1. SQL Basics
Some common queries

2. SELECT
USE AdventureWorks
GO
SELECT ContactID,
Title,
FirstName,
LastName
FROM Person.Contact
Comment:
Person is the Schema and
Contact is the Table
Comment:
Person is the Schema and
Contact is the Table

3. SELECT ALL
USE AdventureWorks
GO
SELECT *
FROM Person.Contact
Comment: Good
practise to
explicity
reference only
necessary
columns
Comment: Good
practise to
explicity
reference only
necessary
columns

4. FILTER
USE AdventureWorks
GO
SELECT Title,
FirstName,
LastName
FROM Person.Contact
WHERE Title = 'Ms.'

5. AND
USE AdventureWorks
GO
SELECT Title,
FirstName,
LastName
FROM Person.Contact
WHERE Title = 'Ms.' AND
LastName = 'Antrim'

6. OR
USE AdventureWorks
GO
SELECT Title,
FirstName,
LastName
FROM Person.Contact
WHERE Title = 'Ms.' OR
LastName = 'Antrim'

7. NOT
USE AdventureWorks
GO
SELECT Title,
FirstName,
LastName
FROM Person.Contact
WHERE NOT Title = 'Ms.'

8. WITHOUT PARENTHESIS
USE AdventureWorks
GO
SELECT Title,
FirstName,
LastName
FROM Person.Contact
WHERE Title = 'Ms.' AND
FirstName = 'Catherine' OR
LastName = 'Adams'

9. With Parentheses
SELECT ContactID,
Title,
FirstName,
MiddleName,
LastName
FROM Person.Contact
WHERE (Title = 'Ms.' AND
FirstName = 'Catherine') OR
LastName = 'Adams'

10. BETWEEN
USE AdventureWorks
GO
SELECT SalesOrderID,
ShipDate
FROM Sales.SalesOrderHeader
WHERE ShipDate BETWEEN '7/28/2002' AND '7/29/2002'

11. Less Than
USE AdventureWorks
GO
SELECT ProductID,
Name,
StandardCost
FROM Production.Product
WHERE StandardCost < 110.0000

12. IS NULL
USE AdventureWorks
GO
SELECT ProductID,
Name,
Weight
FROM Production.Product
WHERE Weight IS NULL

13. IN
USE AdventureWorks
GO
SELECT ProductID,
Name,
Color
FROM Production.Product
WHERE Color IN ('Silver', 'Black', 'Red')

14. LIKE
USE AdventureWorks
GO
SELECT ProductID,
Name
FROM Production.Product
WHERE Name LIKE 'B%'

15. Escape
USE AdventureWorks
GO
SELECT ProductID,
Name
FROM Production.Product
WHERE Name LIKE '%/_%' ESCAPE '/'

16. ORDER BY
USE AdventureWorks
GO
SELECT p.Name,
h.EndDate,
h.ListPrice
FROM Production.Product p
INNER JOIN Production.ProductListPriceHistory h ON
p.ProductID = h.ProductID
ORDER BY p.Name, h.EndDate

17. ORDER - DESC
USE AdventureWorks
GO
SELECT p.Name,
h.EndDate,
h.ListPrice
FROM Production.Product p
INNER JOIN Production.ProductListPriceHistory h ON
p.ProductID = h.ProductID
ORDER BY p.Name DESC, h.EndDate DESC
Comment:
Although queries
sometimes appear
to return data
properly without
an ORDER BY
clause, the natural
ordering of results
is determined by
the physical key
column order in
the clustered index
Comment:
Although queries
sometimes appear
to return data
properly without
an ORDER BY
clause, the natural
ordering of results
is determined by
the physical key
column order in
the clustered index

18. ORDER BY Unselected Column
USE AdventureWorks
GO
SELECT p.Name
FROM Production.Product p

19. TOP
USE AdventureWorks
GO
SELECT TOP 10 v.Name,
v.CreditRating
FROM Purchasing.Vendor v
ORDER BY v.CreditRating DESC, v.Name

20. Variables
USE AdventureWorks
GO
DECLARE @Percentage float
SET @Percentage = 1
SELECT TOP (@Percentage) PERCENT
Name
FROM Production.Product
ORDER BY Name

21. GROUP BY
USE AdventureWorks
GO
SELECT OrderDate,
SUM(TotalDue) TotalDueByOrderDate
FROM Sales.SalesOrderHeader
WHERE OrderDate BETWEEN '7/1/2001' AND '7/31/2001'
GROUP BY OrderDate

