我有以下疑问:
MySQL [Database]> show grants for "user"@"%";
+-----------------------------------------------------------------+
| Grants for user@% |
+-----------------------------------------------------------------+
| GRANT ALL PRIVILEGES ON *.* TO 'user'@'%' |
+-----------------------------------------------------------------+
1 row in set (0.00 sec)
现在我要做的是修改Grants for user@%
的标题,但是下面的语句抛出了一个错误:
show grants for "user"@"%" AS TEST;
show grants AS TEST for "user"@"%";
select * from (show grants AS Test for "user"@"%") as SUB;
我想以某种方式得到以下结果:
MySQL [Database]> show grants for "user"@"%" AS TEST;
+-----------------------------------------------------------------+
| TEST |
+-----------------------------------------------------------------+
| GRANT ALL PRIVILEGES ON *.* TO 'user'@'%' |
+-----------------------------------------------------------------+
1 row in set (0.00 sec)
我想更改标题的原因是,因为我使用的PHP框架将数据库查询转换为对象,而Grants for user@%
将是对象变量/字段,我需要调用它,这非常不方便。
发布于 2019-04-24 03:17:17
如果您不喜欢SHOW GRANTS的结果标题,可以从INFORMATION_SCHEMA中的表中拼凑信息:
我将把编写SQL查询作为练习来完成此任务。
https://stackoverflow.com/questions/55817932
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