SELECT...WHERE EmployeeID NOT IN (...) not working correctly

SELECT...WHERE EmployeeID NOT IN (...) not working correctly

Post by 0to6 » Sat, 26 May 2001 04:01:57



I'm an Access programmer trying to learn SQL Server and I'm experimenting with
Views on the ol' Northwind database that ships with SQL Server.

SELECT FirstName, LastName FROM Employees

obviously selects all the employees.

SELECT FirstName, LastName FROM Employees WHERE EmployeeID IN (SELECT ReportsTo
FROM Employees)

Gives me all the names of the supervisors (that is, all the employees that are
reported to by other employees).

Now, I was figuring that:
SELECT FirstName, LastName FROM Employees WHERE EmployeeID NOT IN (SELECT
ReportsTo FROM Employees)

Would give me all the employees that AREN'T reported to.  Makes sense?  Well, NO
records are returned by this query.  How come?

 
 
 

SELECT...WHERE EmployeeID NOT IN (...) not working correctly

Post by Michael MacGrego » Sat, 26 May 2001 04:10:06


Maybe a repro script would help: DDL for the table involved, example data in
the form of INSERTs and examples of result sets you do get and would like to
get.

Michael MacGregor
Database Architect
SalesDriver

 
 
 

SELECT...WHERE EmployeeID NOT IN (...) not working correctly

Post by Tom Furnes » Sat, 26 May 2001 04:21:58


see
http://www.pinnaclepublishing.com/SQ/SQmag.nsf/0/CEEA6CCC095855708525...
7EC44


I'm an Access programmer trying to learn SQL Server and I'm experimenting
with
Views on the ol' Northwind database that ships with SQL Server.

SELECT FirstName, LastName FROM Employees

obviously selects all the employees.

SELECT FirstName, LastName FROM Employees WHERE EmployeeID IN (SELECT
ReportsTo
FROM Employees)

Gives me all the names of the supervisors (that is, all the employees that
are
reported to by other employees).

Now, I was figuring that:
SELECT FirstName, LastName FROM Employees WHERE EmployeeID NOT IN (SELECT
ReportsTo FROM Employees)

Would give me all the employees that AREN'T reported to.  Makes sense?
Well, NO
records are returned by this query.  How come?

 
 
 

SELECT...WHERE EmployeeID NOT IN (...) not working correctly

Post by Michael MacGrego » Sat, 26 May 2001 04:40:27


Here's another solution, using a modification of the example given in the
article against the Pubs database:

BEGIN TRANSACTION
PRINT '------- Before -------'

SELECT pub_id FROM Publishers AS P
WHERE NOT EXISTS (SELECT * FROM Titles WHERE pub_id = P.pub_id)

UPDATE Titles SET pub_id = NULL
WHERE title_id = 'BU2075'

PRINT '------- After -------'

SELECT pub_id FROM Publishers AS P
WHERE NOT EXISTS (SELECT * FROM Titles WHERE pub_id = P.pub_id)

ROLLBACK TRANSACTION

Using EXISTS or NOT EXISTS overcomes the problem of NULLs.

Michael MacGregor
Database Architect
SalesDriver

 
 
 

SELECT...WHERE EmployeeID NOT IN (...) not working correctly

Post by Joe Celk » Sat, 26 May 2001 05:42:01


The usual example of a tree structure in SQL books is called an adjacency list model and it looks like this:

 CREATE TABLE Personnel
 (emp CHAR(10) NOT NULL PRIMARY KEY,
  boss CHAR(10) DEFAULT NULL REFERENCES Personnel(emp),
  salary DECIMAL(6,2) NOT NULL DEFAULT 100.00);

 Personnel
 emp       boss      salary
 ===========================
 'Albert'  'NULL'    1000.00
 'Bert'    'Albert'   900.00
 'Chuck'   'Albert'   900.00
 'Donna'   'Chuck'    800.00
 'Eddie'   'Chuck'    700.00
 'Fred'    'Chuck'    600.00

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 Personnel table like this, ignoring the left (lft) and right (rgt) columns for now.  This problem is always given with a column for the employee and one for his boss in the textbooks.  This table without the lft and rgt columns is called the adjacency list model, after the graph theory technique of the same name; the pairs of nodes are adjacent to each other.

 CREATE TABLE Personnel
 (emp 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) );

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

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 first 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 and employee columns are the same kind of thing (i.e. names 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, 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 rgt and lft columns (I cannot use the reserved words LEFT and RIGHT in SQL) are what shows the nesting.

