1. 1. SQL Header of tables: –Head(Custoemrs) = (cid, cname, city, discnt) –Head(Orders) = (ordno, month, cid, pid, qty, dollars) –Head(Products) = (pid, pname, city, quantity, price) –Head(Agents) = (aid, aname, city, percent)

2. 1.1 SQL basic- projection List all customer name Select cname from customers Select distinct cname from customers

3. 1.1 SQL basic- where List all customers who live in Dallas –Select cid, cname from customers where city = 'Dallas' List all customers who live in Dallas and discount is bigger than 5. –Select cid, cname from customers where city = ‘Dallas’ and discnt > 5

4. 1.1 SQL basic- join List all customer (cid and cname) who ordered comb –Select distinct c.cid, c.cname from customers c, products p, orders o where c.cid = o.cid and p.pid = o.pid and p.pname = 'comb' List all agents (aid, aname) who have customers who live in ‘Dallas’ –Select distinct a.aid, a.aname from customers c, agents a, orders o where c.cid = o.cid and a.aid = o.aid and c.city = 'Dallas'

5. 1.1 SQL basic- group by List the total order amount (dollars) of each customer (cid) –Select cid, sum(dollars) from orders group by cid List total order amount (dollars) of each customer (cid, cname) –Select o.cid, c.cname, sum(o.dollars) from customers c, orders o where c.cid = o.cid group by o.cid, c.cname

6. 1.1 SQL basic- group by List total order amount (dollars) of each customer (cid, cname) whose total order amount is more than 1500 –Select o.cid, c.cname, sum(o.dollars) as total_sales from customers c, orders o where c.cid = o.cid group by o.cid, c.cname having sum(dollars) > 1500

7. 1.2 SQL advanced - join For each agent (aid, aname) count the number of customers who place orders with them. –Select a.aid, a.aname, count(cid) as number_of_customers from agents a, orders o where a.aid = o.aid group by a.aid, a.aname

8. 1.2 SQL advanced - join For each agent (aid, aname) count the number of customers who place orders with them. If an agent does not have any orders with any customer, his/her info should also be included in the result. –Select a.aid, a.aname, count(cid) as number_of_customers from agents a left join orders o on a.aid = o.aid group by a.aid, a.aname

9. 1.2 SQL advanced – MAX/MIN List agents (aid, aname) who sold most products (dollars). –select a.aid, a.aname from agents a, orders o where a.aid = o.aid group by a.aid, aname having sum(dollars) >= all (select sum(dollars) from orders group by aid )

10. 1.2 SQL advanced – MAX/MIN List agents (aid, aname) who sold most products (dollars) in jan. –select a.aid, a.aname from agents a, orders o where a.aid = o.aid and o.month = 'jan' group by a.aid, aname having sum(dollars) >= all (select sum(dollars) from orders where month = 'jan' group by aid)

11. 1.2 SQL advanced – MAX/MIN List products (pid, pname) that are ordered most (qty) by customers who live in Dallas –select p.pid, p.pname from products p, orders o, customers c –where p.pid = o.pid and c.cid = o.cid and c.city = 'Dallas' –group by p.pid, p.pname –having sum(qty) >= all –(select sum(qty) from orders o, customers c where c.cid = o.cid and c.city = 'Dallas' group by o.pid)

12. 1.2 SQL advanced – All List agents (aid, aname) who place orders with all customers in Dallas –Select a.aid, a.aname from agents a where not exists (select * from customers c where c.city = 'dallas' and not exists (select * from orders o where o.cid = c.cid and o.aid = a.aid))

13. 1.2 SQL advanced – All List customers (cid, cname) who order all the products produced in Newark –select c.cid, c.cname from customers c where not exists (select * from products p where city = 'Newark' and not exists (select * from orders o where o.pid = p.pid and o.cid = c.cid))

14. 2. Advance SQL topics Transact SQL Stored Procedure User Defined Function Cursor Trigger View

15. 2.2 Stored Procedure create proc count_customer_by_city @city char(10), @counter int output as select @counter = count(*) from customers where city = @city return declare @mycounter int exec count_customer_by_city 'Dallas', @mycounter output select @mycounter

16. 2.3 UDF create function total_dollars(@customerid char(4)) returns money as begin declare @sum money select @sum = sum(dollars) from orders where cid = @customerid return @sum End SELECT *, zhoupf.total_dollars(cid) as total_order_amount FROM customers select * from zhoupf.total_dollars('c001') exec zhoupf.total_dollars('c001')

17. 2.3 UDF CREATE FUNCTION total_avg_orders (@customerid char(4) ) RETURNS @total_avg_orders_tab TABLE ( total_order money, avg_order money ) AS BEGIN INSERT @total_avg_orders_tab SELECT sum(dollars), avg(dollars) FROM orders WHERE cid = @customerid RETURN END select * from total_avg_orders('c001') //This is correct because the return value is a table

18. 2.5 Trigger create table Boston_Customers ( cid char(4) not null, cname varchar(15), city varchar(15), discnt decimal(10,2), primary key(cid) );

19. 2.5 Trigger create trigger insert_customer on customers after insert as insert into Boston_customers select * from inserted I where I.city = ‘Boston’ insert into customers values ('c019', 'Tony', 'Boston', 16) select * from boston_customers

20. 3. ER diagram Entity (has attributes) Relationship (can also has attributes) –Many-to-many relationship –Many-to-one relationship –One-to-one relationship

21. 3. ER diagram NU library –Each borrower has a borrower_id, name. –Each library staff has a staff_id, name. –Each (type) book has a ISBN, name, author, number_of_copies. –Each borrower can borrow multiple books. Each (type) of book can be borrowed by multiple borrowers. –When a borrower borrows a book from a staff, the borrow date and return date will be recorded.