Listing Keys with the Same Value in SQL

SQL is a powerful language for managing relational databases, and one of its most fundamental operations is querying data. In this article, we’ll explore how to list keys that have the same value in a database table.

Understanding the Problem Statement

The problem statement begins by describing a table named ABC with two columns: key and val. The key column is of type NUMBER(5), which means it can store integers up to 5 digits, while the val column is also of type NUMBER(5) for consistency. Two rows are inserted into this table:

CREATE TABLE ABC
(
    key NUMBER(5), 
    val NUMBER(5)
);

insert into ABC (key, val) values (1,1);
insert into ABC (key, val) values (1,2);
insert into ABC (key, val) values (1,3);
insert into ABC (key, val) values (2,3);

We’re interested in finding all keys that have the same value ‘val’. This means we want to identify rows where both key and val are equal.

A Delimited List vs. Separate Columns

One approach to solving this problem is to retrieve a delimited list of keys for each unique value in the val column, as long as there’s more than one row with that value. This requires using SQL aggregation functions like listagg().

Using Aggregation and LISTAGG()

The provided solution uses GROUP BY val followed by HAVING count(*) > 1. Here’s a breakdown of this query:

SELECT listagg(key, ',') WITHIN GROUP (ORDER BY key) keys,
       val
FROM abc
GROUP BY val
HAVING count(*) > 1;

Let’s analyze how this works:

• GROUP BY val groups the rows in our table by their values in the val column. This means we’re essentially categorizing each row based on its value.
• count(*) > 1 filters the grouped results to only include those categories with more than one row. Since we want keys that have the same value, these categories represent groups of rows with identical key and val values.
• listagg(key, ',') WITHIN GROUP (ORDER BY key) generates a string containing all keys in each category, separated by commas and sorted alphabetically.

The resulting output will be something like:

keys                 val
1,2,3                  1

This means there’s one row with key values 1, 2, and 3 because it shares the same value in val. Similarly, for category 2, we only have a single key because its corresponding rows only share that specific val.

Alternative Solution: Using Subqueries

Another approach to solving this problem is by using subqueries. Here’s how you could do it:

SELECT k1.key
FROM abc k1
JOIN (
  SELECT val, listagg(key, ',') WITHIN GROUP (ORDER BY key) AS key_list
  FROM abc
  GROUP BY val
  HAVING count(*) > 1
) k2 ON k1.val = k2.val

In this solution:

• We start by selecting rows from abc that share the same value in val using a subquery.
• Within the subquery, we group and aggregate rows with the same value to produce the list of keys.
• Then, we join these aggregated results back onto our original table using matching values between k1.val and k2.val. This allows us to retrieve both the shared value (val) and its corresponding keys.

Handling Special Cases

When working with SQL, it’s often important to consider edge cases that might affect your solution. For instance:

• What happens if there are multiple keys in a single category? In our current solutions, the listagg() function would treat them as separate values for display.
• How do we handle categories without any shared key(s)? Our existing approaches will simply return an empty set of results for these cases.

Best Practices and Additional Tips

When working with SQL queries that involve grouping or aggregating data:

• Test thoroughly: Make sure to run your queries against sample data before applying them to real-world datasets.
• Understand what the functions do: Take a moment to review how aggregation functions like LISTAGG() work, as their behavior can be counterintuitive without understanding their syntax and usage.

In conclusion, listing keys that have the same value in SQL requires some finesse with aggregation functions. By breaking down the problem into manageable pieces and choosing the right tools (in this case, both GROUP BY and LISTAGG()), we can effectively solve it even when dealing with a multitude of unique values.

Additional Examples

To give you more practice with your SQL skills, try answering these questions on your own:

• Given a table called “colors” that contains the names of colors (color_name) and their hex codes (hex_code), write an SQL query to list all colors whose hex code starts with #.

SELECT color_name, hex_code FROM colors WHERE hex_code LIKE '#%';

• Suppose we want to build a table called “scores” that tracks scores over time for athletes in different sports. Write an SQL query to calculate the average score across all games played by each athlete.

SELECT athlete, sport, AVG(score) AS average_score
FROM scores
GROUP BY athlete, sport;

• How can we use SQL to identify which cities have populations that are roughly two standard deviations away from the mean population of their respective states?

This problem might be solved using a combination of GROUP BY, AVG(), and some clever subquerying.

Last modified on 2025-03-04