First use relational selection to keep the Saab cars, then a natural join to find their sales, and finally a relational join to identify the corresponding salespeople. A relational projection givesThe foreign keys ensure that every sale has the referenced car and salesperson. Several qualifying sales by the same person produce just one occurrence of their name under set semantics. Two distinct people with the same name also yield that name only once: this is the requested relational projection onto names.
Construct the required models and the actual customer–model purchases using relational projection and a natural join:The Cartesian product lists every obligation: every customer paired with every model present in
Car. The relational difference consists of unfulfilled obligations. Remove the customer identifiers occurring in this database relation and only then recover names:This is an explicit relational division construction with all customers as the candidate universe. If there are no models, there are no missing obligations, so every customer qualifies. A version starting solely from customers occurring in would incorrectly omit customers with no purchases in that empty-model case.The quantified object here is the
model database attribute, so two makes sharing a model value contribute one required value. If a different application identified a model by the pair (make, model), both the required and purchased relational projections would use that pair. Also, Car contains unsold cars: a model available only in unsold inventory is still a required model, and then nobody qualifies. Testing identifiers before names prevents different customers with the same name from pooling their purchases.An arithmetic mean is database aggregation, which is outside classical relational algebra. Use SQL:For non-null prices of qualifying sales, the result isEach sale contributes once, even when a car was sold on several invoices. Do not use
SELECT AVG(s.price) AS average_sale_price
FROM Sales AS s
JOIN Car AS c ON c.serialnumber = s.serialnumber
WHERE c.make = 'Saab';DISTINCT inside AVG: two different sales at the same price must both count. The car's database primary key makes the relational join match each sale to exactly one car, so the relational join introduces no extra weighting.If there are no qualifying non-null prices,
AVG returns SQL NULL. It ignores null prices if the implementation permits them; it does not replace an absent arithmetic mean with zero.[ref]Aggregate Functions in the PostgreSQL documentation. Section 9.21,
avg and the paragraph on aggregates with no selected rows Articles by others on the same topic
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