这里有一个我在SQLite中成功使用的查询。它创建了属于"Pak"的所有链接的层次结构。
WITH LinkTree(link_id, link_pid, pak_id, link_name, depth)
AS
(
SELECT *, 0 AS depth FROM links
WHERE link_pid = 0
UNION ALL
SELECT l.*, lt.depth+1 AS depth FROM LinkTree lt
JOIN links l ON (lt.link_id = l.link_pid)
)
SELECT * FROM LinkTree WHERE pak_id = 1;
我正试图修改它,使其与Postgres一起使用,但我收到了一个错误,说"深度"令人震惊。
使用了两个简单的表格:
Paks: pak_id, pak_name
Links: link_id, link_pid, link_name, pak_id
除了*_name是varchars之外,所有列都是整数。
有人能帮我吗?
我认为您需要RECURSIVE
关键字:
WITH RECURSIVE LinkTree(link_id, link_pid, pak_id, link_name, depth)
AS
(
SELECT *, 0 AS depth FROM links
WHERE link_pid = 0
UNION ALL
SELECT l.*, lt.depth+1 AS depth FROM LinkTree lt
JOIN links l ON (lt.link_id = l.link_pid)
)
SELECT * FROM LinkTree WHERE pak_id = 1;
编辑:
不要在选择:中使用*
WITH LinkTree(link_id, link_pid, pak_id, link_name, depth)
AS
(
SELECT link_id, link_pid, pak_id, link_name, 0 AS depth
FROM links
WHERE link_pid = 0
UNION ALL
SELECT l.link_id, l.link_pid, l.pak_id, l.link_name, lt.depth+1 AS depth
FROM LinkTree lt
JOIN links l
ON lt.link_id = l.link_pid
)
SELECT *
FROM LinkTree
WHERE pak_id = 1;