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mysql forgets who is logged in: command denied to user ''@'%'

开发者 https://www.devze.com 2023-03-16 13:28 出处:网络
Running show grants; indicates that I am logged in as a user with all privileges on a database. Running show table status; results in an error.And the error does not show the username I am logged in

Running show grants; indicates that I am logged in as a user with all privileges on a database.

Running show table status; results in an error. And the error does not show the username I am logged in as!

It's as if, for this command, mysql forgets who I am. Other select statements work fine. Can anyone explain this? How to fix? Thanks.

Welcome to the MySQL monitor.  Commands end with ; or \g.
Your MySQL connection id is 2
Server version: 5.5.13-log Source distribution

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Type 'help;' or '\h' for help. Type '\c' to clear the current input statement.

mysql> show grants;
+---------------------------------------------------------------------------------------------------------------------+
| Grants for php@localhost                                                                                            |
+-----------------------------------------开发者_StackOverflow社区----------------------------------------------------------------------------+
| GRANT ALL PRIVILEGES ON *.* TO 'php'@'localhost' IDENTIFIED BY PASSWORD '*8F5FF90079BC601F8EA7C148475658E65A0C029D' |
| GRANT ALL PRIVILEGES ON `sunflower_work`.* TO 'php'@'localhost'                                                     |
| GRANT ALL PRIVILEGES ON `news_demo`.* TO 'php'@'localhost'                                                          |
| GRANT ALL PRIVILEGES ON `news_base`.* TO 'php'@'localhost'                                                          |
+---------------------------------------------------------------------------------------------------------------------+
4 rows in set (0.00 sec)

mysql> show table status from sunflower_work;
ERROR 1143 (42000): SELECT command denied to user ''@'%' for column 'uid' in table 'users'
mysql>

update... as suggested by Tomalak, I deleted the user and recreated with fuller privileges and no password. Still the problem persists. Now it looks like this:

mysql> show grants;
+--------------------------------------------------+
| Grants for php@localhost                         |
+--------------------------------------------------+
| GRANT ALL PRIVILEGES ON *.* TO 'php'@'localhost' |
+--------------------------------------------------+
1 row in set (0.00 sec)

mysql> show table status;
ERROR 1143 (42000): SELECT command denied to user ''@'%' for column 'uid' in table 'users'
mysql> 


The issue is probably that you have VIEWS in your database. The views are probably created with specific rights.

As you can tell by your error message, it complains about a different user than the one you are logged in is. This is because for a view you can specify how to determine what rights the view has to look at data.

When you go to your database, try typing:

SHOW FULL TABLES IN sunflower_work WHERE TABLE_TYPE NOT LIKE '%table%';

Then you may wish to look into the rights of the specific views that are there.


The answers here helped me with my specific problem. Many thanks! A view was the culprit as described above.

I got into trouble because the database in question was created from a backup of a remote database which had different users. The 'broken' view was 'defined' by a user I didn't have locally. Even root was unable to run the crashing query.

Changed the view's 'DEFINER' to a valid local user and the problem was solved!

ALTER 
DEFINER = 'a_valid_user'@'localhost' 
VIEW my_view
AS 
SELECT ..... 

Check out ALTER VIEW documentation for MySQL 5.5

Many thanks again!


Take schema backup before proceeding.

If you just imported a dump file in mysql, delete that import and related schema and start again.

open the dump file in a text editor and delete all lines with the following content /*! ~~~~ DEFINER='root' @'%' SQL SECURITY DEFINER */

~ Represents a random number generated by workbench during export

This solution is a quick fix and intended for development environments only.

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