ACTIVITY 4-2 ANSWERS
1. List all client names with branches that use the "Accounting" service (service ID 1).
Breakdown:
outer query: client names (IN)
inner query: branches (IN)
inner query 2: service id
mysql> SELECT cname FROM clients WHERE cid IN (SELECT cid FROM branches WHERE bid IN
(SELECT bid FROM branches_services WHERE sid = 1));
+-----------------------------+
| cname
|
+-----------------------------+
| JV Real Estate
|
| Rabbit Foods Inc
|
| Sharp Eyes Detective Agency |
+-----------------------------+
3 rows in set (0.00 sec)
2. List all clients with individual branch offices having a monthly bill of $2000 or more
Breakdown:
outer query: client names (IN)
inner query: branches (IN)
inner query 2: service fee total $2000 or more (GROUP BY, HAVING, SUM)
mysql> SELECT cname FROM clients WHERE cid IN (SELECT cid FROM branches WHERE bid IN
(SELECT bid FROM branches_services AS bs, services AS s WHERE bs.sid = s.sid GROUP BY
bid HAVING SUM(sfee) >= 2000));
+-----------------------------+
| cname
|
+-----------------------------+
| JV Real Estate
|
| Rabbit Foods Inc
|
+-----------------------------+
2 rows in set (0.00 sec)
3. Rewrite the query above using the NOT operator. You should get the same answer as the query above.
Breakdown:
outer query: client names (IN)
inner query: branches (NOT IN)
inner query 2: branches_service (GROUP BY, HAVING, SUM)
mysql> SELECT cname FROM clients WHERE cid IN (SELECT cid FROM branches WHERE bid NOT
IN (SELECT bid FROM branches_services AS bs, services AS s WHERE bs.sid = s.sid GROUP
BY bid HAVING SUM(sfee) < 2000));
+-----------------------------+
| cname
|
+-----------------------------+
| JV Real Estate
|
| Rabbit Foods Inc
|
+-----------------------------+
2 rows in set (0.00 sec)
4. List the average services each branch office uses.
Breakdown:
outer query:
inner query:
AVG
COUNT branches_services
HINT: The inner query must first be aliased to a table name, or else MySQL will not know how to refer to
columns within it. If you don’t use an alias the result will be:
ERROR 1248 (42000): Every derived table must have its own alias
SELECT AVG(z.stotal) FROM (SELECT bid, COUNT(sid) AS stotal FROM branches_services
GROUP BY bid) AS z;
The correct result should be:
+---------------+
| AVG(z.stotal) |
+---------------+
|
1.7778 |
+---------------+
1 row in set (0.00 sec)
5. Now that you know the average. List the branches where the number of services is above this
average.
mysql> SELECT bid FROM branches_services GROUP BY bid HAVING COUNT(sid) > 1.7778;
+------+
| bid |
+------+
| 1011 |
| 1031 |
| 1041 |
| 1042 |
+------+
4 rows in set (0.00 sec)
6. Now combine the above two queries into one. You should get the same result/output as Question
#5.
Breakdown:
outer query:
inner query:
Query from #5
Query from #4
mysql> SELECT bid FROM branches_services GROUP BY bid HAVING COUNT(sid) > (SELECT
AVG(z.stotal) FROM (SELECT bid, COUNT(bid) AS stotal FROM branches_services GROUP BY
bid) AS z);
+------+
| bid |
+------+
| 1011 |
| 1031 |
| 1041 |
| 1042 |
+------+
4 rows in set (0.00 sec)
7. Now, (just because I’m mean), list the client names corresponding to the branches listed above.
Breakdown: Add an outer query to the query from #6 (DISTINCT)
mysql> SELECT DISTINCT cname FROM clients, branches, (SELECT bid FROM
branches_services GROUP BY bid HAVING COUNT(sid) > (SELECT AVG(z.stotal) FROM (SELECT
bid, COUNT(bid) AS stotal FROM branches_services GROUP BY bid) AS z)) AS x WHERE
clients.cid = branches.cid AND branches.bid = x.bid;
+------------------+
| cname
|
+------------------+
| JV Real Estate
|
| Rabbit Foods Inc |
| DMW Trading
|
+------------------+
3 rows in set (0.00 sec)
8. Delete all branches using the "Recruitment" service (service ID 2).
Breakdown:
outer query:
inner query:
Branches (DELETE, IN)
Branches_services, sid=2
mysql> DELETE FROM branches WHERE bid IN (SELECT bid from branches_services WHERE sid
= 2);
Query OK, 2 rows affected (0.00 sec)
9. Check the branches to make sure the query worked.
mysql> select * from branches;
+------+-----+--------------------------------+------+
| bid | cid | bdesc
| bloc |
+------+-----+--------------------------------+------+
| 1012 | 101 | Accounting Department
| NY
|
| 1013 | 101 | Customer Grievances Department | KA
|
| 1041 | 104 | Branch Office (East)
| MA
|
| 1042 | 104 | Branch Office (West)
| CA
|
| 1101 | 110 | Head Office
| CA
|
| 1032 | 103 | NE Region HO
| CT
|
| 1033 | 103 | NW Region HO
| NY
|
+------+-----+--------------------------------+------+
7 rows in set (0.00 sec)
10. Delete all those customers, any of whose branch offices generate service fee revenues of $500 or less
Breakdown:
outer query:
inner query:
Clients (DELETE, IN)
Branches_services (DISTINCT, GROUP BY, HAVING, SUM)
mysql> DELETE FROM clients WHERE cid IN (SELECT DISTINCT b.cid FROM branches AS b,
branches_services AS bs, services AS s WHERE b.bid = bs.bid AND bs.sid = s.sid GROUP
BY bs.bid HAVING SUM(sfee) <= 500);
Query OK, 1 row affected (0.00 sec)
11. Check the clients table to make sure the query worked.
mysql> select * from clients;
+-----+-----------------------------+
| cid | cname
|
+-----+-----------------------------+
| 101 | JV Real Estate
|
| 102 | ABC Talent Agency
|
| 104 | Rabbit Foods Inc
|
| 110 | Sharp Eyes Detective Agency |
+-----+-----------------------------+
4 rows in set (0.00 sec)
12. Find out which services are in use in 3 or more branch offices
mysql> SELECT sid FROM branches_services GROUP BY sid HAVING COUNT(bid) >= 3;
+-----+
| sid |
+-----+
|
1 |
|
3 |
|
4 |
+-----+
3 rows in set (0.00 sec)
13. Now increase the fee for the above services by 25%.
Breakdown:
outer query:
inner query:
Services, sfee (UPDATE)
Branches_services (GROUP BY, HAVING, COUNT)
mysql> UPDATE services SET sfee = sfee + (sfee * 0.25) WHERE sid IN (SELECT sid FROM
branches_services GROUP BY sid HAVING COUNT(bid) >= 3);
Query OK, 3 rows affected (0.22 sec)
Rows matched: 3 Changed: 3 Warnings: 0
14. Change all branches located in California so that they use the "Security" service instead of the
"Administration" service
Breakdown:
outer query:
inner query:
Branches_services, sfee (UPDATE, SET, IN)
Branches (bloc=’CA’)
mysql> UPDATE branches_services SET sid = 6 WHERE sid = 4 AND bid IN (SELECT bid FROM
branches WHERE bloc = 'CA');
Query OK, 1 row affected (0.00 sec)
Rows matched: 1 Changed: 1 Warnings: 0