Self referencing table - find next manager not self

Wednesday, March 20, 2013 2:39 PM

I have a self referencing table:

CREATE TABLE Staging.tOrgUnit
(
	OrgUnitID int NOT NULL,
	ParentOrgUnitID int NULL,
	Manager char(8) NULL,
	NextManagerNotNull char(8) NULL,
	NextManagerNotSelf char(8) NULL,
	UnitName_EN varchar(40) NOT NULL,
	UnitName_DE varchar(40) NOT NULL,
	CONSTRAINT PK_tOrgUnit
		PRIMARY KEY CLUSTERED (OrgUnitID)
);

This decribes an organizational tree structure, where ParentOrgUnitID ist the self reference to the parent unit. Each unit can have its manager, but manager can also be NULL.

NextManagerNotNull: In the case of Manager = NULL I want to store the information of the manager of the next higher parent unit where Manager NOT NULL.

NextManagerNotSelf: In the case of Manager NOT NULL I want to store the information oft the manger of the next higher parent unit where Manager <> Manager of this unit.

I managed to write the code for NextManagerNotNull:

WITH CTE AS
(
	SELECT 
		O.OrgUnitID,
		O.Manager,
		CAST(NULL AS char(8)) AS ParentManager
	FROM Staging.tOrgUnit O
	WHERE ParentOrgUnitID IS NULL
	
	UNION ALL
	
	SELECT 
		O.OrgUnitID,
		O.Manager,
		CASE WHEN CTE.Manager IS NULL THEN CTE.ParentManager ELSE CTE.Manager END AS ParentManager
	FROM Staging.tOrgUnit O
	INNER JOIN CTE
		ON O.ParentOrgUnitID = CTE.OrgUnitID
)

UPDATE Staging.tOrgUnit	SET NextManagerNotNull = CTE.ParentManager
FROM Staging.tOrgUnit O
INNER JOIN CTE
	ON CTE.OrgUnitID = O.OrgUnitID
WHERE O.Manager IS NULL 
AND O.NextManagerNotNull IS NULL 
AND CTE.ParentManager IS NOT NULL;

This works well but when I tried something similar for NextManagerNotSelf I found out that it was not that easy. This code provides wrong results:

WITH CTE AS
(
	SELECT 
		O.OrgUnitID,
		O.Manager,
		CAST(NULL AS char(8)) AS ParentManager
	FROM Staging.tOrgUnit O
	WHERE ParentOrgUnitID IS NULL
	
	UNION ALL
	
	SELECT 
		O.OrgUnitID,
		O.Manager,
		CASE
			WHEN (CTE.Manager IS NULL) THEN CTE.ParentManager
			WHEN (O.Manager = CTE.Manager) THEN CTE.ParentManager 
			ELSE CTE.Manager 
		END AS ParentManager
	FROM Staging.tOrgUnit O
	INNER JOIN CTE 
		ON O.ParentOrgUnitID = CTE.OrgUnitID
)
SELECT * 
FROM CTE 
WHERE CTE.OrgUnitID < 1000

The results:

Well I understand why the code does not work but I have no clue for a smart solution. Any idea?

Thanks for your help,

Andreas

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Wednesday, March 20, 2013 3:35 PM

Moderator

0

Can you provide some sample data in the form of "insert" statements and the expected result?

AMB

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Wednesday, March 20, 2013 3:42 PM

0
This is a stock posting of mine and it is a bit long. I woudl suggest that you get a copy of TREES & HIERARCHIES IN SQL for even more details.
There are many ways to represent a tree or hierarchy in SQL. This is called an adjacency list model and it looks like this:

CREATE TABLE OrgChart
(emp_name CHAR(10) NOT NULL PRIMARY KEY,
boss_emp_name CHAR(10) REFERENCES OrgChart(emp_name),
salary_amt DECIMAL(6,2) DEFAULT 100.00 NOT NULL,
<< horrible cycle constraints >>);

OrgChart
emp_name boss_emp_name salary_amt
==============================
'Albert' NULL 1000.00
'Bert' 'Albert' 900.00
'Chuck' 'Albert' 900.00
'Donna' 'Chuck' 800.00
'Eddie' 'Chuck' 700.00
'Fred' 'Chuck' 600.00

This approach will wind up with really ugly code -- CTEs hiding recursive procedures, horrible cycle prevention code, etc. The root of your problem is not knowing that rows are not records, that SQL uses sets and trying to fake pointer chains with some vague, magical non-relational "id".

This matches the way we did it in old file systems with pointer chains. Non-RDBMS programmers are comfortable with it because it looks familiar -- it looks like records and not rows.

Another way of representing trees is to show them as nested sets. Since SQL is a set oriented language, this is a better model than the usual adjacency list approach you see in most text books. Let us define a simple OrgChart table like this.

