Vastauksissa käytetty SQL ei ole aina täysin standardien mukaista vaan ratkaisut on tehty Access 97:ssä toimiviksi.
Mallikyselyt ovat saatavilla myös Access 97 -tietokantana:
SELECT SUM(feepaid) AS Summa
FROM Rental
SELECT AVG(feepaid) AS Keskiarvo
FROM Rental
SELECT COUNT(feepaid) AS LKM
FROM Rental
SELECT COUNT(feepaid)
FROM Rental
WHERE memberid = 2
SELECT TapeId, COUNT(tapeId) AS LKM
FROM Rental
GROUP BY Tapeid
SELECT Tape.tapeID, Catalog.title, tape.purchaseprice, catalog.rentalcharge
FROM tape, catalog
WHERE catalog.catalogid = tape.catalogid
ORDER BY Catalog.title
SELECT member.name, Catalog.Title, rental.daterented, feepaid
FROM Tape, Catalog, Rental, Member
WHERE Tape.catalogid = Catalog.catalogid
AND rental.memberid = member.memberid
AND rental.tapeid = tape.tapeid
SELECT Count(*) AS Tommin_vuokraukset
FROM Rental, Member
WHERE member.memberid = Rental.memberid
AND member.name = 'Tommi Lahtonen'
SELECT member.name, COUNT(*) AS LKM, SUM(feepaid) AS Summa
FROM Rental, Member
WHERE member.memberid = Rental.memberid
GROUP by member.name
SELECT member.name, SUM(feepaid) AS Summa
FROM Rental, Member
WHERE member.memberid = Rental.memberid
GROUP BY member.name
HAVING COUNT(*) >= 2
SELECT supplier.suppliername, SUM(rental.feepaid) AS Summa
FROM supplier, tape, rental
WHERE tape.tapeid = rental.tapeid
AND tape.supplierid = supplier.supplierid
GROUP BY supplier.suppliername
SELECT member.name, COUNT(rental.feepaid) AS LKM, SUM(rental.feepaid) as Summa
FROM member, rental
WHERE member.memberid = rental.memberid
GROUP BY member.name
HAVING SUM(rental.feepaid) > 30
ORDER BY SUM(rental.feepaid) ASC
SELECT m1.name, m2.name, m1.address, m2.address
FROM member as m1, member as m2
WHERE m1.address = m2.address
AND m1.name <> m2.name
SELECT *
FROM Member
WHERE MemberID NOT IN (
SELECT MemberID
FROM Rental
)
SELECT SUM(rental.feepaid) AS summa
FROM rental, member
WHERE rental.memberid = member.memberid
AND member.memberid NOT IN (
SELECT m1.memberid
FROM member as m1, member as m2
WHERE m1.address = m2.address
AND m1.name <> m2.name
)
SELECT Name, Year(Now()) - Year (DateJoined) AS Jäsen_vuodet
FROM member
/* ei toimi Accessissa :-( */
SELECT catalog.title, sum(rental.feepaid)
FROM tape LEFT OUTER JOIN rental ON rental.tapeid = tape.tapeid
INNER JOIN catalog ON catalog.catalogid = tape.catalogid
GROUP BY catalog.title
ORDER BY Catalog.title
/* Accessissa toimiva vaihtoehto */
SELECT catalog.title, SUM(rental.feepaid) AS Summa
FROM tape, catalog, rental
WHERE Tape.tapeID = Rental.tapeID
AND Tape.CatalogID = Catalog.CatalogID
GROUP BY catalog.title
UNION
SELECT catalog.title, 0
FROM catalog
WHERE CatalogID NOT IN (
SELECT CatalogID
FROM Tape, Rental
WHERE tape.tapeID = Rental.tapeID
)
ORDER BY Catalog.title