Understanding the Problem and the Solution

In this article, we will explore a common problem in database querying - handling multiple values for an aggregate function. The question provided is about pulling out the top 2 months of sales by customer ID from a given table.

Background and Terminology

To understand the problem, let’s first define some key terms:

• Aggregate Function: An aggregate function is a mathematical operation that takes one or more input values and returns a single output value. Common examples include SUM, MAX, MIN, AVG.
• Partitioning: In the context of SQL queries, partitioning refers to dividing a table into smaller subsets based on certain conditions. This can be used to optimize query performance by reducing the amount of data that needs to be processed.

The Original Query and Error

The original query attempts to achieve the desired result using an aggregate function (SUM) with multiple values. However, the error message “too many values ORA-00913” indicates that the query is trying to return too many rows, which is not allowed in Oracle SQL.

# The Original Query
SELECT sum(net_sales_usd_spot), valid_period, customer_id 
FROM sales_trans_price_output 
WHERE valid_period IN (SELECT valid_period, SUM(net_sales_usd_spot) 
                       FROM sales_trans_price_output 
                       WHERE rank < 2)
GROUP BY valid_period, customer_id

The Correct Solution

The correct solution uses a subquery with row numbering to achieve the desired result.

# The Correct Query
SELECT *
FROM (
  SELECT t.*,
         ROW_NUMBER() OVER (PARTITION BY customer_id ORDER BY sales_amount DESC) rn
  FROM sales_trans_price t
)
WHERE rn <= 2
ORDER BY 1, 2 DESC

How it Works

Let’s break down the correct query step by step:

• ROW_NUMBER(): This function assigns a unique number to each row within a partition of a result set. The numbering starts at 1 and increments sequentially.
• PARTITION BY customer_id: This clause divides the rows into partitions based on the customer_id column.
• ORDER BY sales_amount DESC: This clause sorts the rows within each partition in descending order based on the sales_amount column.
• rn <= 2: This condition filters out the top 2 rows from each partition.

Example Use Case

Suppose we have a table sales_trans_price with the following data:

Running the correct query will return the following result:

Optimizations and Variations

The correct query can be optimized by using indexes on the customer_id and sales_amount columns.

# Optimized Query
CREATE INDEX idx_customer_id ON sales_trans_price (customer_id)
CREATE INDEX idx_sales_amount ON sales_trans_price (sales_amount)

SELECT *
FROM (
  SELECT t.*,
         ROW_NUMBER() OVER (PARTITION BY customer_id ORDER BY sales_amount DESC) rn
  FROM sales_trans_price t
)
WHERE rn <= 2
ORDER BY 1, 2 DESC

Additionally, the query can be modified to return more columns or to filter out rows based on additional conditions.

# Modified Query
SELECT t.*,
         ROW_NUMBER() OVER (PARTITION BY customer_id ORDER BY sales_amount DESC) rn
FROM sales_trans_price t
WHERE valid_period IN ('2', '5', '7')
AND rn <= 2
ORDER BY 1, 2 DESC

Conclusion

In this article, we explored a common problem in database querying - handling multiple values for an aggregate function. We provided the correct solution using a subquery with row numbering and discussed how to optimize it using indexes and additional modifications.

Last modified on 2023-12-20