首先让我提一下，我已经看了很多建议的问题，但没有找到相关的答案。这就是我正在做的。
我连接到Amazon EC2实例。我可以用这个命令登录MySQL根目录:
mysql -u root -p
然后我用host %创建了一个新的用户帐单
CREATE USER 'bill'@'%' IDENTIFIED BY 'passpass';
授予用户bill的所有权限:
grant all privileges on *.* to 'bill'@'%' with grant option;
然后我退出root用户，尝试用bill登录:
mysql -u bill -p
输入正确的密码并得到以下错误:
错误1045(28000):用户“账单”@“localhost”(使用密码:YES)的访问被拒绝
当前回答
我遇到了同样的错误。失败的设置如下:
define("HOSTNAME", "localhost");
define("HOSTUSER", "van");
define("HOSTPASS", "helsing");
define("DBNAME", "crossbow");
$connection = mysqli_connect(HOSTNAME, HOSTUSER, HOSTPASS);
下面编辑的设置是让它工作的一个。注意到区别了吗?
define('HOSTNAME', 'localhost');
define('HOSTUSER', 'van');
define('HOSTPASS', 'helsing');
define('DBNAME', 'crossbow');
$connection = mysqli_connect(HOSTNAME, HOSTUSER, HOSTPASS);
区别在于双引号。与Java相比，它们在PHP中似乎相当重要，它们在转义字符、设置url以及现在向函数传递参数时都有影响。它们更漂亮(我知道)，但总是尽可能多地使用单引号，如果必要的话，可以在其中嵌套双引号。
当我在Linux机器上而不是Windows环境上测试应用程序时，出现了这个错误。
其他回答
这是以下两者之间的区别:
CREATE USER 'bill'@'%' IDENTIFIED BY 'passpass';
and
CREATE USER 'bill'@'localhost' IDENTIFIED BY 'passpass';
检查:
mysql> select user,host from mysql.user;
+---------------+----------------------------+
| user | host |
+---------------+----------------------------+
| bill | % | <=== created by first
| root | 127.0.0.1 |
| root | ::1 |
| root | localhost |
| bill | localhost | <=== created by second
+---------------+----------------------------+
命令
mysql -u bill -p
隐式访问'bill'@'localhost'，而不是'bill'@'%'。
“bill”@“localhost”没有权限
你会得到错误:
ERROR 1045 (28000): Access denied for user 'bill'@'localhost' (using password: YES)
解决问题:
CREATE USER 'bill'@'localhost' IDENTIFIED BY 'passpass';
grant all privileges on . to 'bill'@'localhost' with grant option;
I discovered yet another case that appears on the surface to be an edge case; I can export to the file system, via SELECT INTO .. OUTFILE as root, but not as regular user. While this may be a matter of permissions, I've looked at that, and see nothing especially obvious. All I can say is that executing the query as a regular user who has all permissions on the data base in question returns the access denied error that led me to this topic. When I found the transcript of a successful use of SELECT INTO … OUTFILE in an old project, I noticed that I was logged in as root. Sure enough, when I logged in as root, the query ran as expected.
我希望你在mysql中删除debian-sys- maintenance用户没有造成更大的损害
让你的mysql守护进程正常运行。启动mysql客户端，如下所示
mysql -u debian-sys-maint -p
在另一个终端，cat /etc/mysql/debian.cnf文件。该文件包含密码;当提示输入密码时，粘贴该密码。
http://ubuntuforums.org/showthread.php?t=1836919
这可能只适用于少数人，但事实就是这样。不要用感叹号!在你的密码里。
我做到了，并使用MariaDB得到了上述错误。当我把它简化成数字和字母时，它是可行的。其他字符，如@和$工作得很好-我在同一实例的不同用户中使用了这些字符。
这个地址的第五次回复让我找到了解决办法。
更新:在v8.0.15(可能是这个版本)，PASSWORD()函数不起作用。
你必须:
确保你已经停止MySQL。 使用特权绕过安全模式运行服务器:sudo mysqld_safe——skip-grant-tables 登录:mysql -u root 更新mysql。SET authentication_string=null WHERE user ='root'; mysql> FLUSH PRIVILEGES; mysql >退出; 再次登录:mysql -u root mysql> ALTER USER 'root'@'localhost' IDENTIFIED WITH caching_sha2_password BY 'yourpasswd'
