怎么登录mysql_你真的了解如何登录MySQL么?
昨天同事碰到一个问题,在MySQL上创建了一个用户,host设置为%,本地竟然无法登录。创建一个host为localhost的同名用户后,本地可以登录。感脚很怪异,下面我们重新分析产生这个问题的原因。
1. 现场重现
*root本地登录
Shell>./mysql -uroot
Welcome to the MySQL monitor. Commands end with ; or \g.
Your MySQL connectionid is 7Server version:5.5.17-debug-log Source distribution
Copyright (c)2000, 2011, Oracle and/or its affiliates. All rights reserved.
Oracle is a registered trademark of Oracle Corporation and/or its
affiliates. Other names may be trademarks of their respective
owners.
Type'help;' or '\h' for help. Type '\c' to clear the current input statement.
*切换到mysql库下
mysql>use mysql;
Reading table informationforcompletion of table and column names
You can turn off this feature to get a quicker startup with-A
Database changed
*创建用户
mysql> create user 'u1'@'%' identified by '1111111';
Query OK,0 rows affected (0.00sec)
mysql> selectuser,host from user;+------+-----------+
| user | host |
+------+-----------+
| repl | % |
| u1 | % |
| root | 127.0.0.1 |
| root | ::1 |
| | Ubuntu |
| root | Ubuntu |
| | localhost |
| repl | localhost |
| root | localhost |
+------+-----------+
9 rows in set (0.00 sec)
*本地登录——使用新用户+密码
Shell>./mysql -uu1 --protocol=tcp --port=13000 -p1111111
ERROR1045 (28000): Access denied for user 'u1'@'localhost' (using password: YES)
*本地登录——使用新用户+空密码登录
Shell>./mysql -uu1 --protocol=tcp --port=13000Welcome to the MySQL monitor. Commands end with ; or \g.
Your MySQL connectionid is 11Server version:5.5.17-debug-log Source distribution
Copyright (c)2000, 2011, Oracle and/or its affiliates. All rights reserved.
Oracle is a registered trademark of Oracle Corporation and/or its
affiliates. Other names may be trademarks of their respective
owners.
Type'help;' or '\h' for help. Type '\c' to clearthe current input statement.
mysql> selectcurrent_user;+--------------+
| current_user |
+--------------+
| @localhost |
+--------------+
1 row in set (0.00sec)
mysql> selectuser();+--------------+
| user() |
+--------------+
| u1@localhost |
+--------------+
1 row in set (0.00 sec)
*远程登录——使用新用户+密码
AAA@-ThinkPad:~/mysql-bin/bin$ ./mysql -uu1 --port=13000 -h192.168.1.103 -p1111111
Welcome to the MySQL monitor. Commands end with ; or \g.
Your MySQL connectionid is 13Server version:5.5.17-debug-log Source distribution
Copyright (c)2000, 2011, Oracle and/or its affiliates. All rights reserved.
Oracle is a registered trademark of Oracle Corporation and/or its
affiliates. Other names may be trademarks of their respective
owners.
Type'help;' or '\h' for help. Type '\c' to clearthe current input statement.
