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Daily Archives: March 26, 2015

T-SQL Query | [ Group By on Multiple Columns Puzzle ]

26 Thursday Mar 2015

Posted by in SQL SERVER, T SQL Puzzles

≈ 7 Comments

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T-SQL Query | [ Group By on Multiple Columns Puzzle ]  – In this puzzle we have to count a’s for different Typ’s in columns Value1, Value2 and Value3. E.g. in sample data there are 3 a’s for Typ I and 1 a for Typ o. Please check out the sample input and expected output for details. Sample Input

ID Typ Value1 Value2 Value3
1 I a b
2 O a d f
3 I d b
4 O g l
5 I z g a
6 I z g a

Expected Output

Typ Counts
I 3
O 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 GroupbyMultipleColumns
(
ID INT
,Typ VARCHAR(1)
,Value1 VARCHAR(1)
,Value2 VARCHAR(1)
,Value3 VARCHAR(1)
)
GO

--Insert Data
INSERT INTO GroupbyMultipleColumns(ID,Typ,Value1,Value2,Value3)
VALUES
(1,'I','a','b',''),
(2,'O','a','d','f'),
(3,'I','d','b',''),
(4,'O','g','l',''),
(5,'I','z','g','a'),
(6,'I','z','g','a')

--Verify Data
SELECT ID,Typ,Value1,Value2,Value3 FROM GroupbyMultipleColumns

Update May 14 | Solutions



--


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

SELECT 
       Typ
       ,SUM(CASE Value1 WHEN 'a' THEN 1 ELSE 0 END) 
       +SUM(CASE Value2 WHEN 'a' THEN 1 ELSE 0 END)
       +SUM(CASE Value3 WHEN 'a' THEN 1 ELSE 0 END)
       Counts 
FROM GroupbyMultipleColumns
GROUP BY Typ


--

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T-SQL Query | [ The Football Puzzle ]

26 Thursday Mar 2015

Posted by in SQL SERVER, T SQL Puzzles

≈ 2 Comments

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T-SQL Query | [ The Football Puzzle ]  – In this puzzle we have to find 100th Champion which has most number of championships overall. Please check out the sample input and expected output for details.

Sample Input

club_id club_name championship_year year_100th_champion
1 FENERBAHCE 2007 1
2 GALATASARAY 2006 0
3 BESIKTAS 2003 1
1 FENERBAHCE 2005 0
1 FENERBAHCE 2004 0
2 GALATASARAY 2002 0
2 GALATASARAY 2000 0
2 GALATASARAY 1999 0
2 GALATASARAY 1998 0
2 GALATASARAY 1997 0
1 FENERBAHCE 1996 0
1 FENERBAHCE 2001 0
1 FENERBAHCE 1989 0
1 FENERBAHCE 1985 0

Expected Output

club_name
FENERBAHCE

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 tr_football_league
(
club_id INT,
club_name Varchar(32),
championship_year INT,
year_100th_champion INT
)

--Insert Data
insert into tr_football_league values (1, 'FENERBAHCE', 2007, 1) ;
insert into tr_football_league values (2, 'GALATASARAY', 2006, 0) ;
insert into tr_football_league values (3, 'BESIKTAS', 2003, 1) ;
insert into tr_football_league values (1, 'FENERBAHCE', 2005, 0) ;
insert into tr_football_league values (1, 'FENERBAHCE', 2004, 0) ;
insert into tr_football_league values (2, 'GALATASARAY', 2002, 0) ;
insert into tr_football_league values (2, 'GALATASARAY', 2000, 0) ;
insert into tr_football_league values (2, 'GALATASARAY', 1999, 0) ;
insert into tr_football_league values (2, 'GALATASARAY', 1998, 0) ;
insert into tr_football_league values (2, 'GALATASARAY', 1997, 0) ;
insert into tr_football_league values (1, 'FENERBAHCE', 1996, 0);
insert into tr_football_league values (1, 'FENERBAHCE', 2001, 0) ;
insert into tr_football_league values (1, 'FENERBAHCE', 1989, 0) ;
insert into tr_football_league values (1, 'FENERBAHCE', 1985, 0) ;

--Verify Data
SELECT club_id, club_name , championship_year , year_100th_champion FROM tr_football_league

Update May 14 | Solutions



--


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

;WITH CTE AS
(
	SELECT club_name,
	year_100th_champion,
	COUNT(*) over(PARTITION BY club_name) cnt
	FROM
	tr_football_league
)
,CTE2 AS
(
	SELECT a.club_name,COUNT(a.cnt) Maxy FROM CTE a INNER JOIN tr_football_league b ON a.club_name = b.club_name
	WHERE b.year_100th_champion = 1
	GROUP BY a.club_name
)
SELECT club_name FROM CTE2
WHERE Maxy = (SELECT MAX(Maxy) FROM CTE2)

---------------------------------------
--Sol 2
---------------------------------------


