如何使用 cte 查询和解析相邻列表层次结构？

Question

如何查询此父子层次结构以生成其中级别位于其自己的列中的结果集？样本数据：

SET NOCOUNT ON;
USE Tempdb;

IF OBJECT_ID('dbo.Employees', 'U') IS NOT NULL DROP TABLE dbo.Employees;

CREATE TABLE dbo.Employees
(
  empid   INT         NOT NULL PRIMARY KEY,
  mgrid   INT         NULL     REFERENCES dbo.Employees,
  empname VARCHAR(25) NOT NULL,
  salary  MONEY       NOT NULL,
  CHECK (empid <> mgrid),
  CHECK (empid > 0)
);

CREATE UNIQUE INDEX idx_unc_mgrid_empid ON dbo.Employees(mgrid, empid);

INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES
  (1,  NULL, 'David'  , $10000.00),
  (2,  1,    'Eitan'  ,  $7000.00),
  (3,  1,    'Ina'    ,  $7500.00),
  (4,  2,    'Seraph' ,  $5000.00),
  (5,  2,    'Jiru'   ,  $5500.00),
  (6,  2,    'Steve'  ,  $4500.00),
  (7,  3,    'Aaron'  ,  $5000.00),
  (8,  5,    'Lilach' ,  $3500.00),
  (9,  7,    'Rita'   ,  $3000.00),
  (10, 5,    'Sean'   ,  $3000.00),
  (11, 7,    'Gabriel',  $3000.00),
  (12, 9,    'Emilia' ,  $2000.00),
  (13, 9,    'Michael',  $2000.00),
  (14, 9,    'Didi'   ,  $1500.00);

select * from dbo.Employees
go     

;WITH Tree (empid, mgrid, lv)
AS (

SELECT empid, mgrid, 1
FROM Employees
WHERE mgrid IS NULL 

UNION ALL

SELECT E.empid, E.mgrid, lv + 1
FROM Employees AS E
JOIN Tree
ON E.mgrid= Tree.empid
)
SELECT empid, mgrid, lv
FROM Tree
ORDER BY Lv, empid

生成的表应该具有类似的结构

+-------+-----+--------+--------+--------+--------+--------+
| empid | lvl | level1 | level2 | level3 | level4 | level5 |
+-------+-----+--------+--------+--------+--------+--------+
|     1 |   1 |      1 | NULL   | NULL   | NULL   | NULL   |
|     2 |   2 |      1 | 2      | NULL   | NULL   | NULL   |
|     3 |   2 |      1 | 3      | NULL   | NULL   | NULL   |
|     4 |   3 |      1 | 2      | 4      | NULL   | NULL   |
|     5 |   3 |      1 | 2      | 5      | NULL   | NULL   |
|     6 |   3 |      1 | 2      | 6      | NULL   | NULL   |
|     7 |   3 |      1 | 3      | 7      | NULL   | NULL   |
|     8 |   4 |      1 | 2      | 5      | 8      | NULL   |
|     9 |   4 |      1 | 3      | 7      | 9      | NULL   |
|    10 |   4 |      1 | 2      | 5      | 10     | NULL   |
|    11 |   4 |      1 | 3      | 7      | 11     | NULL   |
|    12 |   5 |      1 | 3      | 7      | 9      | 12     |
|    13 |   5 |      1 | 3      | 7      | 9      | 13     |
|    14 |   5 |      1 | 3      | 7      | 9      | 14     |
+-------+-----+--------+--------+--------+--------+--------+

Answer 1

您的示例数据使问题更加清晰。您可以在下降时收集经理级别：

; with  Tree as
        (
        SELECT  empid
        ,       mgrid
        ,       1 as lv
        ,       1 as level1
        ,       null as level2
        ,       null as level3
        ,       null as level4
        ,       null as level5
        FROM    Employees
        WHERE   mgrid IS NULL 
        UNION ALL
        SELECT  E.empid
        ,       E.mgrid
        ,       T.lv + 1
        ,       T.level1
        ,       case when T.lv = 1 then E.empid else t.level2 end
        ,       case when T.lv = 2 then E.empid else t.level3 end
        ,       case when T.lv = 3 then E.empid else t.level4 end
        ,       case when T.lv = 4 then E.empid else t.level5 end
        FROM    Employees AS E
        JOIN    Tree T
        ON      E.mgrid = T.empid
        )
select  *
from    Tree

Example at SQL Fiddle.

Answer 2

;WITH Tree (empid, level, level1, level2, level3, level4, level5)
AS (

SELECT empid, 1, empid, NULL, NULL, NULL, NULL
FROM Employees
WHERE mgrid IS NULL 

UNION ALL

SELECT E.empid, T.level + 1,
       CASE WHEN T.level+1 = 1 THEN E.empid ELSE T.level1 END,
       CASE WHEN T.level+1 = 2 THEN E.empid ELSE T.level2 END,
       CASE WHEN T.level+1 = 3 THEN E.empid ELSE T.level3 END,
       CASE WHEN T.level+1 = 4 THEN E.empid ELSE T.level4 END,
       CASE WHEN T.level+1 = 5 THEN E.empid ELSE T.level5 END 
FROM Employees AS E
JOIN Tree T
ON E.mgrid= T.empid
)
SELECT empid, level, level1, level2, level3, level4, level5
FROM Tree

Answer 3

有支点？

;WITH Tree (empid, mgrid, lv)
AS (

SELECT empid, mgrid, 1
FROM @Employees
WHERE mgrid IS NULL 

UNION ALL

SELECT E.empid, E.mgrid, lv + 1
FROM @Employees AS E
JOIN Tree
ON E.mgrid= Tree.empid
)
select *
from
Tree
pivot 
(count(empid) for lv in ([1],[2],[3],[4],[5]))p