From 366c2585d7986ba28cd69125f7bb65f08814fe40 Mon Sep 17 00:00:00 2001
From: Joe Heffer <60133133+Joe-Heffer-Shef@users.noreply.github.com>
Date: Fri, 10 Jan 2025 11:26:12 +0000
Subject: [PATCH] Clarify aggregation challenge solution

- Explain the SQL query
- Use full words for the column alias, rather than a single letter
---
 episodes/02-sql-aggregation.md | 7 ++++++-
 1 file changed, 6 insertions(+), 1 deletion(-)

diff --git a/episodes/02-sql-aggregation.md b/episodes/02-sql-aggregation.md
index a8044d69..91aaaec4 100644
--- a/episodes/02-sql-aggregation.md
+++ b/episodes/02-sql-aggregation.md
@@ -226,8 +226,13 @@ Write a query that returns, from the `species` table, the number of
 
 ## Solution
 
+This query counts the number of species records that contain each value of the `taxa` field and names that result `species_count`.
+The `GROUP BY` clause means the query will create an aggregated table with one row for each taxa.
+Only those `taxa` values that have more than ten records will be included because of the `HAVING` clause.
+This filtering is applied _after_ grouping has been done.
+
 ```sql
-SELECT taxa, COUNT(*) AS n
+SELECT taxa, COUNT(*) AS species_count
 FROM species
 GROUP BY taxa
 HAVING n > 10;