SELECT club_name FROM 
(
      SELECT club_name, ROW_NUMBER() OVER (ORDER BY counts desc) ranker
      FROM
      (
            SELECT club_name, max(cnt) counts
            FROM
            (
                  SELECT club_name,
                  year_100th_champion,
                  COUNT(*) over(PARTITION BY club_name) cnt
                  FROM
                  tr_football_league
            ) a
            GROUP BY club_name
            HAVING SUM(year_100th_champion) > 0  
      ) art
) r WHERE ranker = 1




--

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T-SQL Query | [ The Fishbone Salary Puzzle ]

26 Thursday Mar 2015

Posted by in SQL SERVER, T SQL Puzzles

≈ 2 Comments

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T-SQL Query | [ The Fishbone Salary Puzzle ]  – In this puzzle we have to find which employee is getting salary hikes across different months in a fishbone structure. Please check out the sample input and expected output for details.

Sample Input

emp_name sal_date sal_amt
Dick 20-06-1996 500
Harry 20-07-1996 500
Harry 20-09-1996 700
Tom 20-06-1996 500
Tom 20-08-1996 700
Tom 20-10-1996 800
Tom 20-12-1996 900

Expected Output

emp_name 20-06-1996 20-07-1996 20-08-1996 20-09-1996 20-10-1996 20-12-1996
Dick 500 NULL NULL NULL NULL NULL
Harry NULL 500 NULL 700 NULL NULL
Tom 500 NULL 700 NULL 800 900

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 Salaries
(emp_name CHAR(10) NOT NULL,
sal_date DATE NOT NULL,
sal_amt DECIMAL (8,2) NOT NULL,
PRIMARY KEY (emp_name, sal_date));

--Insert Data
INSERT INTO Salaries
VALUES ('Tom', '1996-06-20', 500.00),
('Tom', '1996-08-20', 700.00),
('Tom', '1996-10-20', 800.00),
('Tom', '1996-12-20', 900.00),
('Dick', '1996-06-20', 500.00),
('Harry', '1996-07-20', 500.00),
('Harry', '1996-09-20', 700.00);

--Verify Data
SELECT * FROM Salaries

Update May 14 | Solution


--

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

SELECT Emp_Name,[1996-06-20],[1996-07-20],[1996-08-20],[1996-09-20],[1996-10-20],[1996-12-20] 
FROM Salaries
PIVOT 
(
	MAX(sal_amt) FOR sal_date IN ([1996-06-20],[1996-07-20],[1996-08-20],[1996-09-20],[1996-10-20],[1996-12-20])
)p



--

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T-SQL Query | [ The Double Join Puzzle ]

26 Thursday Mar 2015

Posted by in SQL SERVER, T SQL Puzzles

≈ 1 Comment

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T-SQL Query | [ The Double Join Puzzle ]  – In this puzzle we have to join below 3 input tables and get the desired output. Please check out the sample input and expected output for details.

Sample Input

EmployeeDlts

ID Name
1 Pawan
2 Neeraj
3 Isha

EmployeeProject

1 Microsoft
1 Google
1 HortonWorks
2 Microsoft

EmployeeSkills

ID Skill
1 SQL
1 MSBI
2 SQL
2 SSRS

Expected Output

Id Name Project Skill
1 Pawan Microsoft SQL
1 Pawan Google MSBI
1 Pawan HortonWorks NULL
2 Neeraj Microsoft SQL
2 Neeraj NULL SSRS

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 EmployeeDlts(
ID INT,
Name VARCHAR(10)
)
GO

INSERT INTO EmployeeDlts(ID,Name)
SELECT 1,'Pawan'  UNION ALL
SELECT 2,'Neeraj' UNION ALL
SELECT 3,'Isha'
GO

CREATE TABLE EmployeeSkills(
ID INT,
Skill VARCHAR(10)
)
GO

INSERT INTO EmployeeSkills(ID,Skill)
SELECT 1,'SQL'  UNION ALL
SELECT 1,'MSBI'  UNION ALL
SELECT 2,'SQL' UNION ALL
SELECT 2,'SSRS'

CREATE TABLE EmployeeProject(
ID INT,
Project VARCHAR(15)
)
GO

INSERT INTO EmployeeProject(ID,Project)
SELECT 1,'Microsoft'  UNION ALL
SELECT 1,'Google'  UNION ALL
SELECT 1,'HortonWorks' UNION ALL
SELECT 2,'Microsoft'

Update May 14 | Solution


--

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



;WITH CTE1 AS
(
	SELECT * , ROW_NUMBER() OVER (PARTITION BY ID ORDER BY %%Physloc%%) rnk FROM EmployeeSkills s 
)
,CTE2 AS 
(
	SELECT * , ROW_NUMBER() OVER (PARTITION BY ID ORDER BY %%Physloc%%) rnk FROM EmployeeProject p 
)
SELECT ISNULL(a.ID,b.Id) Id, d.Name , b.Project, a.Skill  FROM CTE1 a FULL OUTER JOIN CTE2 b ON a.ID = b.ID AND a.rnk = b.rnk 
LEFT JOIN EmployeeDlts d ON d.ID = ISNULL(a.ID,b.ID)


--

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T-SQL Query | [ The Manager employee hierarchy Puzzle ]

26 Thursday Mar 2015

Posted by in SQL SERVER, T SQL Puzzles

≈ 1 Comment

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T-SQL Query | [ The Manager employee hierarchy Puzzle ]  – In this puzzle we have to find employees their managers with their hierarchy. Please check out the sample input and expected output for details.

