Hi All,

Please help me with SELECT query, Thanks!

I have a table with below kind of data,

DECLARE @TBL TABLE (ItemId INT IDENTITY(1,1), ItemName NVARCHAR(20), ItemDate DATE, ParentItemName NVARCHAR(20), ItemOrder INT, ReportId INT)
INSERT INTO @TBL (ItemName, ItemDate, ParentItemName, ItemOrder, ReportId)
VALUES ('Plan', '2015-06-01', NULL, 1, 20),('Design', '2015-06-01', NULL, 2, 20),('Test', '2015-06-20', NULL, 3, 20),('Complete', '2015-06-30', NULL, 4, 20),
('Design child A', '2015-06-02', 'Design', 1, 20), ('Design child B', '2015-06-01', 'Design', 2, 20),
('Test child A', '2015-06-10', 'Test', 1, 20), ('Test child B', '2015-06-09', 'Test', 2, 20), ('Test child C', '2015-06-08', 'Test', 3, 20),
('Test grand child A', '2015-06-08', 'Test child B', 1, 20), ('Test grand child B', '2015-06-08', 'Test child B', 2, 20)
select * from @TBL 

Here I want,

1. to display all parent with ORDER BY ItemOrder (no need to sort by ItemDate)

2. display all child row right after their parent (ORDER BY ItemOrder if ItemDate are same, else ORDER BY ItemDate)

3. display all grand child row right after their parent (ORDER BY ItemOrder if ItemDate are same, else ORDER BY ItemDate)

looking for below output,

;With cte As
(Select t.ItemId, t.ItemName, t.ItemDate, t.ParentItemName, t.ItemOrder, t.ReportId, 
    Cast(t.ItemOrder As nvarchar(max)) As SortOrder
From @TBL t
Where t.ParentItemName Is Null
Union All
Select t.ItemId, t.ItemName, t.ItemDate, t.ParentItemName, t.ItemOrder, t.ReportId, 
  c.SortOrder + '-' + Convert(nchar(8), t.ItemDate, 112) + Cast(t.ItemOrder As nvarchar(max)) As SortOrder
From cte c
Inner Join @TBL t On c.ItemName = t.ParentItemName)
Select ItemId, c.ItemName, c.ItemDate, c.ParentItemName, c.ItemOrder, c.ReportId
From cte c
Order By c.SortOrder

Tom

;With cte As
(Select t.ItemId, t.ItemName, t.ItemDate, t.ParentItemName, t.ItemOrder, t.ReportId, 
    Cast(t.ItemOrder As nvarchar(max)) As SortOrder
From @TBL t
Where t.ParentItemName Is Null
Union All
Select t.ItemId, t.ItemName, t.ItemDate, t.ParentItemName, t.ItemOrder, t.ReportId, 
  c.SortOrder + '-' + Convert(nchar(8), t.ItemDate, 112) + Cast(t.ItemOrder As nvarchar(max)) As SortOrder
From cte c
Inner Join @TBL t On c.ItemName = t.ParentItemName)
Select ItemId, c.ItemName, c.ItemDate, c.ParentItemName, c.ItemOrder, c.ReportId
From cte c
Order By c.SortOrder

Tom

• Proposed as answer by gvrspk veni 19 hours 24 minutes ago
• Marked as answer by Yazdani ISTS 17 hours 44 minutes ago

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.