PostgreSQL：如何在线性“祖先后代”中找到最后一个后代。关系 [英] PostgreSQL: How to find the last descendant in a linear "ancestor-descendant" relationship

问题描述

我具有以下数据库结构：

I have the following DB structure:

RELATIONSHIP_TABLE
- id << primary key
- id_ancestor << foreign key to the same table
- id_entry << foreign key to "ENTRY_TABLE"

ENTRY_TABLE
- id
- name
...

表 RELATIONSHIP_TABLE 中的层次结构是线性的。这意味着一条记录最多可以是另一条记录的祖先。示例：

The hierarchy in table "RELATIONSHIP_TABLE" is linear. That means a record can be at most ancestor of one other record. Examples:

1. record1
2. record2 <- record3 <- record4
3. record5 <- record7 <- record9 <- record12

特定层次结构中的每个记录都具有相同的 id_entry 。现在，我想找到带有特定 id_entry 的最后一个后代。以下示例的结果将是：

Every record within a particular hierarchy has the same "id_entry". Now, I would like to find the last descendant with a specific "id_entry". The result of the examples below would be:

1. record1
2. record4
3. record12

有人知道解决方案吗？

Does anyone know a solution?

预先感谢：）

QStormDS

推荐答案

SELECT * 
FROM relationship_table rt
WHERE rt.id_entry = 42
AND NOT EXISTS (
   SELECT * FROM relationship_table nx
   WHERE nx.id_entry = 42      -- you can possibly omit this clause  
   AND nx.id_ancestor = rt.id  -- No children poining to rt ...
   )
 ;

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