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Source  : 2094024
Branch  : main
Author  : Joe Heffer <60133133+Joe-Heffer-Shef@users.noreply.github.com>
Time    : 2026-03-16 01:04:22 +0000
Message : Clarify aggregation challenge solution (#382)

- Add a paragraph to explain the SQL query in the solution
- Use full words for the column alias, rather than a single letter

Author: actions-user

02-sql-aggregation.md

@@ -226,11 +226,16 @@ Write a query that returns, from the `species` table, the number of
 
 ## Solution
 
+This query counts the number of times each value (Bird, Rabbit, Reptile or Rodent) in the `taxa` field occurs, defining a new field named `taxa_count` to hold the result.
+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 taxa_count
 FROM species
 GROUP BY taxa
-HAVING n > 10;
+HAVING taxa_count > 10;
 ```
 
 :::::::::::::::::::::::::

md5sum.txt

@@ -7,7 +7,7 @@
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