CIS 631 – RA/SQL HWK
Dr. H. Assadipour
Queries:
1. Get the names of parts that cost less than 20.00.
SQL> SELECT *
2 FROM part
3 WHERE price<20;
PNO PNAME
QOH
PRICE PLEVEL
---------- ------------------------------ ---------- ---------- ---------10506 Land Before Time I
200
19.99
20
10507 Land Before Time II
156
19.99
20
10508 Land Before Time III
190
19.99
20
10509 Land Before Time IV
60
19.99
20
10701 When Harry Met Sally
120
19.99
30
10800 Dirty Harry
140
14.99
30
6 rows selected.
Re sults ← Π pname ,etc. (σ prics < 20 ( part ))
RA:
2. Get the pairs of customer number values of customers having the same zip code.
SQL> l
1 SELECT c1.cno, c2.cno
2 FROM customer c1, customer c2
3 WHERE c1.zip = c2.zip
4* AND c1.cno < c2.cno
SQL> /
CNO
CNO
---------- ---------1111
2222
RA:
Re sults ← Π c1. zip =c 2. zip (σ c1. zip =c 2. zip (customer _ c1
c1.cno < c 2.cno
customer _ c 2))
c1.cno < c 2.cno
3. Get the names of customers who have ordered parts ONLY from employees living in Wichita.
SQL> SELECT DISTINCT cname
2 FROM customer c, order o, employee e, zipcode z
3 WHERE c.cno = o.cno and o.eno = e.eno and e.zip = z.zip
4 AND z.city ='Wichita';
CNAME
-------------------Barbara
Charles
1
RA:
R1 ← Π ename,eno (σ city −'Wichita ' (employee _ e
e. zip = z . zip
Re sults ← Π c.cname (σ c.cno = o.ono customer _ c
zipcode _ z ))
order _ o
o .eno = R1.eno
R1)
4. Get the names of employees along with their total sales for the year 1995.
To obtain the employee names and sales for all times and years:
SQL> BREAK ON ename;
SQL> SELECT ename, od.qty*p.price
2 FROM employee e, order o, odetail od, part p
3* WHERE e.eno=o.eno and o.ono=od.ono and od.pno=p.pno
ENAME
OD.QTY*P.PRICE
--------------- -------------Jones
19.99
19.99
39.98
59.97
14.99
24.99
99.96
Smith
24.99
19.99
9 rows selected.
To obtain the names of employees along with their total sales for 1995:
SQL> SELECT ename , SUM(od.qty*price)
2 FROM employee e, order o, odetail od, part p
3 WHERE e.eno=o.eno and o.ono=od.ono and od.pno=p.pno, and to_char(received, ‘YY’)=’95’
4* GROUP BY ename
SQL> /
ENAME
SUM(OD.QTY*PRICE)
--------------- ----------------Jones
99.96
Smith
44.98
RA:
Re sults ← Π ename , sum ( qty* price ) (σ YY ='95' (employee _ ee.eno =o.eno
o .ono = od .ono
o det ails _ od
od . pno = p . pno
order _ o)
part _ p ))
5. Get the names of customers who have placed the highest number of orders.
The following query finds the count for customer orders:
SQL> CREATE VIEW maxcounter as
1 SELECT cname, COUNT(*) as counts
2 FROM customer c, order o
3 WHERE c.cno=o.cno
4 GROUP BY cname;
2
SQL> View created.
SQL> SELECT * FROM maxcounter;
CNAME
COUNT(*)
-------------------- ---------Barbara
1
Bertram
1
Charles
2
SQL> SELECT cname, m.counts
2 FROM maxcounter m
3 WHERE m.counts = (SELECT MAX(m1.counts)
4 FROM maxcounter m1);
CNAME
COUNTS
-------------------- ---------Charles
2
RA:
max counter (cno, no_of_orders) ← ((cno
ℑ
COUNT ono) ℑ MAX no_of_orders))
Re sults ← Π cname (σ c.cno= mx.cno max counter _ mx
customer _ c)
6. Get the names of parts that have been ordered the most (in terms of quantity ordered, not
number of orders)
Once again, I have created a view of the total quantity for each part and then used the MAX
function to obtain the maximum value from the totals. This can be combined into one query.
SQL> CREATE VIEW maxqty as
SELECT pname, sum(od.qty) as total
FROM part p, order o, odetail od
WHERE p.pno=od.pno and od.ono=o.ono
GROUP BY pname;
SQL> View created.
SQL> SELECT * FROM maxqty;
PNAME
TOTAL
------------------------------ ---------Dirty Harry
1
Dr. Zhivago
1
Land Before Time I
1
Land Before Time II
1
Land Before Time III
2
Land Before Time IV
3
Sleeping Beauty
5
When Harry Met Sally
1
8 rows selected.
3
SQL> SELECT pname, total
2 FROM maxqty
3 WHERE maxqty.total = (SELECT MAX(total)
4* FROM maxqty)
PNAME
TOTAL
------------------------------ ---------Sleeping Beauty
5
R1( pno, no _ of _ qty ) ← ( pno ℑ SUM (qty )) (ℑ MAX no_of_orders))
Re sults ← Π cname (σ p. pno =T 1. pno R1
part _ p )
4