CREATE TABLE OrgChart
(emp_name CHAR(10) NOT NULL PRIMARY KEY,
lft INTEGER NOT NULL UNIQUE CHECK (lft > 0),
rgt INTEGER NOT NULL UNIQUE CHECK (rgt > 1),
CONSTRAINT order_okay CHECK (lft < rgt));

OrgChart
emp_name lft rgt
======================
'Albert' 1 12
'Bert' 2 3
'Chuck' 4 11
'Donna' 5 6
'Eddie' 7 8
'Fred' 9 10

The (lft, rgt) pairs are like tags in a mark-up language, or parens in algebra, BEGIN-END blocks in Algol-family programming languages, etc. -- they bracket a sub-set. This is a set-oriented approach to trees in a set-oriented language.

The organizational chart would look like this as a directed graph:

Albert (1, 12)
/ \
/ \
Bert (2, 3) Chuck (4, 11)
/ | \
/ | \
/ | \
/ | \
Donna (5, 6) Eddie (7, 8) Fred (9, 10)

The adjacency list table is denormalized in several ways. We are modeling both the Personnel and the Organizational chart in one table. But for the sake of saving space, pretend that the names are job titles and that we have another table which describes the Personnel that hold those positions.

Another problem with the adjacency list model is that the boss_emp_name and employee columns are the same kind of thing (i.e. identifiers of personnel), and therefore should be shown in only one column in a normalized table. To prove that this is not normalized, assume that "Chuck" changes his name to "Charles"; you have to change his name in both columns and several places. The defining characteristic of a normalized table is that you have one fact, one place, one time.

The final problem is that the adjacency list model does not model subordination. Authority flows downhill in a hierarchy, but If I fire Chuck, I disconnect all of his subordinates from Albert. There are situations (i.e. water pipes) where this is true, but that is not the expected situation in this case.

To show a tree as nested sets, replace the nodes with ovals, and then nest subordinate ovals inside each other. The root will be the largest oval and will contain every other node. The leaf nodes will be the innermost ovals with nothing else inside them and the nesting will show the hierarchical relationship. The (lft, rgt) columns (I cannot use the reserved words LEFT and RIGHT in SQL) are what show the nesting. This is like XML, HTML or parentheses.

At this point, the boss_emp_name column is both redundant and denormalized, so it can be dropped. Also, note that the tree structure can be kept in one table and all the information about a node can be put in a second table and they can be joined on employee number for queries.

To convert the graph into a nested sets model think of a little worm crawling along the tree. The worm starts at the top, the root, makes a complete trip around the tree. When he comes to a node, he puts a number in the cell on the side that he is visiting and increments his counter. Each node will get two numbers, one of the right side and one for the left. Computer Science majors will recognize this as a modified preorder tree traversal algorithm. Finally, drop the unneeded OrgChart.boss_emp_name column which used to represent the edges of a graph.

This has some predictable results that we can use for building queries. The root is always (left = 1, right = 2 * (SELECT COUNT(*) FROM TreeTable)); leaf nodes always have (left + 1 = right); subtrees are defined by the BETWEEN predicate; etc. Here are two common queries which can be used to build others:

1. An employee and all their Supervisors, no matter how deep the tree.

SELECT O2.*
FROM OrgChart AS O1, OrgChart AS O2
WHERE O1.lft BETWEEN O2.lft AND O2.rgt
AND O1.emp_name = :in_emp_name;

2. The employee and all their subordinates. There is a nice symmetry here.

SELECT O1.*
FROM OrgChart AS O1, OrgChart AS O2
WHERE O1.lft BETWEEN O2.lft AND O2.rgt
AND O2.emp_name = :in_emp_name;

3. Add a GROUP BY and aggregate functions to these basic queries and you have hierarchical reports. For example, the total salaries which each employee controls:

SELECT O2.emp_name, SUM(S1.salary_amt)
FROM OrgChart AS O1, OrgChart AS O2,
Salaries AS S1
WHERE O1.lft BETWEEN O2.lft AND O2.rgt
AND S1.emp_name = O2.emp_name
GROUP BY O2.emp_name;

4. To find the level and the size of the subtree rooted at each emp_name, so you can print the tree as an indented listing.

SELECT O1.emp_name,
SUM(CASE WHEN O2.lft BETWEEN O1.lft AND O1.rgt
THEN O2.sale_amt ELSE 0.00 END) AS sale_amt_tot,
SUM(CASE WHEN O2.lft BETWEEN O1.lft AND O1.rgt
THEN 1 ELSE 0 END) AS subtree_size,
SUM(CASE WHEN O1.lft BETWEEN O2.lft AND O2.rgt
THEN 1 ELSE 0 END) AS lvl
FROM OrgChart AS O1, OrgChart AS O2
GROUP BY O1.emp_name;

5. The nested set model has an implied ordering of siblings which the adjacency list model does not. To insert a new node, G1, under part G. We can insert one node at a time like this:

BEGIN ATOMIC
DECLARE rightmost_spread INTEGER;

SET rightmost_spread
= (SELECT rgt
FROM Frammis
WHERE part = 'G');
UPDATE Frammis
SET lft = CASE WHEN lft > rightmost_spread
THEN lft + 2
ELSE lft END,
rgt = CASE WHEN rgt >= rightmost_spread
THEN rgt + 2
ELSE rgt END
WHERE rgt >= rightmost_spread;

INSERT INTO Frammis (part, lft, rgt)
VALUES ('G1', rightmost_spread, (rightmost_spread + 1));
COMMIT WORK;
END;

The idea is to spread the (lft, rgt) numbers after the youngest child of the parent, G in this case, over by two to make room for the new addition, G1. This procedure will add the new node to the rightmost child position, which helps to preserve the idea of an age order among the siblings.

