Jpa 使用HQL/Java标准的分层数据

我有一张桌子：

+----+-------------------+------+-------------------+-------+
| id | exType            | exId | depType           | depId |
+----+-------------------+------+-------------------+-------+
| 1  | Ceo               | 1    | Board Of Director | 1     |
| 2  | Board Of Director | 1    | Manager           | 4     |
| 3  | Board Of Director | 1    | Manager           | 5     |
| 4  | Manager           | 4    | TeamLead          | 1     |
| 5  | Manager           | 5    | TeamLead          | 2     |
| 6  | TeamLead          | 1    | Member            | 7     |
| 7  | TeamLead          | 1    | Member            | 8     |
| 8  | TeamLead          | 1    | Member            | 9     |
| 9  | TeamLead          | 1    | Member            | 10    |
| 10 | TeamLead          | 2    | Member            | 19    |
+----+-------------------+------+-------------------+-------+

当我以

exttype

asManager和

exId

as

4

的形式输入时，我需要所有成员的：

Member  7
Member  8
Member  9   
Member  10

Member  7
Member  8
Member  9
Member  10

或者，当我以

exType

作为TeamLead和

exId

作为

1

输入时，我想要所有成员的：

Member  7
Member  8
Member  9   
Member  10

Member  7
Member  8
Member  9
Member  10

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在这里，我只得到一个水平，不完全或请您详细说明以上