如果我有多个对象，任何人都可以映射到多个其他对象。
A -> B, C, D, E
B -> A, E, F, C
C -> B
等
如何将这些实体存储在MySql中并使用JPA + Hibernate来管理它们。例如，如果我搜索“A”，它将返回“B，C，D，E”，如果我搜索“C”，它将返回“B”等。
编辑：
按照建议，我做了以下事情：
[test]> describe relation;
+----------+---------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+----------+---------+------+-----+---------+----------------+
| id | int(11) | NO | PRI | NULL | auto_increment |
| id_start | int(11) | YES | | NULL | |
| id_end | int(11) | YES | MUL | NULL | |
+----------+---------+------+-----+---------+----------------+
3 rows in set (0.00 sec)
MariaDB [test]> describe entity;
+-------+------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+-------+------------+------+-----+---------+----------------+
| id | int(11) | NO | PRI | NULL | auto_increment |
| name | varchar(5) | YES | | NULL | |
+-------+------------+------+-----+---------+----------------+
2 rows in set (0.00 sec)
MariaDB [test]> select * from entity;
+----+------+
| id | name |
+----+------+
| 1 | A |
| 2 | B |
| 3 | C |
| 4 | D |
| 5 | E |
| 6 | F |
+----+------+
6 rows in set (0.00 sec)
MariaDB [test]> select * from relation;
+----+----------+--------+
| id | id_start | id_end |
+----+----------+--------+
| 1 | 3 | 1 |
| 2 | 2 | 1 |
| 3 | 4 | 1 |
| 4 | 5 | 1 |
| 5 | 1 | 2 |
| 6 | 5 | 2 |
| 7 | 6 | 2 |
| 8 | 3 | 2 |
| 9 | 2 | 3 |
+----+----------+--------+
9 rows in set (0.00 sec)
[test]> select E.name, R.id_start from relation R join entity E on E.id=R.id_end where E.name="A";
+------+----------+
| name | id_start |
+------+----------+
| A | 3 |
| A | 2 |
| A | 4 |
| A | 5 |
+------+----------+
问题：
此查询返回相关的“A”ID，这是正确的，但实际上不是B，C，D，E。如何从实体表中获取实际的相关名称？是数据库设计还是SQL语句错误？