CHỈNH SỬA dựa trên bảng của OP:
đây là một ví dụ sử dụng các cột từ định nghĩa bảng của OP, dữ liệu mẫu của tôi như sau:
1-Jerome
|
2-Joe
/ \
3-Paul 6-David
/ \ / \
4-Jack 5-Daniel 7-Ian 8-Helen
--I only included the needed columns from the OP's table here
DECLARE @Contacts table (id varchar(36), first_name varchar(100), reports_to_id varchar(36))
INSERT @Contacts VALUES ('1','Jerome', NULL )
INSERT @Contacts VALUES ('2','Joe' ,'1')
INSERT @Contacts VALUES ('3','Paul' ,'2')
INSERT @Contacts VALUES ('4','Jack' ,'3')
INSERT @Contacts VALUES ('5','Daniel','3')
INSERT @Contacts VALUES ('6','David' ,'2')
INSERT @Contacts VALUES ('7','Ian' ,'6')
INSERT @Contacts VALUES ('8','Helen' ,'6')
DECLARE @Root_id char(4)
--get complete tree---------------------------------------------------
SET @Root_id=null
PRINT '@Root_id='+COALESCE(''''[email protected]_id+'''','null')
;WITH StaffTree AS
(
SELECT
c.id, c.first_name, c.reports_to_id, c.reports_to_id as Manager_id, cc.first_name AS Manager_first_name, 1 AS LevelOf
FROM @Contacts c
LEFT OUTER JOIN @Contacts cc ON c.reports_to_id=cc.id
WHERE [email protected]_id OR (@Root_id IS NULL AND c.reports_to_id IS NULL)
UNION ALL
SELECT
s.id, s.first_name, s.reports_to_id, t.id, t.first_name, t.LevelOf+1
FROM StaffTree t
INNER JOIN @Contacts s ON t.id=s.reports_to_id
WHERE [email protected]_id OR @Root_id IS NULL OR t.LevelOf>1
)
SELECT * FROM StaffTree
--get all below 2---------------------------------------------------
SET @Root_id=2
PRINT '@Root_id='+COALESCE(''''[email protected]_id+'''','null')
;WITH StaffTree AS
(
SELECT
c.id, c.first_name, c.reports_to_id, c.reports_to_id as Manager_id, cc.first_name AS Manager_first_name, 1 AS LevelOf
FROM @Contacts c
LEFT OUTER JOIN @Contacts cc ON c.reports_to_id=cc.id
WHERE [email protected]_id OR (@Root_id IS NULL AND c.reports_to_id IS NULL)
UNION ALL
SELECT
s.id, s.first_name, s.reports_to_id, t.id, t.first_name, t.LevelOf+1
FROM StaffTree t
INNER JOIN @Contacts s ON t.id=s.reports_to_id
WHERE [email protected]_id OR @Root_id IS NULL OR t.LevelOf>1
)
SELECT * FROM StaffTree
--get all below 6---------------------------------------------------
SET @Root_id=6
PRINT '@Root_id='+COALESCE(''''[email protected]_id+'''','null')
;WITH StaffTree AS
(
SELECT
c.id, c.first_name, c.reports_to_id, c.reports_to_id as Manager_id, cc.first_name AS Manager_first_name, 1 AS LevelOf
FROM @Contacts c
LEFT OUTER JOIN @Contacts cc ON c.reports_to_id=cc.id
WHERE [email protected]_id OR (@Root_id IS NULL AND c.reports_to_id IS NULL)
UNION ALL
SELECT
s.id, s.first_name, s.reports_to_id, t.id, t.first_name, t.LevelOf+1
FROM StaffTree t
INNER JOIN @Contacts s ON t.id=s.reports_to_id
WHERE [email protected]_id OR @Root_id IS NULL OR t.LevelOf>1
)
SELECT * FROM StaffTree
ĐẦU RA:
@Root_id=null
id first_name reports_to_id Manager_id Manager_first_name LevelOf
------ ---------- ------------- ---------- ------------------ -----------
1 Jerome NULL NULL NULL 1
2 Joe 1 1 Jerome 2
3 Paul 2 2 Joe 3
6 David 2 2 Joe 3
7 Ian 6 6 David 4
8 Helen 6 6 David 4
4 Jack 3 3 Paul 4
5 Daniel 3 3 Paul 4
(8 row(s) affected)
@Root_id='2 '
id first_name reports_to_id Manager_id Manager_first_name LevelOf
------ ---------- ------------- ---------- ------------------ -----------
2 Joe 1 1 Jerome 1
3 Paul 2 2 Joe 2
6 David 2 2 Joe 2
7 Ian 6 6 David 3
8 Helen 6 6 David 3
4 Jack 3 3 Paul 3
5 Daniel 3 3 Paul 3
(7 row(s) affected)
@Root_id='6 '
id first_name reports_to_id Manager_id Manager_first_name LevelOf
------ ---------- ------------- ---------- ------------------ -----------
6 David 2 2 Joe 1
7 Ian 6 6 David 2
8 Helen 6 6 David 2
(3 row(s) affected)
