Retrieving the Most Expensive Movie and Its Neighbors in Oracle SQL

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In this article, we’ll explore different approaches to retrieve the most expensive movie and its neighboring records from an Oracle database. We’ll delve into various techniques, including using ORDER BY conditions, ranking columns, and utilizing subqueries.

Introduction

The question at hand is to find the most expensive movie in a collection of movies with their corresponding purchase prices. However, instead of simply retrieving the record with the highest price, we want to get the top 2 records, including the most expensive one and its neighboring values.

Background: Oracle SQL Basics

Before diving into the solutions, let’s quickly review some basic concepts in Oracle SQL:

• Subqueries: A subquery is a query nested inside another query. We’ll use subqueries to retrieve the maximum purchase price.
• Join: In this case, we’re working with a single table (movie), but in more complex scenarios, joins are used to combine data from multiple tables.

Solution 1: Using ORDER BY Condition

One straightforward approach is to use an ORDER BY condition to sort the records by purchase price in descending order and then fetch the top 2 rows:

SELECT *
FROM movie
ORDER BY purchase_price DESC;
FETCH FIRST 2 ROWS ONLY;

This query works as follows:

1. The SELECT * statement retrieves all columns from the movie table.
2. The ORDER BY purchase_price DESC clause sorts the records in descending order based on the purchase_price column.
3. Finally, the FETCH FIRST 2 ROWS ONLY statement limits the output to the top 2 rows.

Solution 2: Ranking Columns

Another approach is to use a ranking function (such as RANK() or DENSE_RANK()) to assign a rank to each record based on its purchase price, and then fetch the top 2 records:

SELECT *
FROM (
    SELECT *
    FROM movie
    ORDER BY purchase_price DESC
)
WHERE RANK() OVER (ORDER BY purchase_price DESC) <= 2;

This query works as follows:

1. The inner SELECT * statement retrieves all columns from the movie table, sorts them in descending order by purchase_price, and assigns a rank to each record.
2. The outer SELECT * statement filters the output to include only records with a rank less than or equal to 2.

Solution 3: Using Subqueries

We can also use a subquery to find the maximum purchase price and then fetch all records with a price greater than or equal to this value:

SELECT *
FROM movie
WHERE purchase_price >= (
    SELECT MAX(purchase_price) FROM movie);

This query works as follows:

1. The inner SELECT MAX(purchase_price) statement finds the maximum purchase price in the movie table.
2. The outer SELECT * statement filters the output to include only records with a purchase_price greater than or equal to the maximum value found.

Solution 4: Using Between Max and Min

Another approach is to use a condition that checks if the purchase price is between the minimum and maximum values:

SELECT *
FROM movie
WHERE purchase_price BETWEEN (SELECT MIN(purchase_price) FROM movie)
                         AND (SELECT MAX(purchase_price) FROM movie);

This query works as follows:

1. The inner SELECT MIN() and MAX() statements find the minimum and maximum purchase prices in the movie table.
2. The outer SELECT * statement filters the output to include only records with a purchase_price between these two values.

Conclusion

In conclusion, there are multiple ways to retrieve the most expensive movie and its neighboring records from an Oracle database. We’ve explored four different approaches using ORDER BY conditions, ranking columns, subqueries, and using between max and min values. Depending on the specific requirements of your use case, you may find that one approach is more suitable than others.

Further Reading

For a deeper dive into Oracle SQL concepts, we recommend checking out:

• Oracle’s Official Documentation: This is the official source of documentation for Oracle Database.
• SQL Server Tutorial by Tutorials Point: This tutorial covers SQL Server concepts, including ordering and ranking data.

Last modified on 2023-10-18