Question

我正在使用PostgreSQL并且有一个表格，其路径列的类型为ltree。

我想解决的问题是：给定整个树结构，除了根之外，父母的孩子最多。

示例数据如下所示：

path column = ; has a depth of 0 and has 11 children its id is 1824 # dont want this one because its the root
path column = ; has a depth of 0 and has 1 children its id is 1823
path column = 1823; has a depth of 1 and has 1 children its id is 1825
path column = 1823.1825; has a depth of 2 and has 1 children its id is 1826
path column = 1823.1825.1826; has a depth of 3 and has 1 children its id is 1827
path column = 1823.1825.1826.1827; has a depth of 4 and has 1 children its id is 1828
path column = 1824.1925.1955.1959.1972.1991; has a depth of 6 and has 5 children its id is 2001
path column = 1824.1925.1955.1959.1972.1991.2001; has a depth of 7 and has 1 children its id is 2141
path column = 1824.1925.1955.1959.1972.1991.2001; has a depth of 7 and has 0 children its id is 2040
path column = 1824.1925.1955.1959.1972.1991.2001; has a depth of 7 and has 1 children its id is 2054
path column = 1824.1925.1955.1959.1972.1991.2001; has a depth of 7 and has 0 children its id is 2253
path column = 1824.1925.1955.1959.1972.1991.2001; has a depth of 7 and has 1 children its id is 2166
path column = 1824.1925.1955.1959.1972.1991.2001.2054; has a depth of 8 and has 0 children its id is 2205
path column = 1824.1925.1955.1959.1972.1991.2001.2141; has a depth of 8 and has 0 children its id is 2161
path column = 1824.1925.1955.1959.1972.1991.2001.2166; has a depth of 8 and has 1 children its id is 2389
path column = 1824.1925.1955.1959.1972.1991.2001.2166.2389; has a depth of 9 and has 0 children its id is 2402
path column = 1824.1925.1983; has a depth of 3 and has 1 children its id is 2135
path column = 1824.1925.1983.2135; has a depth of 4 and has 0 children its id is 2239
path column = 1824.1926; has a depth of 2 and has 5 children its id is 1942
path column = 1824.1926; has a depth of 2 and has 11 children its id is 1928 # this is the row I am after
path column = 1824.1926; has a depth of 2 and has 2 children its id is 1933
path column = 1824.1926; has a depth of 2 and has 2 children its id is 1989
path column = 1824.1926.1928; has a depth of 3 and has 3 children its id is 2051
path column = 1824.1926.1928; has a depth of 3 and has 0 children its id is 2024
path column = 1824.1926.1928; has a depth of 3 and has 2 children its id is 1988

因此，在此示例中，id 1824 （根）的行有11个子节点，id 1928 的行有11个子节点，深度为2;这是我追求的那一行。

我是ltree和sql的新手。

（这是一个修订过的问题，在Ltree find parent with most children postgresql关闭后添加了样本数据）。

Answer 1

解决方案

要查找具有最多子节点的节点：

SELECT subpath(path, -1, 1), count(*) AS children
FROM   tbl
WHERE  path <> ''
GROUP  BY 1
ORDER  BY 2 DESC
LIMIT  1;

...并排除根节点：

SELECT *
FROM  (
   SELECT ltree2text(subpath(path, -1, 1))::int AS tbl_id, count(*) AS children
   FROM   tbl
   WHERE  path <> ''
   GROUP  BY 1
   ) ct
LEFT   JOIN (
   SELECT tbl_id
   FROM   tbl
   WHERE  path = ''
   ) x USING  (tbl_id)
WHERE  x.tbl_id IS NULL
ORDER  BY children DESC
LIMIT  1

假设根节点有一个空的ltree（''）作为路径。可能是NULL。然后使用path IS NULL ...

您示例中的获胜者实际上是2001，有5个孩子。

-> SQLfiddle

如何？

使用subpath(...)提供的功能the additional module ltree。
在负偏移量的路径中获取最后一个节点，这是该元素的直接父级。
计算父母出现的频率，排除根节点并获取最高计数的剩余节点。
使用ltree2text()从ltree中提取值。
如果多个节点拥有最多的子节点，则在示例中选择任意一个节点。

测试用例

这是我为了获得有用的测试用例而做的工作（在修剪一些噪音之后）：

请参阅SQLfiddle。

换句话说：请记得下次提供一个有用的测试用例。

其他栏目

回答评论。
首先，扩展测试用例：

ALTER TABLE tbl ADD COLUMN postal_code text
              , ADD COLUMN whatever serial;
UPDATE tbl SET postal_code = (1230 + whatever)::text;

看看：

SELECT * FROM tbl;

只需JOIN结果给基表中的父级：

SELECT ct.*, t.postal_code
FROM  (
   SELECT ltree2text(subpath(path, -1, 1))::int AS tbl_id, count(*) AS children
   FROM   tbl
   WHERE  path <> ''
   GROUP  BY 1
   ) ct
LEFT   JOIN (
   SELECT tbl_id
   FROM   tbl
   WHERE  path = ''
   ) x USING  (tbl_id)
JOIN  tbl t USING (tbl_id)
WHERE  x.tbl_id IS NULL
ORDER  BY children DESC
LIMIT  1;

Postgresql ltree查询找到大多数孩子的父母;排除根

1 个答案:

解决方案

如何？

测试用例

其他栏目