Move a challenge from 03-sql-joins to 02-sql-aggregation ?

[MathildeMousset]

In 03-sql-joins, I am confused by the inclusion of the the challenge within the JOINING episode:

Count the number of records in the surveys table that have a NULL value in the species_id column.

SELECT COUNT(*)
FROM surveys
WHERE species_id IS NULL;

As far as I understand, this challenge is less about joining than about counting, and dealing with counting nulls. My suggestion is to include it in the COUNT part of 02-sql-aggregation episode

Additional suggestions

• I would add information on the existence of IS NOT NULL command at this point, which is used in an example at the end of the episode, but not explicitly mentionned before

• I would perhaps add the following code, that can be used to highlight the difference of behaviour between COUNT(*) and COUNT(species_id):

SELECT COUNT(*) - COUNT(species_id)
FROM surveys