mysql> selectcurrent_user();+----------------+
| current_user() |
+----------------+
| u1@% |
+----------------+
1 row in set (0.01 sec)
*远程登录——使用新用户+空密码
AAA@-ThinkPad:~/mysql-bin/bin$ ./mysql -uu1 --port=13000 -h192.168.1.103ERROR1045 (28000): Access denied for user 'u1'@'-ThinkPad' (using password: NO)
tips:
user与current_user的区别:
user:client提供给server的用户名和密码
current_user:连接到server上的真正的用户名和密码
2. 现场分析
上面进行了本地登录和远程登录的实验,远程登录与预期一致,本地登录出现了问题。
本地登录使用密码登录时竟然失败,不使用时竟然成功,即使成功但是current_user竟然是‘@localhost’,也就是说根本不是u1用户登录成功,而是‘@localhost’登录成功。
我们在上面的user表的信息可以看到确实存在一个user为空,host为localhost的用户。这是个新库创建时自带的用户。
下面就有两个疑问了:
为嘛本地登录使用密码不能成功?为嘛不用密码登录成功但登录用户却是‘@localhost’呢?结合代码来看看真正的认证过程吧。
3. 代码分析
关于用户认证的代码基本都在sql/sql_acl.cc文件中,acl即access control list(访问控制列表),MySQL在系统启动时会调用acl_load,将mysql.user表中的信息全部加载到系统中,这里不涉及加载,只为寻找真相。为了方便用户自行跟踪,给一个身份验证的堆栈。
(gdb) bt
#0 compare_hostname (host=0x1d80580, hostname=0xb45e52 "localhost", ip=0x1dc8430 "127.0.0.1")
at/home/loushuai/src/mysql-server/mysql-5.5/sql/sql_acl.cc:2041#1 0x00000000005990dc in find_mpvio_user (mpvio=0x7fffe8165530) at /home/loushuai/src/mysql-server/mysql-5.5/sql/sql_acl.cc:8215#2 0x000000000059a651 in parse_client_handshake_packet (mpvio=0x7fffe8165530, buff=0x7fffe8165418, pkt_len=58)
at/home/loushuai/src/mysql-server/mysql-5.5/sql/sql_acl.cc:8746#3 0x000000000059adba in server_mpvio_read_packet (param=0x7fffe8165530, buf=0x7fffe8165418)
at/home/loushuai/src/mysql-server/mysql-5.5/sql/sql_acl.cc:8970#4 0x000000000059c338 in native_password_authenticate (vio=0x7fffe8165530, info=0x7fffe8165548)
at/home/loushuai/src/mysql-server/mysql-5.5/sql/sql_acl.cc:9547#5 0x000000000059b350 in do_auth_once (thd=0x1d5e470, auth_plugin_name=0xfd1280, mpvio=0x7fffe8165530)
at/home/loushuai/src/mysql-server/mysql-5.5/sql/sql_acl.cc:9133#6 0x000000000059b7fe in acl_authenticate (thd=0x1d5e470, connect_errors=0, com_change_user_pkt_len=0)
at/home/loushuai/src/mysql-server/mysql-5.5/sql/sql_acl.cc:9269#7 0x00000000006d4ed9 in check_connection (thd=0x1d5e470) at /home/loushuai/src/mysql-server/mysql-5.5/sql/sql_connect.cc:524#8 0x00000000006d5034 in login_connection (thd=0x1d5e470) at /home/loushuai/src/mysql-server/mysql-5.5/sql/sql_connect.cc:582#9 0x00000000006d5500 in thd_prepare_connection (thd=0x1d5e470) at /home/loushuai/src/mysql-server/mysql-5.5/sql/sql_connect.cc:716#10 0x00000000006d599a in do_handle_one_connection (thd_arg=0x1d5e470)
at/home/loushuai/src/mysql-server/mysql-5.5/sql/sql_connect.cc:782#11 0x00000000006d54c9 in handle_one_connection (arg=0x1d5e470) at /home/loushuai/src/mysql-server/mysql-5.5/sql/sql_connect.cc:708#12 0x00007ffff6ee3efc in start_thread () from /lib/x86_64-linux-gnu/libpthread.so.0#13 0x00007ffff6c1e59d in clone () from /lib/x86_64-linux-gnu/libc.so.6#14 0x0000000000000000 in ?? ()
堆栈上的#4 native_password_authenticate便是真正的身份认证函数,感兴趣的同学可以自己仔细看,我们具体看下find_mpvio_user函数:
static bool find_mpvio_user(MPVIO_EXT *mpvio)
{
DBUG_ENTER("find_mpvio_user");
DBUG_PRINT("info", ("entry: %s", mpvio->auth_info.user_name));
DBUG_ASSERT(mpvio->acl_user == 0);
mysql_mutex_lock(&acl_cache->lock);for (uint i=0; i < acl_users.elements; i++)
{
ACL_USER*acl_user_tmp= dynamic_element(&acl_users, i, ACL_USER*);if ((!acl_user_tmp->user ||
!strcmp(mpvio->auth_info.user_name, acl_user_tmp->user)) &&compare_hostname(&acl_user_tmp->host, mpvio->host, mpvio->ip))
{
mpvio->acl_user= acl_user_tmp->copy(mpvio->mem_root);if (acl_user_tmp->plugin.str == native_password_plugin_name.str ||acl_user_tmp->plugin.str ==old_password_plugin_name.str)
mpvio->acl_user_plugin= acl_user_tmp->plugin;elsemake_lex_string_root(mpvio->mem_root,&mpvio->acl_user_plugin,
acl_user_tmp->plugin.str,
acl_user_tmp->plugin.length, 0);break;
}
....