Sample Input

EmpID EmpName ReportsTo
1 Jacob NULL
2 Rui NULL
3 Jacobson NULL
4 Jess 1
5 Steve 1
6 Bob 1
7 Smith 2
8 Bobbey 2
9 Steffi 3
10 Bracha 3
11 John 5
12 Michael 6
13 Paul 6
14 Lana 7
15 Johnson 7
16 Mic 8
17 Stev 8
18 Paulson 9
19 Jessica 10

Expected Output

Hierarchy Hierarchy EmpName EmpId Level FullyQualifiedName
Jacob 1 Jacob 1 0 .Jacob.
    Bob 6 Bob 6 1 .Jacob..Bob.
        Michael 12 Michael 12 2 .Jacob..Bob..Michael.
        Paul 13 Paul 13 2 .Jacob..Bob..Paul.
    Jess 4 Jess 4 1 .Jacob..Jess.
    Steve 5 Steve 5 1 .Jacob..Steve.
        John 11 John 11 2 .Jacob..Steve..John.
Jacobson 3 Jacobson 3 0 .Jacobson.
    Bracha 10 Bracha 10 1 .Jacobson..Bracha.
        Jessica 19 Jessica 19 2 .Jacobson..Bracha..Jessica.
    Steffi 9 Steffi 9 1 .Jacobson..Steffi.
        Paulson 18 Paulson 18 2 .Jacobson..Steffi..Paulson.
Rui 2 Rui 2 0 .Rui.
    Bobbey 8 Bobbey 8 1 .Rui..Bobbey.
        Mic 16 Mic 16 2 .Rui..Bobbey..Mic.
        Stev 17 Stev 17 2 .Rui..Bobbey..Stev.
    Smith 7 Smith 7 1 .Rui..Smith.
        Johnson 15 Johnson 15 2 .Rui..Smith..Johnson.
        Lana 14 Lana 14 2 .Rui..Smith..Lana.

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 Employees  (EmpID INT, EmpName VARCHAR(20), ReportsTo INT)

--Insert Data
INSERT INTO Employees(EmpID, EmpName, ReportsTo)
SELECT 1, 'Jacob', NULL UNION ALL
SELECT 2, 'Rui', NULL UNION ALL
SELECT 3, 'Jacobson', NULL UNION ALL
SELECT 4, 'Jess', 1 UNION ALL
SELECT 5, 'Steve', 1 UNION ALL
SELECT 6, 'Bob', 1 UNION ALL
SELECT 7, 'Smith', 2 UNION ALL
SELECT 8, 'Bobbey', 2 UNION ALL
SELECT 9, 'Steffi', 3 UNION ALL
SELECT 10, 'Bracha', 3 UNION ALL
SELECT 11, 'John', 5 UNION ALL
SELECT 12, 'Michael', 6 UNION ALL
SELECT 13, 'Paul', 6 UNION ALL
SELECT 14, 'Lana', 7 UNION ALL
SELECT 15, 'Johnson', 7 UNION ALL
SELECT 16, 'Mic', 8 UNION ALL
SELECT 17, 'Stev', 8 UNION ALL
SELECT 18, 'Paulson', 9 UNION ALL
SELECT 19, 'Jessica', 10

--Verify Data
SELECT * FROM Employees E

UPDATE – 24-Apr-2015 – Solution 1 


--

;WITH CTE(EmpName , EmpId, Level,FullyQualifiedName) AS (
     Select E.EmpName, E.EmpID, 0 Level
     , Cast('.'+E.EmpName+'.' as Varchar(MAX)) FullyQualifiedName         
     From Employees E Where E.ReportsTo IS NULL
     UNION ALL
     Select E.EmpName, E.EmpID, c.Level + 1 , c.FullyQualifiedName+'.'+E.EmpName+'.' FullyQualifiedName  
     From Employees E INNER JOIN CTE c on c.EmpID = e.ReportsTo 
)
SELECT SPACE(LEVEL*4) + H.EmpName Hierarchy,SPACE(LEVEL*4) + CAST(H.EmpID AS VARCHAR(MAX)) Hierarchy , * FROM CTE H 
ORDER BY H.FullyQualifiedName
`
--

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