Generate N Level Product List Using CTE

Hello Friends,
Here I show you example which will generate navigation list for product through N level. Please check below query for generate N Level product list.

Query:-
-----------------------------------------------

DECLARE @Category TABLE
(
CategoryID INT
,Name VARCHAR(100)
,ParentID INT
)

INSERT INTO @Category(CategoryID,Name,ParentID) VALUES ( 1,'Mobile',0)
INSERT INTO @Category(CategoryID,Name,ParentID) VALUES ( 2,'Nokia',1)
INSERT INTO @Category(CategoryID,Name,ParentID) VALUES ( 3,'Samsung',1)
INSERT INTO @Category(CategoryID,Name,ParentID) VALUES ( 4,'Headphone',0)
INSERT INTO @Category(CategoryID,Name,ParentID) VALUES ( 5,'Nokia N70',2)
INSERT INTO @Category(CategoryID,Name,ParentID) VALUES ( 6,'Nokia N72',2)
INSERT INTO @Category(CategoryID,Name,ParentID) VALUES ( 7,'IBall',4)

;WITH CTE(CategoryID,Name,ParentID,level,ParentName)
AS
(
SELECT CategoryID,Name,ParentID,0,cast('.' + Name + '.' AS VARCHAR(MAX)) FROM @Category WHERE ParentID=0
UNION ALL
SELECT C.CategoryID,C.Name,CT.ParentID,level + 1,CAST (CT.ParentName + '.' + C.Name AS VARCHAR(MAX)) FROM @Category C
INNER JOIN CTE CT ON CT.CategoryID=C.ParentID
)
SELECT SPACE(level * 4) + Name AS Products FROM CTE ORDER BY ParentName

Output
-------------------

Common Table Expression Example

Hello Friends,
Here I put the sample code example using CTE.

--Generate Fibonacii Series using CTE


;WITH FibonaciiSeries(N,F1,F2,F3)
AS
(
SELECT CAST(1 AS BIGINT), CAST(1 AS BIGINT),CAST(0 AS BIGINT),CAST(1+0 AS BIGINT)
UNION ALL
SELECT N+1, F2,F3,F2+F3 FROM FibonaciiSeries WHERE N<92
)
SELECT * FROM FibonaciiSeries

--OUTPUT
-------------------------



--Find Factorial

;WITH Factorial(N,Number)
AS
(
SELECT CAST (1 AS BIGINT), CAST (1 AS BIGINT)
UNION ALL
SELECT N+1,CAST ((N+1)*(Number) AS BIGINT) FROM Factorial WHERE N<10
)

SELECT * FROM Factorial

--OUTPUT
-------------------------

Find the second highest salary for each department

Hello Friends,
Here I post one TSQL Challenge. I hopes it will help you.

TSQL Beginner’s Challenge #1 – Find the second highest salary for each department

http://beyondrelational.com/blogs/tcb/archive/2009/10/13/tsql-beginner-s-challenge-1-find-the-second-highest-salary-for-each-department.aspx

TSQL Challenge Solved-

TSQL:-
SET NOCOUNT ON
DECLARE @Employees TABLE(
EmployeeID INT IDENTITY,
EmployeeName VARCHAR(15),
Department VARCHAR(15),
Salary NUMERIC(16,2)
)

INSERT INTO @Employees(EmployeeName, Department, Salary)
VALUES('T Cook','Finance', 40000)
INSERT INTO @Employees(EmployeeName, Department, Salary)
VALUES('D Michael','Finance', 25000)
INSERT INTO @Employees(EmployeeName, Department, Salary)
VALUES('A Smith','Finance', 25000)
INSERT INTO @Employees(EmployeeName, Department, Salary)
VALUES('D Adams','Finance', 15000)

INSERT INTO @Employees(EmployeeName, Department, Salary)
VALUES('M Williams','IT', 80000)
INSERT INTO @Employees(EmployeeName, Department, Salary)
VALUES('D Jones','IT', 40000)
INSERT INTO @Employees(EmployeeName, Department, Salary)
VALUES('J Miller','IT', 50000)
INSERT INTO @Employees(EmployeeName, Department, Salary)
VALUES('L Lewis','IT', 50000)

INSERT INTO @Employees(EmployeeName, Department, Salary)
VALUES('A Anderson','Back-Office', 25000)
INSERT INTO @Employees(EmployeeName, Department, Salary)
VALUES('S Martin','Back-Office', 15000)
INSERT INTO @Employees(EmployeeName, Department, Salary)
VALUES('J Garcia','Back-Office', 15000)
INSERT INTO @Employees(EmployeeName, Department, Salary)
VALUES('T Clerk','Back-Office', 10000)


SELECT * FROM @Employees

;WITH SecondHighestSalary
AS
(
SELECT EmployeeID,EmployeeName,Department,SALARY AS SALARY,RANK() over (partition by Department order by salary) AS SalaryNumber FROM @Employees
)
SELECT EmployeeID,EmployeeName,Department,SALARY FROM SecondHighestSalary WHERE SalaryNumber=2



OUTPUT:-

Using common table expressions (CTE) to generate sequences

Hello frnds,
One of the best enhancements in T-SQL with SQL Server 2005 was Common Table Expressions(CTEs). CTEs are very helpful in writing more readable and manageable queries. The good things don’t end here; self-referencing CTEs are a very powerful method of implementing recursion in SQL queries. In this post, I will present a few examples of generating sequences using CTEs.
The following statements create a number sequence from 1 to 10.

Example:-

   1:  Declare @start int, @end int

   2:  Select @start=1, @end=10

   3:   

   4:   

   5:  ;WITH CTESequence(number)

   6:  AS

   7:  (

   8:  SELECT @start

   9:  UNION ALL

  10:  SELECT number +1 FROM CTESequence

  11:  WHERE number<@end

  12:  )

  13:  SELECT * FROM CTESequence

Output is :-

TSQL Challenge 3 to Reverse string without using Reverse() function

I have accept TSQL Challenge 3 from site and try to solve it.
Here is a URL of TSQL Challenge
http://beyondrelational.com/blogs/tc/archive/2009/05/24/tsql-challenge-3.aspx

Solved TSQL Challenge-3

DECLARE @t TABLE( ID INT IDENTITY, data VARCHAR(20))
INSERT INTO @t(data) SELECT 'Jacob'
INSERT INTO @t(data) SELECT 'Sebastian'

SET NOCOUNT ON

;WITH ReverseName AS
(
SELECT T.ID,T.data,CAST('' AS VARCHAR) AS ReverseData ,TotalLength=LEN(T.data),position=0 FROM @t T
UNION ALL
SELECT ReverseName.ID,ReverseName.data , CAST((ReverseName.ReverseData + substring(ReverseName.data,len(ReverseName.data)-position,1)) AS VARCHAR ) as ReverseData ,ReverseName.TotalLength,position=position+1 FROM ReverseName
where ReverseName.TotalLength>=position
)


SELECT ID,ReverseData AS [Reverse String] FROM ReverseName where position>TotalLength order by ID desc

OUTPUT:-