}
acl_users即为缓存mysql.user表中数据的动态数组。函数的基本逻辑是逐个遍历acl_users中的每个user,首先判断acl_user_tmp的用户名为空或者用户名和登录的用户名相同,然后比较host的值,由于‘@localhost’在列表中比‘u1@%’靠前,而且我们是本地登录,这就导致acl_user_tmp为空且host比较成功,就返回了‘@localhost’。这就是为什么current_user为‘@localhost’的原因。
那么为嘛加了密码就不行了呢?这是由于在获得了内存中的user后,会进行密码的验证。
static int native_password_authenticate(MYSQL_PLUGIN_VIO *vio,
MYSQL_SERVER_AUTH_INFO*info)
{
....if (pkt_len == 0) /*no password*/DBUG_RETURN(mpvio->acl_user->salt_len != 0 ?CR_ERROR : CR_OK);
info->password_used=PASSWORD_USED_YES;if (pkt_len ==SCRAMBLE_LENGTH)
{if (!mpvio->acl_user->salt_len)
DBUG_RETURN(CR_ERROR);
DBUG_RETURN(check_scramble(pkt, mpvio->scramble, mpvio->acl_user->salt) ?CR_ERROR : CR_OK);
}
....
}
如果输入了密码,那么就会进入pkt_len==SCRAMBLE_LEN的分支,而此时的用户为系统默认的用户'@localhost',所以mpvio->acl_user->salt_len必然为0,
故返回ERROR。
为嘛‘@localhost’在列表中比‘u1@%’靠前,这就涉及到acl_users的排序问题了,先给出一个堆栈:
#0 get_sort (count=1) at /home/loushuai/src/mysql-server/mysql-5.5/sql/sql_acl.cc:1266#1 0x0000000000580427 in acl_load (thd=0x1db0c50, tables=0x7fffffffc970)
at/home/loushuai/src/mysql-server/mysql-5.5/sql/sql_acl.cc:872#2 0x00000000005815bf in acl_reload (thd=0x1db0c50) at /home/loushuai/src/mysql-server/mysql-5.5/sql/sql_acl.cc:1174#3 0x000000000057f8b6 in acl_init (dont_read_acl_tables=false) at /home/loushuai/src/mysql-server/mysql-5.5/sql/sql_acl.cc:644#4 0x000000000055973b in mysqld_main (argc=11, argv=0x12d6f38) at /home/loushuai/src/mysql-server/mysql-5.5/sql/mysqld.cc:4551#5 0x0000000000552154 in main (argc=2, argv=0x7fffffffe0c8) at /home/loushuai/src/mysql-server/mysql-5.5/sql/main.cc:25
#1 现在执行到
acl_user.sort=get_sort(2,acl_user.host.hostname,acl_user.user);
用于计算当前acl_user的sort值,用于后面进行重新排序。
我们看下具体的排序函数:
static ulong get_sort(uintcount,...)
{
va_list args;
va_start(args,count);ulong sort=0;/*Should not use this function with more than 4 arguments for compare.*/DBUG_ASSERT(count<= 4);while (count--)
{char *start, *str= va_arg(args,char*);uint chars= 0;uint wild_pos= 0; /*first wildcard position*/
if ((start=str))
{for (; *str ; str++)
{if (*str == wild_prefix && str[1])
str++;//如果碰到%或者_,则记录wild_pos
else if (*str == wild_many || *str ==wild_one)
{
wild_pos= (uint) (str - start) + 1;break;
}
chars= 128; //Marker that chars existed
}
}
sort= (sort << 8) + (wild_pos ? min(wild_pos, 127) : chars);
}
va_end(args);returnsort;
}
从上面可以看出,根据host和user连个字段的值进行排序(host作为高位区分(sort<<8)),当存在%时,使用%的位置作为sort,否则就用128.
下面我们就具体看下一下三个用户的排序:
no host user1. localhost u12. localhost3. % u1
首先比较host,显然1,2 大于3,因为%用的是wild_pos,而1,2走chars=128.
然后比较1,2,显然1的优先级高,因为2为空。
故在acl_users中的顺序为1 2 3,也就是说,如果你创建了一个u1@localhost,那么使用u1就可以登录成功,注意,这时候就需要创建时的密码了。
终于找到原因了,往往细微的地方,我们往往拿捏不住,不断的发觉这些细微现象的真相,才能有所提高。
真相只有一个:-)
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