-5.5.16-linux2.6-x86_64# ./mysql -ubill -ppass --socket=/tmp/mysql-5.5.sock
Welcome to the MySQL monitor (...)
mysql> SELECT user, host FROM mysql.user;
+------+-----------+
| user | host |
+------+-----------+
| bill | % |
| root | 127.0.0.1 |
| root | ::1 |
| root | localhost |
+------+-----------+
4 rows in set (0.00 sec)
mysql> SELECT USER(), CURRENT_USER();
+----------------+----------------+
| USER() | CURRENT_USER() |
+----------------+----------------+
| bill@localhost | bill@% |
+----------------+----------------+
1 row in set (0.02 sec)
mysql> SHOW VARIABLES LIKE 'skip_networking';
+-----------------+-------+
| Variable_name | Value |
+-----------------+-------+
| skip_networking | ON |
+-----------------+-------+
1 row in set (0.00 sec)
Edit 2
Exact same setup, except I re-activated networking, and I now create an anonymous user ''@'localhost'.
root@myhost:/home/mysql-5.5.16-linux2.6-x86_64# ./mysql
Welcome to the MySQL monitor (...)
mysql> CREATE USER ''@'localhost' IDENTIFIED BY 'anotherpass';
Query OK, 0 rows affected (0.00 sec)
mysql> Bye
root@myhost:/home/mysql-5.5.16-linux2.6-x86_64# ./mysql -ubill -ppass \
--socket=/tmp/mysql-5.5.sock
ERROR 1045 (28000): Access denied for user 'bill'@'localhost' (using password: YES)
root@myhost:/home/mysql-5.5.16-linux2.6-x86_64# ./mysql -ubill -ppass \
-h127.0.0.1 --protocol=TCP
ERROR 1045 (28000): Access denied for user 'bill'@'localhost' (using password: YES)
root@myhost:/home/mysql-5.5.16-linux2.6-x86_64# ./mysql -ubill -ppass \
-hlocalhost --protocol=TCP
ERROR 1045 (28000): Access denied for user 'bill'@'localhost' (using password: YES)
Edit 3
Same situation as in edit 2, now providing the anonymous user's password.
root@myhost:/home/mysql-5.5.16-linux2.6-x86_64# ./mysql -ubill -panotherpass -hlocalhost
Welcome to the MySQL monitor (...)
mysql> SELECT USER(), CURRENT_USER();
+----------------+----------------+
| USER() | CURRENT_USER() |
+----------------+----------------+
| bill@localhost | @localhost |
+----------------+----------------+
1 row in set (0.01 sec)
Conclusion 1, from edit 1: One can authenticate as 'bill'@'%'through a socket.
Conclusion 2, from edit 2: Whether one connects through TCP or through a socket has no impact on the authentication process (except one cannot connect as anyone else but 'something'@'localhost' through a socket, obviously).
Conclusion 3, from edit 3: Although I specified -ubill, I have been granted access as an anonymous user. This is because of the "sorting rules" advised above. Notice that in most default installations, a no-password, anonymous user exists (and should be secured/removed).
Best Answer
Try to login via the terminal using the following command:
It will then prompt for your password. If this fails, then definitely the username or password is incorrect. If this works, then your database's password needs to be enclosed in quotes: