SQL Video Archives - SQL Training Online

September 27, 2017 by Joey Blue Leave a Comment

SQL Joins Tutorial for Beginners – Inner Join, Left Join, Right Join, Full Outer Join – SQL Training Online

How to write SQL Joins in 10 minutes. (*Plus another 8 minutes of bonus material.)

Learn how to create SQL Joins. The first 10 minutes teach you the basics. Inner Join, Left Outer Join, Right Outer Join, and Full Outer Join. The second 10 minutes show you are few techniques that will help you as you start building joins.

Video: SQL Join Tutorial

How to write SQL Joins in 10 minutes.

The following Scripts are used in the training:

--Create Customer Table
CREATE TABLE [dbo].[Customer](
	[CustomerId] [int] NOT NULL,
	[CityId] [int] NULL,
	[CustomerName] [varchar](50) NOT NULL
) ON [PRIMARY]
GO
--Insert Customer Records
INSERT [dbo].[Customer] ([CustomerId], [CityId], [CustomerName]) VALUES (1, 1, N'Bob Smith')
GO
INSERT [dbo].[Customer] ([CustomerId], [CityId], [CustomerName]) VALUES (2, 1, N'Sally Smith')
GO
INSERT [dbo].[Customer] ([CustomerId], [CityId], [CustomerName]) VALUES (3, 2, N'Tom Smith')
GO
INSERT [dbo].[Customer] ([CustomerId], [CityId], [CustomerName]) VALUES (4, NULL, N'Mary Smith')
GO
--Create City Table
CREATE TABLE [dbo].[City](
	[CityId] [int] NOT NULL,
	[CityName] [varchar](50) NOT NULL
) ON [PRIMARY]
GO
--Insert City Records
INSERT [dbo].[City] ([CityId], [CityName]) VALUES (1, N'Kansas City')
GO
INSERT [dbo].[City] ([CityId], [CityName]) VALUES (2, N'New York')
GO
INSERT [dbo].[City] ([CityId], [CityName]) VALUES (3, N'Houston')
GO

See the original SQL Join Tutorial for Beginners video.

September 13, 2017 by Joey Blue Leave a Comment

Learn Basic SQL – 1 Hour Training Course – SQL Training Online

How to write SQL from scratch in 1 hour.

**Full Beginner’s Course**

In this video I show you how to write SQL using SQL Server and SQL Server Management Studio. We go through Creating a Database, Creating Tables, Inserting, Updating, Deleting, Selecting, Grouping, Summing, Indexing, Joining, and other basics you need to get starting writing SQL.

Video: Learn Basic SQL

How to write SQL from scratch in 1 hour.

The following Scripts are used in the training:

--Create Database
Create Database Customer;

--Create Customer Table
create table Customer
(
	FirstName varchar(50),
	LastName varchar(50),
	Age int
);

--Insert Initial Data
insert into Customer (FirstName,LastName,[Age]) values ('Joey','Blue',40);
insert into Customer (FirstName,LastName,[Age]) values ('Barry','Bonds',50);
insert into Customer (FirstName,LastName,[Age]) values ('Mike','Schmidt',60);
insert into Customer (FirstName,LastName,[Age]) values ('Mike','Schmidt1',61);
insert into Customer (FirstName,LastName,[Age]) values ('Mike','Schmidt2',62);
insert into Customer (FirstName,LastName,[Age]) values ('Mike','Schmidt3',63);
insert into Customer (FirstName,LastName,[Age]) values ('Mike','Schmidt4',64);
insert into Customer (FirstName,LastName,[Age]) values ('Mike','Schmidt5',65);

--This deletes mike schmidt
/*
DELETE Customer
where FirstName='Mike'
and LastName Like 'Schmidt_';
*/

--Select Mike Schmidt
select *
from Customer
where FirstName='Mike'
and LastName Like 'Schmidt_';

--Update City on Barry Bonds
update Customer
Set City='Goddard'
where FirstName='Barry'
and LastName = 'Bonds' ;

--Add City to the Customer Table
alter table customer
add City varchar(50);

--Drop the Customer Table
drop table customer;

--Create Customer Table with Primary Key
create table Customer
(
	Id int Primary Key identity(1,1),
	FirstName varchar(50),
	LastName varchar(50),
	Age int,
	City varchar(50)
)

--Insert Customer Data
insert into Customer 
(FirstName,LastName,[Age],City) values ('Joey','Blue',40, 'Goddard');
insert into Customer 
(FirstName,LastName,[Age],City) values ('Barry','Bonds',50, 'San Francisco');
insert into Customer 
(FirstName,LastName,[Age],City) values ('Mike','Schmidt',60, 'KC');

--Create Products Table
create table Products
(
	id int primary key identity(1,1),
	ProductName varchar(50)
)

--Add Price to Products Table
alter table Products
add Price float;

--Select everything from Products table
select * from Products

--Insert Products
insert into Products (ProductName, Price) values ('Baseball', 5.95);
insert into Products (ProductName, Price) values ('Bat', 195.99);

--Create Orders Table
Create table Orders
(
	OrderId int primary key identity(1,1),
	OrderDate Datetime,
	CustomerID int,
	ProductID int
)

--Insert Orders
insert into Orders (OrderDate,CustomerID,ProductID) values (GetDate(),1,1);
insert into Orders (OrderDate,CustomerID,ProductID) values (GetDate(),1,1);
insert into Orders (OrderDate,CustomerID,ProductID) values (GetDate(),2,1);
insert into Orders (OrderDate,CustomerID,ProductID) values (GetDate(),2,1);
insert into Orders (OrderDate,CustomerID,ProductID) values (GetDate(),2,2);
insert into Orders (OrderDate,CustomerID,ProductID) values (GetDate(),2,2);
insert into Orders (OrderDate,CustomerID,ProductID) values (GetDate(),2,2);
insert into Orders (OrderDate,CustomerID,ProductID) values (GetDate(),2,2);
insert into Orders (OrderDate,CustomerID,ProductID) values (GetDate(),3,2);
insert into Orders (OrderDate,CustomerID,ProductID) values (GetDate(),3,2);
insert into Orders (OrderDate,CustomerID,ProductID) values (GetDate(),4,2); --Bad Record

--Select from Tables
select * from Orders;
select * from Products;
select * from Customer;

--Delete Bad Order  **Your OrderId may be different
delete Orders
where OrderId=21;

--Add Foreign Keys to the Orders table
alter table orders
add foreign key (CustomerId) references Customer(Id);

alter table orders
add foreign key (ProductId) references Products(Id);

--Find the Total and Average Price per City
select c.City,sum(p.Price),AVG(p.price) Total
from Orders o 
inner join Products p on o.ProductID=p.id
inner join Customer c on o.CustomerID=c.Id
group by c.City

See the original SQL Training Course for Beginners video.

November 1, 2012 by Joey Blue Leave a Comment

How to Test SQL Server Functions

Video: How to Test SQL Server Functions – SQL Training Online
In this video, I show you a trick on how to test SQL Functions in SQL Server 2012.

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How do you test a SQL Server Function?

Testing a new function is easier to do when you don’t actually try to query a table to do it.

I will show you how to trim off the front part of an email address by first testing the functions, then putting them into the actual SQL query.

Let’s start with the following SQL query that is written against the Adventure Works database.

SELECT
s.[BusinessEntityID]
,s.[Name]
,ct.[Name] AS [ContactType]
,p.[Title]
,p.[FirstName]
,p.[MiddleName]
,p.[LastName]
,p.[Suffix]
,pp.[PhoneNumber]
,pnt.[Name] AS [PhoneNumberType]
,ea.[EmailAddress]
,p.[EmailPromotion]
FROM [Sales].[Store] s
INNER JOIN [Person].[BusinessEntityContact] bec
ON bec.[BusinessEntityID] = s.[BusinessEntityID]
INNER JOIN [Person].[ContactType] ct
ON ct.[ContactTypeID] = bec.[ContactTypeID]
INNER JOIN [Person].[Person] p
ON p.[BusinessEntityID] = bec.[PersonID]
LEFT OUTER JOIN [Person].[EmailAddress] ea
ON ea.[BusinessEntityID] = p.[BusinessEntityID]
LEFT OUTER JOIN [Person].[PersonPhone] pp
ON pp.[BusinessEntityID] = p.[BusinessEntityID]
LEFT OUTER JOIN [Person].[PhoneNumberType] pnt
ON pnt.[PhoneNumberTypeID] = pp.[PhoneNumberTypeID];

This query joins multiple tables and can be intimidating to the beginner. The main thing to look at is that it returns an email address.

I want to trim off the front part of the email address.

So, I will copy an email address to work on as an example.

‘[email protected]’

Now, SQL Server allows you to run a Select query without a FROM clause.

We will use that capability to test our function. Here is an example.

select '[email protected]'

Now, I want to find the ‘@’ character in the string, so I will use the CharIndex function. Now it is time to test the SQL Server Function:

select charindex('@','[email protected]')

This returns the number 9, since the ‘@’ sign is the 9^th character.

Now, we need to use the substring to pull off all of the character until the ‘@’ sign. So, let’s test the substring function:

select charindex('@','[email protected]dventure-works.com')
,substring('[email protected]',1,9-1)

This does what we want, so now we need to test both SQL Server Functions together.

select substring('[email protected]',1,charindex('@','[email protected]')-1)

This is exactly what we want. So, we just need to put the function into the main query.

SELECT
s.[BusinessEntityID]
,s.[Name]
,ct.[Name] AS [ContactType]
,p.[Title]
,p.[FirstName]
,p.[MiddleName]
,p.[LastName]
,p.[Suffix]
,pp.[PhoneNumber]
,pnt.[Name] AS [PhoneNumberType]
,ea.[EmailAddress]
,substring(ea.[EmailAddress] ,1,charindex('@',ea.[EmailAddress] )-1)
,p.[EmailPromotion]
FROM [Sales].[Store] s
INNER JOIN [Person].[BusinessEntityContact] bec
ON bec.[BusinessEntityID] = s.[BusinessEntityID]
INNER JOIN [Person].[ContactType] ct
ON ct.[ContactTypeID] = bec.[ContactTypeID]
INNER JOIN [Person].[Person] p
ON p.[BusinessEntityID] = bec.[PersonID]
LEFT OUTER JOIN [Person].[EmailAddress] ea
ON ea.[BusinessEntityID] = p.[BusinessEntityID]
LEFT OUTER JOIN [Person].[PersonPhone] pp
ON pp.[BusinessEntityID] = p.[BusinessEntityID]
LEFT OUTER JOIN [Person].[PhoneNumberType] pnt
ON pnt.[PhoneNumberTypeID] = pp.[PhoneNumberTypeID];

There it is. That is how you Test SQL Server Functions.

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