Question

表tree是PostgreSQL 8.3+中带有祖先数组的示例表：

----+-----------
 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}

要获取后代的每个id计数，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)

并得到以下结果：

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

我猜应该存在较短或更智能的查询语句才能获得相同的结果，这可能吗？也许像WITH RECURSIVE或使用带有循环的函数来生成查询？

Answer 1

乍一看，

看起来就像一个递归查询的案例，但这更简单：
只是嵌套，分组并计数：

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，要包括根本没有任何后代的祖先，请加入LEFT JOIN。

有一个隐式LATERAL联接到集合返回函数unnest()。参见：

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

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

Find unmatched rows between two tables dynamically

Answer 2

您的工会实际上只是自我联接...

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

最好的方法来计数祖先数组的后代数量？

2 个答案: