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Improving my SQL BI Skills

Improving my SQL BI Skills

Daily Archives: March 25, 2015

T-SQL Query | [ The work order puzzle ]

25 Wednesday Mar 2015

Posted by Pawan Kumar Khowal in SQL SERVER, T SQL Puzzles

≈ 6 Comments

Tags

Complex SQL Challenges, Complex TSQL Challenge, Interview Qs.SQL SERVER Questions, Interview Questions on SQL, InterviewQuestions, InterviewQuestions for SQL, Learn complex SQL, Learn SQL, Learn T-SQL, PL/SQL Challenges, Puzzles, Queries for SQL Interview, SQL, SQL 2012, SQL 2014, SQL 2014 Interview Questions, SQL Challenge, SQL Challenges, SQL pl/sql puzzles, SQL Puzzles, SQL SERVER Interview questions, SQL SERVER2005/2008, SQL Sudoku, SQLSERVER, T SQL Puzzles, T-SQL Challenge, Tough SQL Challenges, Tough SQL Puzzles, TSQL, TSQL Challenge, TSQL Challenges, TSQL Interview questions, TSQL Queries


T-SQL Query | [ The work order puzzle ]  – In this puzzle we have to find out orders where the StepId = zero and Status = ‘C AND all the remaining rows for that OrderId Should have Status= ‘W’, For example, the query should return only ‘AA100’ in the sample data. Please check out the sample input and expected output for details.

Sample Input

WorkOrderID STEP_NBR STEP_STATUS
AA100 0 C
AA100 1 W
AA100 2 W
AA200 0 W
AA200 1 W
AA300 0 C
AA300 1 C

Expected Output

WorkOrderID
AA100

Rules/Restrictions

  • The solution should be should use “SELECT” statement or “CTE”.
  • Add your solution(s) in the comments section or send you solution(s) to pawankkmr@gmail.com

Script

Use the below script to generate the source table and fill them up with the sample data.


--Create Table

CREATE TABLE WorkOrders
(
WorkOrderID CHAR(5) NOT NULL,
STEP_NBR INTEGER NOT NULL CHECK (step_nbr BETWEEN 0 AND 1000),
STEP_STATUS CHAR(1) NOT NULL CHECK (step_status IN ('C', 'W')), -- complete, waiting
)
GO

--Insert Data
INSERT INTO WorkOrders(WorkOrderID,STEP_NBR,STEP_STATUS) VALUES
('AA100', 0, 'C'),
('AA100', 1, 'W'),
('AA100', 2, 'W'),
('AA200', 0, 'W'),
('AA200', 1, 'W'),
('AA300', 0, 'C'),
('AA300', 1, 'C')
GO


SELECT WorkOrderID,STEP_NBR,STEP_STATUS FROM WorkOrders

Update May 14 | Solutions



--


---------------------------------------
--Sol 1
---------------------------------------

SELECT workorder_id
FROM WorkOrders
GROUP BY workorder_id
HAVING COUNT(*) =    COUNT(CASE WHEN step_nbr <> 0 AND step_status = 'W' THEN 1 ELSE NULL END) 
                   + COUNT(CASE WHEN step_nbr = 0 AND step_status = 'C' THEN 1 ELSE NULL END)


---------------------------------------
--Sol 2
---------------------------------------
				   
SELECT workorder_id
FROM Projects
GROUP BY workorder_id
HAVING COUNT(step_nbr) = SUM
							(
							CASE
							WHEN step_nbr <> 0 AND step_status = 'W' THEN 1
							WHEN step_nbr = 0 AND step_status = 'C' THEN 1
							ELSE 0 END
							)




--

Add a comment if you have any other solution in mind. We all need to learn.

Keep Learning

http://MSBISkills.com

T-SQL Query | [ The Friday Salary Puzzle ]

25 Wednesday Mar 2015

Posted by Pawan Kumar Khowal in SQL SERVER, T SQL Puzzles

≈ 4 Comments

Tags

Complex SQL Challenges, Complex TSQL Challenge, Interview Qs.SQL SERVER Questions, Interview Questions on SQL, InterviewQuestions, InterviewQuestions for SQL, Learn complex SQL, Learn SQL, Learn T-SQL, PL/SQL Challenges, Puzzles, Queries for SQL Interview, SQL, SQL 2012, SQL 2014, SQL 2014 Interview Questions, SQL Challenge, SQL Challenges, SQL pl/sql puzzles, SQL Puzzles, SQL SERVER Interview questions, SQL SERVER2005/2008, SQL Sudoku, SQLSERVER, T SQL Puzzles, T-SQL Challenge, Tough SQL Challenges, Tough SQL Puzzles, TSQL, TSQL Challenge, TSQL Challenges, TSQL Interview questions, TSQL Queries


T-SQL Query | [ The Friday Salary Puzzle ]  – In this puzzle we have to find the last friday of the DOJ(Month) of each person. If the DOJ is greater than 15 then we have find the last friday for next month. Please check out the sample input and expected output for details.

Sample Input

ID Name Salary DOJ
1 A 100 02-Oct-14
2 B 200 16-Mar-13
3 C 300 02-Jan-14
4 D 400 17-Feb-12
5 E 500 08-Feb-12

Expected Output

ID NAME Salary DOJ
1 A 100 31-Oct-14
2 B 200 26-Apr-13
3 C 300 31-Jan-14
4 D 400 30-Mar-12
5 E 500 24-Feb-12

Rules/Restrictions

  • The solution should be should use “SELECT” statement or “CTE”.
  • Add your solution(s) in the comments section or send you solution(s) to pawankkmr@gmail.com

Script

Use the below script to generate the source table and fill them up with the sample data.


--Create Table
CREATE TABLE [dbo].[EmployeeSalary]
(
[ID] [int] NOT NULL,
[Name] [varchar](50) NULL,
[Salary] [float] NULL,
[DOJ] [datetime] NULL
)
GO

--Insert Data
INSERT [dbo].[EmployeeSalary] ([ID], [Name], [Salary], [DOJ]) VALUES (1, N'A', 100, CAST(0x0000A3B800000000 AS DateTime))
GO
INSERT [dbo].[EmployeeSalary] ([ID], [Name], [Salary], [DOJ]) VALUES (2, N'B', 200, CAST(0x0000A18300000000 AS DateTime))
GO
INSERT [dbo].[EmployeeSalary] ([ID], [Name], [Salary], [DOJ]) VALUES (3, N'C', 300, CAST(0x0000A2A700000000 AS DateTime))
GO
INSERT [dbo].[EmployeeSalary] ([ID], [Name], [Salary], [DOJ]) VALUES (4, N'D', 400, CAST(0x00009FFA00000000 AS DateTime))
GO
INSERT [dbo].[EmployeeSalary] ([ID], [Name], [Salary], [DOJ]) VALUES (5, N'E', 500, CAST(0x00009FF100000000 AS DateTime))
GO

--Verify Data
SELECT * FROM [EmployeeSalary]

UPDATE – 20-Apr-2015 – Solution 1


--


--Solution 1



;WITH CTE AS
(
     SELECT ID , NAME , Salary , DOJ , 1 Value FROM [EmployeeSalary] WHERE DAY(DOJ) > 15
     UNION ALL
     SELECT r.ID , r.NAME , r.Salary , DATEADD(M,1,r.DOJ) DOJ , Value + 1 FROM [EmployeeSalary] r INNER JOIN CTE on r.ID = CTE.ID
     WHERE Value <= 1 AND DAY(r.DOJ) > 15
)
,CTE2 AS 
(
     SELECT  ID , NAME , Salary , MAX(DOJ) DOJ
     FROM CTE 
     GROUP BY ID,Name,Salary
     UNION ALL SELECT * FROM [EmployeeSalary] WHERE DAY(DOJ) < 15
)
,CTE3 AS
(
     SELECT ID , NAME , Salary , DOJ , 1 Value FROM CTE2 a
     UNION ALL
     SELECT b.ID , b.NAME , b.Salary , DATEADD(D,1,b.DOJ) DOJ , Value + 1 FROM CTE3 b INNER JOIN CTE2 a ON a.ID = b.ID
     WHERE DAY(b.DOJ) < DAY(EOMONTH(b.DOJ))     
)
,CTE4 AS
(
     SELECT * , ROW_NUMBER() OVER (PARTITION BY ID ORDER BY VALUE DESC) rnks FROM CTE3 WHERE DATENAME(DW,DOJ) = 'Friday' 
)
SELECT ID , NAME , Salary , DOJ FROM CTE4 WHERE rnks = 1



--

Add a comment if you have any other solution in mind. We all need to learn.

Keep Learning

http://MSBISkills.com

T-SQL Query | [ Fruit Count Puzzle ]

25 Wednesday Mar 2015

Posted by Pawan Kumar Khowal in SQL SERVER, T SQL Puzzles

≈ 5 Comments

Tags

Complex SQL Challenges, Complex TSQL Challenge, Interview Qs.SQL SERVER Questions, Interview Questions on SQL, InterviewQuestions, InterviewQuestions for SQL, Learn complex SQL, Learn SQL, Learn T-SQL, PL/SQL Challenges, Puzzles, Queries for SQL Interview, SQL, SQL 2012, SQL 2014, SQL 2014 Interview Questions, SQL Challenge, SQL Challenges, SQL pl/sql puzzles, SQL Puzzles, SQL SERVER Interview questions, SQL SERVER2005/2008, SQL Sudoku, SQLSERVER, T SQL Puzzles, T-SQL Challenge, Tough SQL Challenges, Tough SQL Puzzles, TSQL, TSQL Challenge, TSQL Challenges, TSQL Interview questions, TSQL Queries


T-SQL Query | [ Fruit Count Puzzle ]  – In this puzzle we have to count fruits per person. Please check out the sample input and expected output for details.

Sample Input

Name Fruit
Neeraj MANGO
Neeraj MANGO
Neeraj MANGO
Neeraj APPLE
Neeraj ORANGE
Neeraj LICHI
Neeraj LICHI
Neeraj LICHI
Isha MANGO
Isha MANGO
Isha APPLE
Isha ORANGE
Isha LICHI
Gopal MANGO
Gopal MANGO
Gopal APPLE
Gopal APPLE
Gopal APPLE
Gopal ORANGE
Gopal LICHI
Mayank MANGO
Mayank MANGO
Mayank APPLE
Mayank APPLE
Mayank ORANGE
Mayank LICHI

Expected Output

Name MangoCount APPLECount LICHICount ORANGECount
Gopal 2 3 1 1
Isha 2 1 1 1
Mayank 2 2 1 1
Neeraj 3 1 3 1

Rules/Restrictions

  • The solution should be should use “SELECT” statement or “CTE”.
  • Add your solution(s) in the comments section or send you solution(s) to pawankkmr@gmail.com

Script

Use the below script to generate the source table and fill them up with the sample data.


--Create table
CREATE TABLE FruitCount
(
Name VARCHAR(20)
,Fruit VARCHAR(25)
)
GO

--Insert Data
INSERT INTO FruitCount(Name,Fruit) VALUES
('Neeraj'    ,'MANGO'),
('Neeraj'    ,'MANGO'),
('Neeraj'    ,'MANGO'),
('Neeraj'    ,'APPLE'),
('Neeraj'    ,'ORANGE'),
('Neeraj'    ,'LICHI'),
('Neeraj'    ,'LICHI'),
('Neeraj'    ,'LICHI'),
('Isha'     ,'MANGO'),
('Isha'     ,'MANGO'),
('Isha'     ,'APPLE'),
('Isha'     ,'ORANGE'),
('Isha'     ,'LICHI'),
('Gopal' ,'MANGO'),
('Gopal' ,'MANGO'),
('Gopal' ,'APPLE'),
('Gopal' ,'APPLE'),
('Gopal' ,'APPLE'),
('Gopal' ,'ORANGE'),
('Gopal' ,'LICHI'),
('Mayank'  ,'MANGO'),
('Mayank'  ,'MANGO'),
('Mayank'  ,'APPLE'),
('Mayank'  ,'APPLE'),
('Mayank'  ,'ORANGE'),
('Mayank'  ,'LICHI')

--Verify Data
SELECT Name,Fruit FROM FruitCount

UPDATE – 24-Apr-2015 – Solution 1


--


SELECT 
	 Name
	,SUM(CASE WHEN Fruit='MANGO' THEN 1 ELSE 0 END) MangoCount 
	,SUM(CASE WHEN Fruit='APPLE' THEN 1 ELSE 0 END) APPLECount 
	,SUM(CASE WHEN Fruit='LICHI' THEN 1 ELSE 0 END) LICHICount 
	,SUM(CASE WHEN Fruit='ORANGE' THEN 1 ELSE 0 END) ORANGECount  
FROM FruitCount
GROUP BY Name

`
--

Add a comment if you have any other solution in mind. We all need to learn.

Keep Learning

http://MSBISkills.com

T-SQL Query | [ Group By XML Path Puzzle ]

25 Wednesday Mar 2015

Posted by Pawan Kumar Khowal in SQL SERVER, T SQL Puzzles

≈ Leave a comment

Tags

Complex SQL Challenges, Complex TSQL Challenge, Interview Qs.SQL SERVER Questions, Interview Questions on SQL, InterviewQuestions, InterviewQuestions for SQL, Learn complex SQL, Learn SQL, Learn T-SQL, PL/SQL Challenges, Puzzles, Queries for SQL Interview, SQL, SQL 2012, SQL 2014, SQL 2014 Interview Questions, SQL Challenge, SQL Challenges, SQL pl/sql puzzles, SQL Puzzles, SQL SERVER Interview questions, SQL SERVER2005/2008, SQL Sudoku, SQLSERVER, T SQL Puzzles, T-SQL Challenge, Tough SQL Challenges, Tough SQL Puzzles, TSQL, TSQL Challenge, TSQL Challenges, TSQL Interview questions, TSQL Queries


T-SQL Query | [ Group By XML Path Puzzle ]  – In this puzzle we have to show distinct students and the courses & instructors comma separated. Please note that you have to use XML Path to solve this puzzle. Please check out the sample input and expected output for details.

Sample Input

StudentName Course Instructor RoomNo
Mark Algebra Dr. James 101
Mark Maths Dr. Jones 201
Joe Algebra Dr. James 101
Joe Science Dr. Ross 301
Joe Geography Dr. Lisa 401
Jenny Algebra Dr. James 101

Expected Output

StudentName Taught by
Jenny Algebra by Dr. James in Room No 101
Joe Algebra by Dr. James in Room No 101, Science by Dr. Ross in Room No 301, Geography by Dr. Lisa in Room No 401
Mark Algebra by Dr. James in Room No 101, Maths by Dr. Jones in Room No 201

Rules/Restrictions

  • The solution should be should use “SELECT” statement or “CTE”.
  • Add your solution(s) in the comments section or send you solution(s) to pawankkmr@gmail.com

Script

Use the below script to generate the source table and fill them up with the sample data.

 


--CREATE TABLE

CREATE TABLE TestTable 
(
  StudentName VARCHAR(100)
, Course VARCHAR(100)
, Instructor VARCHAR(100)
, RoomNo VARCHAR(100)
)
GO

-- Populate table

INSERT INTO TestTable (StudentName, Course, Instructor, RoomNo)
SELECT 'Mark', 'Algebra', 'Dr. James', '101'
UNION ALL
SELECT 'Mark', 'Maths', 'Dr. Jones', '201'
UNION ALL
SELECT 'Joe', 'Algebra', 'Dr. James', '101'
UNION ALL
SELECT 'Joe', 'Science', 'Dr. Ross', '301'
UNION ALL
SELECT 'Joe', 'Geography', 'Dr. Lisa', '401'
UNION ALL
SELECT 'Jenny', 'Algebra', 'Dr. James', '101'
GO

-- Check orginal data

SELECT StudentName, Course, Instructor, RoomNo
FROM TestTable
GO

Update May 14 | Solution


--

/************   SOLUTION 1    | Pawan Kumar Khowal     ****************/

SELECT b.StudentName 
			, STUFF 
				((
				SELECT ', ' + Course + ' by ' + CAST(Instructor AS VARCHAR(MAX)) + ' in Room No ' + CAST(RoomNo AS VARCHAR(MAX))
				FROM TestTable a
				WHERE ( a.StudentName = b.StudentName )
				FOR XML PATH('')
				) ,1,2,'') 
				AS cusr
FROM TestTable b
GROUP BY b.StudentName

--

Add a comment if you have any other solution in mind. We all need to learn.

Keep Learning

http://MSBISkills.com

T-SQL Query | [ Normalize (Divide) Amount between Months ]

25 Wednesday Mar 2015

Posted by Pawan Kumar Khowal in SQL SERVER, T SQL Puzzles

≈ 6 Comments

Tags

Complex SQL Challenges, Complex TSQL Challenge, Interview Qs.SQL SERVER Questions, Interview Questions on SQL, InterviewQuestions, InterviewQuestions for SQL, Learn complex SQL, Learn SQL, Learn T-SQL, PL/SQL Challenges, Puzzles, Queries for SQL Interview, SQL, SQL 2012, SQL 2014, SQL 2014 Interview Questions, SQL Challenge, SQL Challenges, SQL pl/sql puzzles, SQL Puzzles, SQL SERVER Interview questions, SQL SERVER2005/2008, SQL Sudoku, SQLSERVER, T SQL Puzzles, T-SQL Challenge, Tough SQL Challenges, Tough SQL Puzzles, TSQL, TSQL Challenge, TSQL Challenges, TSQL Interview questions, TSQL Queries


T-SQL Query | [ Normalize (Divide) Amount between Months ]  – In this puzzle we have Normalize (Divide) Amount between Months. Please check out the sample input and expected output for details.

Sample Input

Start End Amount
14-Apr-14 13-May-14 200
15-May-14 16-Jun-14 320

Expected Output

Start End Amount
14-Apr-14 30-Apr-14 100
01-May-14 13-May-14 100
15-May-14 31-May-14 160
01-Jun-14 16-Jun-14 160

Rules/Restrictions

  • The solution should be should use “SELECT” statement or “CTE”.
  • Add your solution(s) in the comments section or send you solution(s) to pawankkmr@gmail.com

Script

Use the below script to generate the source table and fill them up with the sample data.


--Create Table

CREATE TABLE TestSplitData
(
 Start DATETIME
,EndDt DATETIME
,Amount INT
)
GO

--Insert Data

INSERT INTO TestSplitData(Start,EndDt,Amount)
VALUES
('14-Apr-14','13-May-14',200),
('15-May-14','16-Jun-14',320)

--Verify Data

SELECT Start,EndDt,Amount FROM TestSplitData

Please leave a comment if you need solution to the above puzzle

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