如何在变量中使用GRANT？

我在GRANT和变量一起在MySql中有一些麻烦。如何在变量中使用GRANT？

SET @username := 'user123', @pass := 'pass123'; 

GRANT USAGE ON *.* TO @[email protected]'%' IDENTIFIED BY @pass; 
GRANT INSERT (header1, header2, headern) ON `data` TO @[email protected]'%'; 
GRANT SELECT (header1, header2) ON `data2` TO @[email protected]'%';

我希望把用户名和密码的变量在脚本的开头，再后来在GRANT使用它们

所以不是这样的：

GRANT USAGE ON *.* TO 'user123'@'%' IDENTIFIED BY 'pass123';

我会喜欢使用这样的东西：

GRANT USAGE ON *.* TO @[email protected]'%' IDENTIFIED BY pass;

我真的很感激，如果有人能告诉我适当的声明秒。谢谢你的声音！

来源

2013-02-12 Pho3nixHun

应该不是被'IDENTIFIED BY @ pass'，你错过了'@' – asifsid88 2013-02-12 10:37:31

谢谢，我纠正它，但这不是解决的主要问题。 :( – Pho3nixHun 2013-02-12 10:40:10

变量只能用于SQL允许表达式的地方，GRANT语句中没有表达式，你可以用动态SQL做。 – Barmar 2013-02-12 10:53:35

SET @object = '*.*'; 
SET @user = '''user1''@''localhost'''; 

SET @query = CONCAT('GRANT UPDATE ON ', @object, ' TO ', @user); 
PREPARE stmt FROM @query; 
EXECUTE stmt; 
DEALLOCATE PREPARE stmt;

DROP PROCEDURE IF EXISTS `test`.`spTest`$$ 

CREATE DEFINER=`root`@`localhost` PROCEDURE `spTest`(varLogin char(16), varPassword char(64)) 
BEGIN 
    DECLARE varPasswordHashed CHAR(41); 
    SELECT PASSWORD(varPassword) INTO varPasswordHashed; 

    # Any of the following 3 lines will cause the creation to fail 
    CREATE USER [email protected]'localhost' IDENTIFIED BY varPassword; 
    GRANT USAGE ON test.* TO [email protected]'localhost' IDENTIFIED BY varPassword; 
    GRANT USAGE ON test.* TO [email protected]'localhost' IDENTIFIED BY PASSWORD varPasswordHashed; 

    ## The following 3 lines won't cause any problem at create time 
    CREATE USER [email protected]'localhost' IDENTIFIED BY 'AnyPassordString'; 
    GRANT USAGE ON test.* TO [email protected]'localhost' IDENTIFIED BY 'AnyPassordString'; 
    GRANT USAGE ON test.* TO [email protected]'localhost' IDENTIFIED BY PASSWORD 'AnyPassordString'; 
END$$ 

DELIMITER;

来源

2013-02-12 10:56:16 user2001117

非常感谢！它就像一个魅力！:) – Pho3nixHun 2013-02-12 12:23:26

如何在变量中使用GRANT？

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