计算祖先数组后代数的最佳方法？

Question

表tree是一个示例表，其中包含PostgreSQL 8.3+中的祖先数组：

----+-----------
 id | ancestors 
----+-----------
  1 | {}
  2 | {1}
  3 | {1,2}
  4 | {1}
  5 | {1,2}
  6 | {1,2}
  7 | {1,4}
  8 | {1}
  9 | {1,2,3}
 10 | {1,2,5}

为了获得后代的每个id计数数，我可以这样做：

SELECT 1 AS id, COUNT(id) AS descendant_count FROM tree WHERE 1 = ANY(ancestors)
  UNION
SELECT 2 AS id, COUNT(id) AS descendant_count FROM tree WHERE 2 = ANY(ancestors)
  UNION
SELECT 3 AS id, COUNT(id) AS descendant_count FROM tree WHERE 3 = ANY(ancestors)
  UNION
SELECT 4 AS id, COUNT(id) AS descendant_count FROM tree WHERE 4 = ANY(ancestors)
  UNION
SELECT 5 AS id, COUNT(id) AS descendant_count FROM tree WHERE 5 = ANY(ancestors)
  UNION
SELECT 6 AS id, COUNT(id) AS descendant_count FROM tree WHERE 6 = ANY(ancestors)
  UNION
SELECT 7 AS id, COUNT(id) AS descendant_count FROM tree WHERE 7 = ANY(ancestors)
  UNION
SELECT 8 AS id, COUNT(id) AS descendant_count FROM tree WHERE 8 = ANY(ancestors)
  UNION
SELECT 9 AS id, COUNT(id) AS descendant_count FROM tree WHERE 9 = ANY(ancestors)
  UNION
SELECT 10 AS id, COUNT(id) AS descendant_count FROM tree WHERE 10 = ANY(ancestors)

并得到如下结果：

----+------------------
 id | descendant_count
----+------------------
  1 | 9
  2 | 5
  3 | 1
  4 | 1
  5 | 1
  6 | 0
  7 | 0
  8 | 0
  9 | 0
 10 | 0

我想应该存在一个更短或更智能的查询语句来获得相同的结果，可以吗？也许喜欢WITH RECURSIVE或者用循环创建函数来生成查询？

Answer 1

你的一组工会实际上只是一个自我连接..。

SELECT
    tree.id,
    COUNT(descendant.id) AS descendant_count
FROM
    tree
LEFT JOIN
    tree   AS descendant
        ON tree.id = ANY(descendant.ancestors)
GROUP BY
    tree.id

Answer 2

乍一看，这是递归查询的一个例子，但这个例子更简单：

只是休息、分组和数数：

SELECT id AS ancestor, COALESCE (a1.id, 0) AS descendants_count
FROM   tree
LEFT   JOIN (
   SELECT a.id, count(*) AS descendant_count
   FROM   tree t, unnest(t.ancestors) AS a(id)
   GROUP  BY 1
   ) a1 USING (id)
ORDER  BY 1;

并且，为了包括没有任何后代的祖先，请加入LEFT JOIN。

在集合返回函数unnest()中有一个隐式unnest()连接。请参见：

• What is the difference between LATERAL and a subquery in PostgreSQL?

撇开：

如果您最终不得不使用多个UNION子句，那么请考虑使用UNION ALL。请参见：

• Find unmatched rows between two tables dynamically