22. GROUP BY ALL
USE AdventureWorks
GO
SELECT OrderDate,
SUM(TotalDue) TotalDueByOrderDate
FROM Sales.SalesOrderHeader
WHERE OrderDate BETWEEN '7/1/2001' AND '7/31/2001'
GROUP BY ALL OrderDate

23. HAVING
USE AdventureWorks
GO
SELECT s.Name,
COUNT(w.WorkOrderID) Cnt
FROM Production.ScrapReason s
INNER JOIN Production.WorkOrder w ON
s.ScrapReasonID = w.ScrapReasonID
GROUP BY s.Name
HAVING COUNT(*)>50

24. DISTINCT
USE AdventureWorks
GO
SELECT DISTINCT HireDate
FROM HumanResources.Employee

25. AVG
USE AdventureWorks
GO
SELECT AVG(ListPrice)
FROM Production.Product

26. AVG and DISTINCT
USE AdventureWorks
GO
SELECT AVG(DISTINCT ListPrice)
FROM Production.Product

27. Column ALIASES
USE AdventureWorks
GO
SELECT Color AS 'Grouped Color',
AVG(DISTINCT ListPrice) AS 'Average Distinct List Price',
AVG(ListPrice) 'Average List Price'
FROM Production.Product
GROUP BY Color

28. INFORMATION SCHEMA
USE AdventureWorks
GO
SELECT column_name + ' IS NULL AND '
FROM INFORMATION_SCHEMA.columns
WHERE table_name = 'Employee'
ORDER BY ORDINAL_POSITION

29. String concatenation
USE AdventureWorks
GO
SELECT 'The ' +
p.name +
' is only ' +
CONVERT(varchar(25),p.ListPrice) +
'!'
FROM Production.Product p
WHERE p.ListPrice between 100 AND 120
ORDER BY p.ListPrice

30. Comma Delimited List
USE AdventureWorks
GO
DECLARE @Shifts varchar(20)
SET @Shifts = ''
SELECT @Shifts = @Shifts + s.Name + ','
FROM HumanResources.Shift s
ORDER BY s.EndTime
SELECT @Shifts

31. SELECT INTO
USE AdventureWorks
GO
SELECT CustomerID,
Name,
SalesPersonID,
Demographics
INTO Store_Archive
FROM Sales.Store
Comment:
This creates a
table called
Store_Archive
Comment:
This creates a
table called
Store_Archive

32. Create a schema with SELECT
USE AdventureWorks
GO
SELECT CustomerID,
Name,
SalesPersonID,
Demographics
INTO Store_Archive2
FROM Sales.Store
WHERE 1=0

33. Puzzle
I've got The Customers table : T1 .. has two columns :
ID(int) , IsDeleted(bit)
I want in a one sql query to return :
NumberOfAllCustomers --
NumberOfDeletedCustomers --
NumberOfExistingCustomers

34. Prepare data for puzzle
USE tempdb;
IF OBJECT_ID('dbo.T1') IS NOT NULL
DROP TABLE dbo.T1;
CREATE TABLE dbo.t1
(
ID INT NOT NULL,
IsDeleted bit
);
INSERT INTO dbo.T1(ID, IsDeleted) VALUES (1,0);
INSERT INTO dbo.T1(ID, IsDeleted) VALUES (2,1);
INSERT INTO dbo.T1(ID, IsDeleted) VALUES (3,0);
INSERT INTO dbo.T1(ID, IsDeleted) VALUES (4,1);
INSERT INTO dbo.T1(ID, IsDeleted) VALUES (5,0);
GO

35. Check expected answers
SELECT Count(ID) AS NumberOfAllExistingCustomers
FROM T1
SELECT Count(ID) AS NumberOfDeletedCustomers
FROM T1 WHERE IsDeleted = 1
SELECT Count(ID) AS NumberOfExistingCustomers
FROM T1 WHERE IsDeleted = 0

36. Answer One
SELECT COUNT(*) AS NumberOfAllCustomers,
SUM(CASE IsDeleted WHEN 1 THEN 1 END)
AS NumberOfDeletedCustomers,
SUM(CASE IsDeleted WHEN 0 THEN 1 END)
AS NumberOfExistingCustomers
FROM T1
GO

37. Answer Two
select numberOfAllCustomers, numberOfDeletedCustomers,
numberOfAllCustomers - numberOfDeletedCustomers
AS
NumberOfExistingCustomers
from (
select
(select count(id) from t1) numberOfAllCustomers,
(select count(id) from t1 where isdeleted=cast(1 as bit))
numberOfDeletedCustomers
) t;