How to write a query that returns the median value in SQL?
Sep 23, 2025 am 03:52 AM
The median in SQL is calculated using ROW_NUMBER() and COUNT() or PERCENTILE_CONT(0.5). First, assign row numbers and get total count via window functions. Then determine middle positions: if count is odd, pick the central value; if even, average the two middle values. Use a CTE to compute positions and filter rows accordingly. For databases supporting PERCENTILE_CONT (like SQL Server, PostgreSQL), directly use SELECT PERCENTILE_CONT(0.5) WITHIN GROUP (ORDER BY value) for simpler syntax. MySQL 8.0 supports window functions but lacks PERCENTILE_CONT; earlier versions require manual method. SQLite does not support percentile functions, so use the ROW_NUMBER approach. Prefer PERCENTILE_CONT when available, otherwise simulate with ranking and arithmetic.
To get the median value in SQL, you need to use window functions and some conditional logic because there's no built-in MEDIAN() function in most SQL databases (except Oracle and some newer versions of others). The method varies slightly depending on your database system, but here’s a standard approach using ROW_NUMBER(), COUNT(), and arithmetic to find the middle value(s).
Using ROW_NUMBER() and COUNT()
This method works across many SQL platforms like PostgreSQL, SQL Server, and MySQL (8.0 ).Assume you have a table called numbers with a column value. Here's how to compute the median:
WITH sorted_values AS (
SELECT
value,
ROW_NUMBER() OVER (ORDER BY value) AS row_num,
COUNT(*) OVER () AS total_count
FROM numbers
),
median_positions AS (
SELECT
CASE
WHEN total_count % 2 = 1 THEN (total_count 1) / 2
ELSE total_count / 2
END AS pos1,
CASE
WHEN total_count % 2 = 1 THEN (total_count 1) / 2
ELSE (total_count / 2) 1
END AS pos2
FROM sorted_values
LIMIT 1
)
SELECT AVG(value * 1.0) AS median_value
FROM sorted_values, median_positions
WHERE row_num IN (pos1, pos2);How it works:
- The first CTE assigns a row number to each value in sorted order and gets the total count.
- The second CTE calculates the one or two middle positions:
- If count is odd: both positions are the same (middle).
- If even: take the average of the two center values.
- The final query selects values at those positions and averages them to handle both cases uniformly.
Alternative: Using PERCENTILE_CONT()
If your database supports it (e.g., SQL Server, PostgreSQL, Oracle), this is simpler.SELECT PERCENTILE_CONT(0.5) WITHIN GROUP (ORDER BY value) AS median_value FROM numbers;
This directly computes the 50th percentile (median) and handles even/odd counts automatically.
Notes by Database
-
MySQL: No
PERCENTILE_CONTbefore 8.0; use the first method. - PostgreSQL: Supports both methods.
-
SQL Server: Full support for
PERCENTILE_CONT. - SQLite: No window function support for percentiles; use manual ranking.
Basically, use PERCENTILE_CONT(0.5) if available. Otherwise, simulate it using row numbers and count.
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