6. To convert a nested sets model into an adjacency list model:

SELECT B.emp_name AS boss_emp_name, E.emp_name
FROM OrgChart AS E
LEFT OUTER JOIN
OrgChart AS B
ON B.lft
= (SELECT MAX(lft)
FROM OrgChart AS S
WHERE E.lft > S.lft
AND E.lft < S.rgt);

7. To find the immediate parent of a node:

SELECT MAX(P2.lft), MIN(P2.rgt)
FROM Personnel AS P1, Personnel AS P2
WHERE P1.lft BETWEEN P2.lft AND P2.rgt
AND P1.emp_name = @my_emp_name;

I have a book on TREES & HIERARCHIES IN SQL which you can get at Amazon.com right now. It has a lot of other programming idioms for nested sets, like levels, structural comparisons, re-arrangement procedures, etc.

--CELKO-- Books in Celko Series for Morgan-Kaufmann Publishing: Analytics and OLAP in SQL / Data and Databases: Concepts in Practice Data / Measurements and Standards in SQL SQL for Smarties / SQL Programming Style / SQL Puzzles and Answers / Thinking in Sets / Trees and Hierarchies in SQL
- Proposed As Answer by Kalman TothMicrosoft Community Contributor, Moderator Wednesday, March 20, 2013 6:45 PM
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Wednesday, March 20, 2013 6:49 PM

Moderator

0

Take a look at the orgchart of AdventureWorks Cycles fictional company:

http://www.sqlusa.com/bestpractices2005/organizationtree/

Same chart using hierarchyid data type:

http://www.sqlusa.com/bestpractices2008/orgchart/

Kalman Toth Database & OLAP Architect sqlusa.com
Paperback / Kindle: Pass SQL Exam 70-461 & Job Interview: Programming SQL Server 2012
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Thursday, March 21, 2013 8:24 AM

Hello Hunchback!

Testdata:

INSERT INTO Staging.tOrgUnit
(OrgUnitID, ParentOrgUnitID, Manager, UnitName_EN, UnitName_DE)
VALUES
(200, NULL, 'Big Boss', '', ''),
(202, 200,  NULL,       '', ''),
(203, 202,  NULL,       '', ''),
(204, 203,  NULL,       '', ''),
(205, 204,  '--Boss--', '', ''),
(206, 205,  NULL,       '', ''),
(207, 206,  NULL,       '', ''),
(208, 207,  '--Boss--', '', ''),
(209, 208,  '--Boss--', '', ''),
(210, 209,  '--Boss--', '', ''),
(211, 210,  '--Boss--', '', ''),
(212, 211,  NULL,       '', ''),
(213, 212,  NULL,       '', ''),
(214, 213,  NULL,       '', '')

Query:

WITH CTE AS
(
	SELECT 
		O.OrgUnitID,
		O.Manager,
		CAST(NULL AS char(8)) AS ParentManager
	FROM Staging.tOrgUnit O
	WHERE ParentOrgUnitID IS NULL
	
	UNION ALL
	
	SELECT 
		O.OrgUnitID,
		O.Manager,
		CASE
			WHEN (CTE.Manager IS NULL) THEN CTE.ParentManager
			WHEN (O.Manager = CTE.Manager) THEN CTE.ParentManager 
			ELSE CTE.Manager 
		END AS ParentManager
	FROM Staging.tOrgUnit O
	INNER JOIN CTE 
		ON O.ParentOrgUnitID = CTE.OrgUnitID
)
SELECT * 
FROM CTE 
WHERE CTE.OrgUnitID < 1000

Result:

OrgUnitID   Manager    ParentManager
200         Big Boss   NULL
202         NULL       Big Boss
203         NULL       Big Boss
204         NULL       Big Boss
205         --Boss--   Big Boss
206         NULL       --Boss--
207         NULL       --Boss--
208         --Boss--   --Boss--
209         --Boss--   --Boss--
210         --Boss--   --Boss--
211         --Boss--   --Boss--
212         NULL       --Boss--
213         NULL       --Boss--
214         NULL       --Boss--

Expected Result

OrgUnitID   Manager    ParentManager
200         Big Boss   NULL
202         NULL       Big Boss
203         NULL       Big Boss
204         NULL       Big Boss
205         --Boss--   Big Boss
206         NULL       --Boss--
207         NULL       --Boss--
208         --Boss--   Big Boss
209         --Boss--   Big Boss
210         --Boss--   Big Boss
211         --Boss--   Big Boss
212         NULL       --Boss--
213         NULL       --Boss--
214         NULL       --Boss--

Thanks for your help,

Andreas

Self referencing table - find next manager not self

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