这是一个示例表来帮助说明我的问题:
mysql> select * from test;
+----+--------------+--------+
| id | type | siteid |
+----+--------------+--------+
| 1 | First Visit | 100 |
| 2 | Second Visit | 100 |
| 3 | First Visit | 300 |
| 4 | First Visit | 400 |
| 5 | Second Visit | 500 |
| 6 | Second Visit | 600 |
+----+--------------+--------+
我正在尝试将表自身连接起来,以将具有相同 siteid 值的行放在一起。这是我的尝试:
mysql> select * from test T1
-> LEFT OUTER JOIN test T2 on T1.siteid = T2.siteid and T1.id <> T2.id;
+----+--------------+--------+------+--------------+--------+
| id | type | siteid | id | type | siteid |
+----+--------------+--------+------+--------------+--------+
| 1 | First Visit | 100 | 2 | Second Visit | 100 |
| 2 | Second Visit | 100 | 1 | First Visit | 100 |
| 3 | First Visit | 300 | NULL | NULL | NULL |
| 4 | First Visit | 400 | NULL | NULL | NULL |
| 5 | Second Visit | 500 | NULL | NULL | NULL |
| 6 | Second Visit | 600 | NULL | NULL | NULL |
+----+--------------+--------+------+--------------+--------+
这基本上是我正在寻找的结果,除了第一 2 行。我想消除其中之一。所以,我尝试了以下方法:
mysql> select * from test T1
-> LEFT OUTER JOIN test T2 on T1.siteid = T2.siteid and T1.id <> T2.id
-> GROUP BY T1.siteid;
+----+--------------+--------+------+--------------+--------+
| id | type | siteid | id | type | siteid |
+----+--------------+--------+------+--------------+--------+
| 1 | First Visit | 100 | 2 | Second Visit | 100 |
| 3 | First Visit | 300 | NULL | NULL | NULL |
| 4 | First Visit | 400 | NULL | NULL | NULL |
| 5 | Second Visit | 500 | NULL | NULL | NULL |
| 6 | Second Visit | 600 | NULL | NULL | NULL |
+----+--------------+--------+------+--------------+--------+
这正是我正在寻找的输出。但是,我了解到这不是使用 GROUP BY 的标准方式,并且上述语句在 ORACLE 上失败,给了我一个
一般 SQL 错误。
ORA-00979: 不是 GROUP BY 表达式
任何人都可以就如何获得最后一张表以及与 ORACLE 一起使用的结果提供一些帮助吗?