If that mental model does not work, then imagine a little worm crawling anti-clockwise along the tree.  Every time he gets to the left or right side of a node, he numbers it.  The worm stops when he gets all the way around the tree and back to the top.

This is a natural way to model a parts explosion, since a final assembly is made of physically nested assemblies that final break down into separate parts.

At this point, the boss 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 Personnel.boss 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 P2.*
   FROM Personnel AS P1, Personnel AS P2
  WHERE P1.lft BETWEEN P2.lft AND P2.rgt
    AND P1.emp = :myemployee;

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

 SELECT P2.*
   FROM Personnel AS P1, Personnel AS P2
  WHERE P1.lft BETWEEN P2.lft AND P2.rgt
    AND P2.emp = :myemployee;

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 P2.emp, SUM(S1.salary)
   FROM Personnel AS P1, Personnel AS P2,
        Salaries AS S1
  WHERE P1.lft BETWEEN P2.lft AND P2.rgt
    AND P1.emp = S1.emp
  GROUP BY P2.emp;

4. To find the level of each node, so you can print the tree as an indented listing.

 SELECT COUNT(P2.emp) AS indentation, P1.emp
   FROM Personnel AS P1, Personnel AS P2
  WHERE P1.lft BETWEEN P2.lft AND P2.rgt
  GROUP BY P1.emp
  ORDER BY P1.lft;

5. The nested set model has an implied ordering of siblings which the adjacency list model does not.  To insert a new node as the rightmost sibling.

BEGIN
DECLARE right_most_sibling INTEGER;

SET right_most_sibling
    = (SELECT rgt
         FROM Personnel
        WHERE emp = :your_boss);

UPDATE Personnel
   SET lft = CASE WHEN lft > right_most_sibling
                  THEN lft + 2
                  ELSE lft END,
       rgt = CASE WHEN rgt >= right_most_sibling
                  THEN rgt + 2
                  ELSE rgt END
 WHERE rgt >= right_most_sibling;

INSERT INTO Personnel (emp, lft, rgt)
VALUES ('New Guy', right_most_sibling, (right_most_sibling + 1))
END;

6. To convert an adjacency list model into a nested set model, use a push down stack algorithm.  Assume that we have these tables:

-- Tree holds the adjacency model
CREATE TABLE Tree
(emp CHAR(10) NOT NULL,
 boss CHAR(10));

INSERT INTO Tree
SELECT emp, boss FROM Personnel;

-- Stack starts empty, will holds the nested set model
CREATE TABLE Stack
(stack_top INTEGER NOT NULL,
 emp CHAR(10) NOT NULL,
 lft INTEGER,
 rgt INTEGER);

BEGIN ATOMIC
DECLARE counter INTEGER;
DECLARE max_counter INTEGER;
DECLARE current_top INTEGER;

SET counter = 2;
SET max_counter = 2 * (SELECT COUNT(*) FROM Tree);
SET current_top = 1;

INSERT INTO Stack
SELECT 1, emp, 1, NULL
  FROM Tree
 WHERE boss IS NULL;

DELETE FROM Tree
 WHERE boss IS NULL;

WHILE counter <= (max_counter - 2)
LOOP IF EXISTS (SELECT *
                   FROM Stack AS S1, Tree AS T1
                  WHERE S1.emp = T1.boss
                    AND S1.stack_top = current_top)
     THEN
     BEGIN -- push when top has subordinates and set lft value
       INSERT INTO Stack
       SELECT (current_top + 1), MIN(T1.emp), counter, NULL
         FROM Stack AS S1, Tree AS T1
        WHERE S1.emp = T1.boss
          AND S1.stack_top = current_top;

        DELETE FROM Tree
         WHERE emp = (SELECT emp
                        FROM Stack
                       WHERE stack_top = current_top + 1);

        SET counter = counter + 1;
        SET current_top = current_top + 1;
     END
     ELSE
     BEGIN  -- pop the stack and set rgt value
       UPDATE Stack
          SET rgt = counter,
              stack_top = -stack_top -- pops the stack
        WHERE stack_top = current_top
       SET counter = counter + 1;
       SET current_top = current_top - 1;
     END IF;
 END LOOP;
END;

This approach will be two to three orders of magnitude faster than the adjacency list model for subtree and aggregate operations.

For details, see the chapter in my book JOE CELKO'S SQL FOR SMARTIES (Morgan-Kaufmann, 1999, second edition)

--CELKO--

SQL guru at Trilogy
===========================
Please post DDL, so that people do not have to guess what the keys, constraints, Declarative Referential Integrity, datatypes, etc. in your schema are.

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