问题描述
首先让我提一下，我已经浏览了许多建议的问题，但没有找到相关的答案.这就是我正在做的事情.
我已连接到我的 Amazon EC2 实例.我可以使用以下命令使用 MySQL root 登录:
mysql -u root -p然后我用主机 % 创建了一个新的用户账单
CREATE USER 'bill'@'%' IDENTIFIED BY 'passpass';授予用户 bill 的所有权限:
使用授权选项将*.*的所有权限授予'bill'@'%'；然后我从 root 用户退出并尝试使用 bill 登录:
mysql -u bill -p输入正确的密码，但出现此错误:
<块引用>ERROR 1045 (28000): 用户 'bill'@'localhost' 访问被拒绝(使用密码:YES)
解决方案您可能有一个匿名用户
''@'localhost'或''@'127.0.0.1'.根据手册:
<块引用>当可能有多个匹配项时，服务器必须确定哪个匹配项他们使用.它解决了这个问题如下:(...)
- 当客户端尝试连接时，服务器按排序顺序查看[表 mysql.user] 的行.
- 服务器使用与客户端主机名和用户名匹配的第一行.
(...)服务器使用排序规则，将具有最具体的 Host 值的行排在第一位.文字主机名[例如'localhost']和IP地址是最具体的.
因此，当从 localhost 连接时，这样的匿名用户会屏蔽"任何其他用户，例如 '[any_username]'@'%'.
'bill'@'localhost' 确实匹配 'bill'@'%'，但会匹配(例如)''@'localhost' 事先.
推荐的解决方案是删除此匿名用户(无论如何，这通常是一件好事).
<小时>下面的编辑大多与主要问题无关.这些只是为了回答在此线程中的其他评论中提出的一些问题.
编辑 1
通过套接字验证为 'bill'@'%'.
编辑 2
完全相同的设置，除了我重新激活网络，我现在创建一个匿名用户 ''@'localhost'.
编辑 3
与编辑 2 中的情况相同，现在提供匿名用户的密码.
<前>root@myhost:/home/mysql-5.5.16-linux2.6-x86_64# ./mysql -ubill -panotherpass -hlocalhost欢迎使用 MySQL 监视器 (...)mysql> SELECT USER(), CURRENT_USER();+----------------+----------------+|用户() |CURRENT_USER() |+----------------+----------------+|bill@localhost |@本地主机 |+----------------+----------------+1 行(0.01 秒)结论 1，来自编辑 1:可以通过套接字验证为 'bill'@'%'.
结论 2，来自编辑 2:无论是通过 TCP 连接还是通过套接字连接对身份验证过程没有影响(除了不能作为其他任何人连接，但 'something'@'localhost' 通过显然是一个套接字).
结论 3，来自编辑 3:虽然我指定了 -ubill，但我已被授予匿名用户访问权限.这是因为上面建议的排序规则".请注意，在大多数默认安装中，存在无密码的匿名用户(并且应该被固定/移除).
First let me mention that I've gone through many suggested questions and found no relevent answer. Here is what I'm doing.
I'm connected to my Amazon EC2 instance. I can login with MySQL root with this command:
mysql -u root -p
Then I created a new user bill with host %
CREATE USER 'bill'@'%' IDENTIFIED BY 'passpass';
Granted all the privileges to user bill:
grant all privileges on *.* to 'bill'@'%' with grant option;
Then I exit from root user and try to login with bill:
mysql -u bill -p
entered the correct password and got this error:
ERROR 1045 (28000): Access denied for user 'bill'@'localhost' (using password: YES)
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).
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