Get all Parent Child data of a table
HP
I have a table that have two column one is ParentID and other is ID .
Such that
ID Parent ID
1 1
2 1
3 1
4 2
5 2
6 4
I need to do waht is : If i said i need all parent, child for ID 1 than it
should return 1,2,3,4,5,6
If i say to get data for ID 5 than it should return 5
If i say to get data for ID 2 than it should return 2,4,5,6
Table having more than 5000 records.
Thanks,
HP
John Bell
Hi
Look at CTEs to get the hierarchy
http://msdn.microsoft.com/en-us/library/ms175972.aspx
If you are wanting to concatenate rows then look at
http://databases.aspfaq.com/general/how-do-i-concatenate-strings-from-a-column-into-a-single-row.html
John
HP
But I am working on sql server 2000.
"John Bell" <jbellne...@hotmail.com> wrote in message
news:m1mu76p0vbj28fhm1...@4ax.com...
Dan Guzman
Time to upgrade :-)
Since CTEs are not available in earlier versions, you'll need to perform
recursion manually. The script below uses a table variable for this purpose
and assumes no recursive relationships are defined. Note that ID 1 of your
data has the same value for both the parent and child. I think it is more
proper to specify a NULL ParentID for root nodes in the hierarchy.
DECLARE @ID int;
SET @ID = 2;
DECLARE @AncestorsAndDecendents TABLE(
ID int NOT NULL PRIMARY KEY
,ParentID int NULL);
--get self
INSERT INTO @AncestorsAndDecendents
(ID, ParentID)
SELECT
ID, ParentID
FROM dbo.Foo
WHERE ID = @ID;
--get all decendents
WHILE @@ROWCOUNT > 0
BEGIN
INSERT INTO @AncestorsAndDecendents
SELECT c.ID, c.ParentID
FROM @AncestorsAndDecendents ad
JOIN dbo.Foo c ON
c.ParentID = ad.ID
WHERE
NOT EXISTS(
SELECT *
FROM @AncestorsAndDecendents ad2
WHERE ad2.ID = c.ID
);
END
SELECT ID FROM @AncestorsAndDecendents;
--
Hope this helps.
Dan Guzman
SQL Server MVP
http://weblogs.sqlteam.com/dang/
"HP" <H...@text.com> wrote in message
news:O3jvsgnS...@TK2MSFTNGP05.phx.gbl...
--CELKO--
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 constaints >>);
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 = :myemployee;
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 = :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 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 O1.emp_name = S1.emp_name
GROUP BY O2.emp_name;
4. To find the level of each emp_name, so you can print the tree as an
indented listing.
SELECT T1.node,
SUM(CASE WHEN T2.lft <= T1.lft THEN 1 ELSE 0 END
+ CASE WHEN T2.rgt < T1.lft THEN -1 ELSE 0 END) AS lvl
FROM Tree AS T1, Tree AS T2
WHERE T2.lft <= T1.lft
GROUP BY T1.node;
An untested version of this using OLAP functions might be better able
to use the ordering.
SELECT T1.node,
SUM(CASE WHEN T2.lft <= T1.lft THEN 1 ELSE 0 END
+ CASE WHEN T2.rgt < T1.lft THEN -1 ELSE 0 END)
OVER (ORDER BY T1.lft
RANGE BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS
lvl
FROM Tree AS T1, Tree AS T2
WHERE T2.lft <= T1.lft;
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;
8. To convert an adjacency list to a nested set model, use a push down
stack. Here is version with a stack in SQL/PSM.
-- Tree holds the adjacency model
CREATE TABLE Tree
(node CHAR(10) NOT NULL,
parent CHAR(10));
-- Stack starts empty, will holds the nested set model
CREATE TABLE Stack
(stack_top INTEGER NOT NULL,
node CHAR(10) NOT NULL,
lft INTEGER,
rgt INTEGER);
CREATE PROCEDURE TreeTraversal ()
LANGUAGE SQL
DETERMINISTIC
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;
--clear the stack
DELETE FROM Stack;
-- push the root
INSERT INTO Stack
SELECT 1, node, 1, max_counter
FROM Tree
WHERE parent IS NULL;
-- delete rows from tree as they are used
DELETE FROM Tree WHERE parent IS NULL;
WHILE counter <= max_counter- 1
DO IF EXISTS (SELECT *
FROM Stack AS S1, Tree AS O1.
WHERE S1.node = O1.parent
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(O1.node), counter, NULL
FROM Stack AS S1, Tree AS O1.
WHERE S1.node = O1.parent
AND S1.stack_top = current_top;
-- delete rows from tree as they are used
DELETE FROM Tree
WHERE node = (SELECT node
FROM Stack
WHERE stack_top = current_top + 1);
-- housekeeping of stack pointers and counter
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;
END IF;
END WHILE;
-- SELECT node, lft, rgt FROM Stack;
-- the top column is not needed in the final answer
-- move stack contents to new tree table
END;
I have a book on TREES & HIERARCHIES IN SQL which you can get at
Amazon.com right now.
For a good article on using CTEs and recursion, see:
http://www.sqlservercentral.com/columnists/fBROUARD/recursivequeriesinsql1999andsqlserver2005.asp
HP
"--CELKO--" <jcel...@earthlink.net> wrote in message
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