WHERE first.city = second.city
AND first.sname < second.sname;
Псевдонимам нет необходимости иметь именно такие имена.
SELECT cname, first.onum, second.onum
FROM Orders first, Orders second, Customers
WHERE first.cnum = second.cnum
AND first.cnum = Customers.cnum
AND first.onum < second.onum;
Ваш вывод может иметь некоторые отличия, но в вашем ответе все логические
компоненты должны быть такими же.
SELECT a.cname, a.city
FROM Customers a, Customers b
WHERE a.rating = b.rating
AND b.cnum = 2001;
Глава 10
1. SELECT *
FROM Orders
WHERE cnum =
(SELECT cnum
FROM Customers
WHERE cname = 'Cisneros');
Или
SELECT *
FROM Orders
WHERE cnum IN
(SELECT cnum
FROM Customers
WHERE cname = 'Cisneros');
SELECT DISTINCT cname, rating
FROM Customers, Orders
WHERE amt >
(SELECT AVG (amt)
FROM Orders)
AND Orders.cnum = Customers.cnum;
SELECT snum, SUM (amt)
FROM Orders
GROUP BY snum
HAVING SUM (amt) >
(SELECT MAX (amt)
FROM Orders);
Глава 11
SELECT cnum, cname
FROM Customers outer
WHERE rating =
(SELECT MAX (rating)
FROM Customers inner
WHERE inner.city = outer.city);
2. Решение с помощью соотнесенног7о подзапроса:
SELECT snum, sname
FROM Salespeople main
WHERE city IN
(SELECT city
FROM Customers inner
WHERE inner.snum < > main.snum);
Решение с помощью объединения:
SELECT DISTINCT first.snum, sname
