How can you output a list of all members, including the individual who recommended them (if any), without using any joins? Ensure that there are no duplicates in the list, and that each firstname + surname pairing is formatted as a column and ordered.
Schema reminder
DB schema

Expected Results

member recommender
Anna Mackenzie Darren Smith
Anne Baker Ponder Stibbons
Burton Tracy
Charles Owen Darren Smith
Darren Smith
David Farrell
David Jones Janice Joplette
David Pinker Jemima Farrell
Douglas Jones David Jones
Erica Crumpet Tracy Smith
Florence Bader Ponder Stibbons
Gerald Butters Darren Smith
Henrietta Rumney Matthew Genting
Henry Worthington-Smyth Tracy Smith
Hyacinth Tupperware
Jack Smith Darren Smith
Janice Joplette Darren Smith
Jemima Farrell
Joan Coplin Timothy Baker
John Hunt Millicent Purview
Matthew Genting Gerald Butters
Millicent Purview Tracy Smith
Nancy Dare Janice Joplette
Ponder Stibbons Burton Tracy
Ramnaresh Sarwin Florence Bader
Tim Boothe Tim Rownam
Tim Rownam
Timothy Baker Jemima Farrell
Tracy Smith

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Answers and Discussion Show

select distinct mems.firstname || ' ' ||  mems.surname as member,
	(select recs.firstname || ' ' || recs.surname as recommender 
		from cd.members recs 
		where recs.memid = mems.recommendedby
		cd.members mems
order by member;          

This exercise marks the introduction of subqueries. Subqueries are, as the name implies, queries within a query. They're commonly used with aggregates, to answer questions like 'get me all the details of the member who has spent the most hours on Tennis Court 1'.

In this case, we're simply using the subquery to emulate an outer join. For every value of member, the subquery is run once to find the name of the individual who recommended them (if any). A subquery that uses information from the outer query in this way (and thus has to be run for each row in the result set) is known as a correlated subquery.

Answering this question correctly requires the use of a subquery

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