Question

我有下一张表tree：

 id, name, boss, group
 1, Boss 1, 9, false
 2, Boss 2, 9, false
 3, Group 1, 1, true
 4, Group 2, 2, true
 5, Employee 1, 3, false
 6, Employee 2, 3, false
 7, Employee 3, 3, false
 8, Employee 4, 4, false
 9, Boss 0, null, false

必须以下一种方式表示：

Boss 0 
|___  Boss 1 
|    |-- Group 1
|    |________ Employee 1
|    |________ Employee 2
|    |________ Employee 3
|___ Boss 2
     |-- Group 2
     |________ Employee 4

我可以得到这个结果：

id, name, level 
9, Boss 0, 1
1, Boss 1, 2
2, Boss 2, 2
3, Group 1, 3
4, Group 2, 3
5, Employee 1, 4
6, Employee 2, 4
7, Employee 3, 4
8, Employee 4, 4

使用下一个查询：

WITH RECURSIVE t(id, name, boss, level, group) AS
(
 SELECT
     p1.id,
     p1.name,
     p1.boss,
     1 as level,
     p1.group
     FROM tree as p1
     WHERE p1.boss is null
 UNION ALL
 SELECT p2.id,
     p2.name,
     p2.boss,
     CASE WHEN p2.group = true THEN level + 1
         WHEN p2.group is null THEN level
     END,
     p2.group
 FROM tree as p2
     INNER JOIN t on p2.boss = t.id
 )
 SELECT * FROM t WHERE t.group is null

但是，我需要得到的是下一个信息：每个员工下面有多少人直接和间接？例如：

Boss 0:
   2 Direct
   4 Indirect

也就是说，我要找的是这样的：

id, name, level 
9, Boss 0, 1
1, Boss 1, 2
2, Boss 2, 2
3, Group 1, 3
4, Group 2, 3
5, Employee 1, 3
6, Employee 2, 3
7, Employee 3, 3
8, Employee 4, 3

在这种情况下我该怎么办？您是否认为使用嵌套集模型更好地解决此类问题？

Answer 1

您没有指定RDBMS，所以我使用SQL Server：

<强> SqlFiddleDemo

WITH t(id, name, boss, [level], [group]) AS
(
  SELECT
     p1.id,
     p1.name,
     p1.boss,
     1 as [level],
     p1.[group]
   FROM tree as p1
   WHERE p1.boss IS NULL
   UNION ALL
   SELECT 
     p2.id,
     p2.name,
     p2.boss,
     CASE WHEN t.[group] = 0 THEN [level] + 1
         ELSE [level]
     END,
     p2.[group]
   FROM tree as p2
   JOIN t 
     ON p2.boss = t.id
 )
SELECT * 
FROM t

如何在SQL中使用两种节点创建分层树？

1 个答案: