sql - PostgreSQL递归聚合节点

我在postgres中的表格如下所示。将数组中的这些值解释为在有向图中连接的节点的ID。我要获取的是可能路径的列表(将每行的最后一个ID与其他行的第一个ID匹配)

数据:

  foo
-------
 {1}
 {2,7}
 {3,4}
 {4,6}
 {5}
 {6,8}
 {7}
 {8}

预期结果:

{1}
{2,7}
{3,4,6,8}
{5}

我尝试使用递归查询和窗口函数，但是它没有按预期工作。

最佳答案

您是否正在寻找这样的东西:

WITH RECURSIVE x AS (
        -- choose first level - without more connections
        SELECT id, id AS full_id, 1 AS level
          FROM foo
         WHERE NOT EXISTS (
               SELECT 1
                 FROM foo AS foo2
                WHERE foo.id != foo2.id
                  AND foo.id[1] = foo2.id[array_length(foo2.id, 1)])
        -- add tail
        UNION ALL
        SELECT x.id, x.full_id || foo.id[2:array_length(foo.id, 1)], level + 1
          FROM x
          JOIN foo ON (
               foo.id != x.id
               AND foo.id[1] = x.full_id[array_length(x.full_id, 1)]
               AND array_length(foo.id, 1) != 1)
), z AS (
   -- looks for maximum length
   SELECT max(level) OVER (PARTITION BY id), * FROM x
)
-- choose only with maximum length
SELECT full_id FROM z WHERE max = level

关于sql - PostgreSQL递归聚合节点，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/41997988/