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SQL Week 3 Practice Problems

Practice Topics

  • Advanced Joins
  • Subqueries
  • Set Operations

Advanced Joins

Overview

In this Week 2 post, you were introduced to a basic join. In that post, an INNER JOIN was used. In this post, you will learn about the difference between INNER JOIN and LEFT JOIN. These are the only two types of joins that you will need to understand.

INNER JOIN: An INNER JOIN will successfully display rows and fields from two tables when they each contain matching values found in the field(s) used to connect both tables together. Tables are connected by specifying common fields in the join condition within the FROM statement.

Let’s revisit the example where we display customer FirstName and LastName from the AdventureWorks2019.Person.Person table alongside the CustomerID and TotalDue from the AdventureWorks2019.Sales.SalesOrderHeader table. Reminder, we have to leverage the AdventureWorks2019.Sales.Customer table as a bridge between the Person and SalesOrderHeader tables because there isn’t a common field that connects the SalesOrderHeader and Person tables.

In the event that a CustomerID found in the SalesOrderHeader table is missing in the Customer table, the entire record spanning the SalesOrderHeader, Customer, and Person tables would be eliminated from the query results. This is due to the fact that INNER JOINs require matching values to be present across all connected tables.

LEFT JOIN: A LEFT JOIN will always display rows and fields from the first table listed in the FROM statement (ie. the “LEFT” table). If the second table listed in the FROM statement (ie. the “RIGHT” table) contains a matching value in the common field(s) specified in the join, then rows and fields from the “RIGHT” table will also be successfully displayed. Similar to an INNER JOIN, the common field(s) connecting two tables together are specified in the join condition within the FROM statement.

Let’s again revisit the example where we display customer FirstName and LastName from the AdventureWorks2019.Person.Person table alongside the CustomerID and TotalDue from the AdventureWorks2019.Sales.SalesOrderHeader table. Reminder, we have to leverage the AdventureWorks2019.Sales.Customer table as a bridge between the Person and SalesOrderHeader tables because there isn’t a common field that connects the SalesOrderHeader and Person tables.

If a CustomerID found in the SalesOrderHeader table is missing in the Customer table, we will still be able to display the record from the SalesOrderHeader table. However, we will not be able to display the corresponding record from the Customer table. We will see the rows and fields in the “LEFT” table because LEFT JOINs do not require matching values to be present in both connected tables.

Example: List the ProductID and ProductName fields from the AdventureWorks2019.Production.Product table along with the Rating and Comments fields from the AdventureWorks2019.Production.ProductReview table. Please only display products that have a review.

SQL Code Highlighting
SELECT
    a.ProductID,
    a.Name,
    b.Rating,
    b.Comments
FROM
    AdventureWorks2019.Production.Product a
INNER JOIN
    AdventureWorks2019.Production.ProductReview b
    ON a.ProductID = b.ProductID;

Example: List the ProductID and ProductName fields from the AdventureWorks2019.Production.Product table along with the Rating and Comments fields from the AdventureWorks2019.Production.ProductReview table. Please display all products, even if they don’t have a review.

SQL Code Highlighting
SELECT
    a.ProductID,
    a.Name,
    b.Rating,
    b.Comments
FROM
    AdventureWorks2019.Production.Product a
LEFT JOIN
    AdventureWorks2019.Production.ProductReview b
    ON a.ProductID = b.ProductID;

Example: List the ProductID and ProductName fields from the AdventureWorks2019.Production.Product table that do not have a review in the AdventureWorks2019.Production.ProductReview table. Hint: Use a LEFT JOIN, but add a condition in the WHERE statement that only retains records when the Rating field IS NULL. I call this a negative join because you are using a LEFT JOIN to identify records from your “LEFT” table that do not have a corresponding record in the “RIGHT” table.

SQL Code Highlighting
SELECT
    a.ProductID,
    a.Name,
    b.Rating,
    b.Comments
FROM
    AdventureWorks2019.Production.Product a
LEFT JOIN
    AdventureWorks2019.Production.ProductReview b
    ON a.ProductID = b.ProductID
WHERE
    B.Rating IS NULL;

Self Practice

  • Write a query to perform an inner join between Sales.SalesOrderHeader and Sales.Customer on CustomerID and select SalesOrderID, OrderDate, and CustomerID.
  • Write a query to perform a left join between Production.Product and Production.ProductReview on ProductID and select all columns from Product and Comments from ProductReview.
SQL Code Highlighting
SELECT
    soh.SalesOrderID,
    soh.OrderDate,
    soh.CustomerID
FROM
    AdventureWorks2019.Sales.SalesOrderHeader soh
INNER JOIN
    AdventureWorks2019.Sales.Customer c
    ON soh.CustomerID = c.CustomerID;
SQL Code Highlighting
SELECT
    p.*,
    pr.Comments
FROM
    AdventureWorks2019.Production.Product p
LEFT JOIN
    AdventureWorks2019.Production.ProductReview pr
    ON p.ProductID = pr.ProductID;

Subqueries

Overview

Sometimes, you need to embed a query within a query. This is called a subquery. There are several reasons for this:

  • Occasionally, you need to create dynamic lookups based on live data within your database.
  • Sometimes, you need to return the latest record within a defined group. For example, maybe you want to identify what a customer ordered on his/her most recent purchase.
  • When creating a metric summary, you may need to mix data at different levels of granularity. For example, you may need to calculate the distribution of sales for each product.
  • More complex queries often require logic to be processed sequentially. In other words, you may need to preprocess data before aggregating it.
A subquery can be used in the SELECT, FROM, and WHERE statements. Most of the time, it is used in the FROM statement and the output of a subquery is treated exactly like a table. This means that you can SELECT fields from the results of a subquery. You can even join other tables to the results of a subquery. In the “Self-Practice” section, you will practice using subqueries in the WHERE statement.

Example: Using the AdventureWorks2019.Sales.SalesOrderDetail table, calculate the % of Total Sales for each ProductID per SalesOrderID. Total sales can be determined by using aggregation within a subquery. In the output, include the Name from the AdventureWorks2019.Production.Product table and give it an alias name of ProductName. You haven’t yet learned how to do math in SQL. To determine the % of Total Sales, you can divide the LineTotal by the TotalSales that you will derive in your subquery. Division works exactly the same as it does in Excel; use /.

SQL Code Highlighting
SELECT
    a.SalesOrderID,
    a.ProductID,
    b.Name as ProductName,
    c.TotalSales,
    a.LineTotal / c.TotalSales as "Percent of Total Sales"
FROM
    AdventureWorks2019.Sales.SalesOrderDetail a
INNER JOIN
    AdventureWorks2019.Production.Product b
    ON a.ProductID = b.ProductID
INNER JOIN
    (
    SELECT
        SalesOrderID,
        SUM(LineTotal) as TotalSales
    FROM
        AdventureWorks2019.Sales.SalesOrderDetail
    GROUP BY
        SalesOrderID
    ) c
    ON a.SalesOrderID = c.SalesOrderID;
    

Self Practice

  • Write a query to select FirstName and LastName from the Person.Person table where BusinessEntityID is in a subquery that selects BusinessEntityID from the HumanResources.Employee table.
  • Write a query to select all products from the Production.Product table where ListPrice is greater than the average ListPrice of all products.
  • Write a query that returns every product Name purchased for each customer’s most recent order. Use the MAX(SalesOrderID) for each CustomerID to identify each customer’s most recent order. Display the SalesOrderID, CustomerID, and product Name
SQL Code Highlighting
SELECT
    FirstName,
    LastName
FROM
    AdventureWorks2019.Person.Person
WHERE
    BusinessEntityID IN (
        SELECT BusinessEntityID
        FROM AdventureWorks2019.HumanResources.Employee
    );
SQL Code Highlighting
SELECT
    *
FROM
    AdventureWorks2019.Production.Product
WHERE
    ListPrice > (
        SELECT AVG(ListPrice)
        FROM AdventureWorks2019.Production.Product
    );
SQL Code Highlighting
SELECT
    a.SalesOrderID,
    c.CustomerID,
    b.Name as ProductName
FROM
    AdventureWorks2019.Sales.SalesOrderDetail a
INNER JOIN
    AdventureWorks2019.Production.Product b
    ON a.ProductID = b.ProductID
INNER JOIN
    (
    SELECT
        CustomerID,
        MAX(SalesOrderID) as LastOrderID
    FROM
        AdventureWorks2019.Sales.SalesOrderHeader
    GROUP BY
        CustomerID
    ) c
    ON a.SalesOrderID = c.LastOrderID;
    

Set Operations

Overview

The UNION operator in SQL is used to combine the results of two or more SELECT queries into a single result set. The key points about UNION are:

  • Distinct Results: UNION removes duplicate rows from the combined result set. If you want to include duplicates, you can use UNION ALL.
  • Same Number of Columns: Each SELECT query within the UNION must have the same number of columns in the result sets.
  • Compatible Data Types: The columns must have compatible data types in each SELECT query.

Example: Create a BusinessEntityID list of people with contact information stored in either the AdventureWorks2019.Person.Address or AdventureWorks2019.Person.EmailAddress tables. Ensure that the list is unique.

SQL Code Highlighting
SELECT 
    BusinessEntityID
FROM
    AdventureWorks2019.Person.Address
UNION
SELECT
    BusinessEntityID
FROM
    AdventureWorks2019.Person.EmailAddress;

The INTERSECT operator in SQL is used to return only the rows that are common to the result sets of two SELECT queries. The key points about INTERSECT are:

  • Distinct Results: INTERSECT automatically removes duplicate rows from the result set.
  • Same Number of Columns: Each SELECT query within the INTERSECT must have the same number of columns in the result sets.
  • Compatible Data Types: The columns must have compatible data types in each SELECT query.

Example: Find common CustomerID values from AdventureWorks2019.Sales.SalesOrderHeader and BusinessEntityID values from AdventureWorks2019.Person.BusinessEntityAddress to see which active customers have an address on file.

SQL Code Highlighting
SELECT
    CustomerID FROM
    AdventureWorks2019.Sales.SalesOrderHeader
INTERSECT
SELECT
    BusinessEntityID AS CustomerID FROM
    AdventureWorks2019.Person.BusinessEntityAddress;

Self Practice

  • Write a query to find all BusinessEntityID that are either in the HumanResources.Employee table or the Sales.Customer table.
  • Write a query to find all ProductID that are in both the Production.Product table and the Sales.SalesOrderDetail table.
SQL Code Highlighting
SELECT
    BusinessEntityID
FROM
    AdventureWorks2019.HumanResources.Employee
UNION
SELECT
    CustomerID AS BusinessEntityID
FROM
    AdventureWorks2019.Sales.Customer;
SQL Code Highlighting
SELECT
    ProductID
FROM
    AdventureWorks2019.Production.Product
INTERSECT
SELECT
    ProductID
FROM
    AdventureWorks2019.Sales.SalesOrderDetail;

Conclusion

As we wrap up this week’s SQL homework assignments, we’ve delved into even more advanced aspects of SQL, including complex joining, subqueries, and set operations. These exercises are designed to build your confidence and proficiency in SQL, equipping you with the skills necessary for more advanced queries in the future.

Remember, consistent practice is key to mastering SQL. If you find yourself needing extra help or wanting to deepen your understanding, don’t hesitate to sign up for our private or group training sessions. These sessions offer personalized guidance and a collaborative learning environment to further enhance your SQL skills. Stay tuned for next week’s assignments, and keep pushing your boundaries as you continue on your journey to becoming an SQL expert!

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