使用递归查询

SQL Server 中的分层与递归查询

Jasmin Ludolf

Content Developer

IT 组织的层级

该组织包含：

ID - 员工编号
员工 Name
公司内的 JobTitle
公司内的 Department
公司内的 Supervisor

描述层级的字段：

ID
Supervisor

员工 2 的上级是 1

IT 组织

+-----------------------+---------------------+------------+------------+
|ID | Name              | Position            | Department | Supervisor |    
|-----------------------|---------------------|------------|------------|
| 1 | Heinz Griesser    | IT Director         | IT         | 0          |
| 2 | Andreas Sitter    | Security Manager    | IT         | 1          |
| 3 | Thomas Bergman    | Innovation Manager  | IT         | 1          |
| 4 | Hannes Berg       | Operation Manager   | IT         | 1          |
| 5 | Anna Kruggel      | Administrator       | IT         | 4          |
| 6 | Karin Pacher      | Developer           | IT         | 4          |
+-----------------------+---------------------+------------+------------+

层级数据的常见任务

获取记录的层级

你的上级是谁？

获取层级深度

获取组织的层级深度

将递归结果合并为一列

我有哪些上级？

获取层级

WITH hierarchy AS (
    SELECT ID,Supervisor
        FROM employee 
        WHERE supervisor = 0

    UNION ALL
    SELECT emp.ID, emp.Supervisor
        FROM employee emp

    JOIN hierarchy
      ON emp.Supervisor = hierarchy.ID)
SELECT *
    FROM hierarchy

获取层级

+---+-----------+
|ID |Supervisor |
|---|-----------|
|1  | 0         |
|2  | 1         |
|3  | 2         |
+---+-----------+

获取层级深度

WITH hierarchy AS (
    SELECT ID, Supervisor, 1 as LEVEL
        FROM employee
           WHERE Supervisor = 0

    UNION ALL
    SELECT emp.ID, emp.Supervisor, LEVEL + 1
        FROM employee emp

    JOIN hierarchy
    ON emp.Supervisor = hierarchy.ID
)
SELECT *
    FROM hierarchy

获取层级深度

+---+-----------+-------|
|ID |Supervisor | Level |
|---|-----------|-------|
|1  | 0         | 0     |
|2  | 1         | 1     |
+---+-----------+-------+

将递归结果合并为一列

WITH hierarchy AS (
    SELECT ID, Supervisor, CAST('0' AS VARCHAR(MAX)) as PATH
        FROM employee
        WHERE Supervisor = 0

    UNION ALL
    SELECT emp.ID, emp.Supervisor, Path + '->' + CAST(emp.Supervisor AS VARCHAR(MAX))
        FROM employee emp
    INNER JOIN hierarchy
    ON emp.Supervisor = hierarchy.ID
)
SELECT *
    FROM hierarchy

将递归结果合并为一列

+-----+--------+
| PATH         |
|--------------|
| 0 -> 1 -> 4  |
+-----+--------+

查询 IT 组织

SQL Server 中的分层与递归查询