MYSQL：加入包含存储函数的 2 个选择语句时出错

Posted 2023-03-27

【中文标题】MYSQL：加入包含存储函数的 2 个选择语句时出错

【英文标题】：MYSQL: Error joining 2 select statements that include a stored function

【发布时间】：2017-01-30 15:33:30

【问题描述】：

下午好，

我创建了一个存储函数来模拟 mysql 中“SHOW GRANTS FOR”命令的行为。脚本定义如下：

CREATE FUNCTION get_grants(p_user text, p_host text, p_type text, p_obj text) 
   RETURNS text 
   NOT DETERMINISTIC 
   READS SQL DATA
BEGIN
...
END;

如果我在 select 语句中使用该函数，它会正确执行。例如：

SELECT User as USER, Host as HOST, 
      'GLOBAL' as PRIV_TYPE, NULL as DATABASE_NAME,   
       mydatabase.get_grants(User, Host, 'GLOBAL', '') as PRIVILEGES
  FROM mysql.user

但是，当我使用 UNION 或 UNION ALL 语句来加入 2 个调用具有不同参数值的函数的选择语句时，如下所示：

SELECT User as USER, Host as HOST, 
       'GLOBAL' as PRIV_TYPE, NULL as DATABASE_NAME,   
       mydatabase.get_grants(User, Host, 'GLOBAL', '') as PRIVILEGES
  FROM mysql.user
UNION ALL
SELECT User as USER, Host as HOST,
       'DB' as PRIV_TYPE, Db as DATABASE_NAME,
       mydatabase.get_grants(User, Host, 'DB', Db) as PRIVILEGES
  FROM mysql.db
 ORDER BY USER, HOST, DATABASE_NAME IS NULL, DATABASE_NAME;

我收到以下错误：#1222 - 使用的 SELECT 语句的列数不同。

谁能解释这里发生了什么以及如何解决这个问题？

提前致谢。

【参考方案1】：

我认为是 UNION ALL 导致了这个问题。 通常，如果您使用 UNION ALL，列数是否不同并不重要，但 UNION 必须相等。 因此，请尝试使用 UNION ALL 并将 NULL 删除为 DATABASE_NAME 或仅使用 UNION 并将 NULL 保留为 DATABASE_NAME

SELECT User as USER, Host as HOST, 
       'GLOBAL' as PRIV_TYPE, NULL as DATABASE_NAME,   
       mydatabase.get_grants(User, Host, 'GLOBAL', '') as PRIVILEGES
  FROM mysql.user
UNION
SELECT User as USER, Host as HOST,
       'DB' as PRIV_TYPE, Db as DATABASE_NAME,
       mydatabase.get_grants(User, Host, 'DB', Db) as PRIVILEGES
  FROM mysql.db
 ORDER BY USER, HOST, DATABASE_NAME IS NULL, DATABASE_NAME;

或

SELECT User as USER, Host as HOST, 
       'GLOBAL' as PRIV_TYPE,   
       mydatabase.get_grants(User, Host, 'GLOBAL', '') as PRIVILEGES
  FROM mysql.user
UNION ALL
SELECT User as USER, Host as HOST,
       'DB' as PRIV_TYPE, Db as DATABASE_NAME,
       mydatabase.get_grants(User, Host, 'DB', Db) as PRIVILEGES
  FROM mysql.db
 ORDER BY USER, HOST, DATABASE_NAME IS NULL, DATABASE_NAME;

【参考方案2】：

在几乎让我的大脑试图找到问题之后，事实证明实际问题与 MySQL 提供的错误消息没有任何关系（至少与文档中提供的有关此错误消息的描述无关） .

在这种情况下，错误出现在存储函数的实现中。在 SELECT...INTO 命令中，我在 INTO 子句中的两个变量之间缺少一个“，”。选择的列数与 INTO 子句中的变量数相对应，问题已解决，我设法在上述查询中使用该函数，没有任何问题。

作为参考，有问题的查询是：

SELECT Select_priv, Insert_priv, Update_priv, Delete_priv,
       Create_priv, Alter_priv, Index_priv, Drop_priv,
       Create_tmp_table_priv, Create_view_priv, Show_view_priv,
       Create_routine_priv, Alter_Routine_priv, Execute_priv, 
       Event_priv, Trigger_priv, Grant_priv, Lock_tables_priv, 
       References_priv
  INTO v_select_priv, v_insert_priv, v_update_priv, v_delete_priv,
       v_create_priv, v_alter_priv, v_index_priv, v_drop_priv,
       v_tmptables_priv, v_createview_priv, v_showview_priv, 
       v_createroutine_priv, v_alterroutine_priv, v_execute_priv,
       v_event_priv, v_trigger_priv, v_grant_priv, v_locktables_priv,
       v_references_priv
  FROM mysql.db
 WHERE User = p_user
   AND Host = p_host
   AND Db = p_obj;