SUBQUERIES
ACTIVITY 4-1
• Create a database called Subexamples
• Start a logfile called ACT4-1
• Save the subtext.txt file
– Copy the contents of the file
– Paste into MYSQL
– This will create the tables we will use for all Activities
this week
• Make sure that all tables have been created and
that there is data in each
– SERVICES, CLIENTS, BRANCHES, BRANCHES_SERVICES
List all clients with branch offices in California only
mysql> SELECT cname, bdesc, bloc FROM clients, branches WHERE
clients.cid = branches.cid AND branches.bloc = 'CA';
+-----------------------------+----------------------+------+
| cname
| bdesc
| bloc |
+-----------------------------+----------------------+------+
| JV Real Estate
| Corporate HQ
| CA
|
| Rabbit Foods Inc
| Branch Office (West) | CA
|
| Sharp Eyes Detective Agency | Head Office
| CA
|
+-----------------------------+----------------------+------+
3 rows in set (0.03 sec)
List of all the branch offices belonging to "Rabbit Foods Inc".
You could run two SELECT statements:
mysql> SELECT cid FROM clients WHERE cname = 'Rabbit Foods
Inc';
+-----+
| cid |
+-----+
| 104 |
+-----+
1 row in set (0.00 sec)
AND
mysql> SELECT bdesc FROM branches WHERE cid = 104;
+----------------------+
| bdesc
|
+----------------------+
| Branch Office (East) |
| Branch Office (West) |
+----------------------+
2 rows in set (0.00 sec)
OR run one subquery:
mysql> SELECT bdesc FROM branches WHERE cid = (SELECT cid FROM
clients WHERE cname = 'Rabbit Foods Inc');
+----------------------+
| bdesc
|
+----------------------+
| Branch Office (East) |
| Branch Office (West) |
+----------------------+
2 rows in set (0.02 sec)
• A subquery makes it possible to combine two
or more queries into a single statement
• Uses the results of one query in the
conditional clause of the other
• Subqueries are usually regular SELECT
statements, and are separated from their
parent query by parentheses
You can nest subqueries to any depth, so long as the basic
rules on the previous slide are followed.
List the services used by Sharp Eyes Detective Agency
mysql> SELECT sname FROM services WHERE sid = (SELECT sid FROM
branches_services WHERE bid = (SELECT bid FROM branches WHERE
cid = (SELECT cid FROM clients WHERE cname = 'Sharp Eyes
Detective Agency')));
+------------+
| sname
|
+------------+
| Accounting |
+------------+
1 row in set (0.00 sec)
Subqueries are usually preceded by a conditional WHERE clause, which can
contain any of the following comparison and logical operators
= values are equal
<> values are unequal
<= value on left is less than or equal to value on right
>= value on left is greater than or equal to value on right
< value on left is less than value on right
> value on left is greater than value on right
BETWEEN value on left lies between values on right
NOT logical NOT
AND logical AND
OR logical OR
List of all those clients with exactly two branch offices.
First, you need to figure out a way to obtain the number of branch
offices per client
mysql> SELECT cid, COUNT(bid) FROM branches GROUP BY cid;
+-----+------------+
| cid | COUNT(bid) |
+-----+------------+
| 101 |
3 |
| 103 |
3 |
| 104 |
2 |
| 110 |
1 |
+-----+------------+
4 rows in set (0.02 sec)
Then filter out those with just two offices with a HAVING clause
mysql> SELECT cid, COUNT(bid) FROM branches GROUP BY cid
HAVING COUNT(bid) = 2;
+-----+------------+
| cid | COUNT(bid) |
+-----+------------+
| 104 |
2 |
+-----+------------+
1 row in set (0.00 sec)
Then hand the client ID over to the "clients" table in order to
get the client name
mysql> SELECT cname FROM clients WHERE cid = 104;
+------------------+
| cname
|
+------------------+
| Rabbit Foods Inc |
+------------------+
1 row in set (0.00 sec)
This subquery will take care of the three steps:
mysql> SELECT cname FROM clients WHERE cid = (SELECT cid FROM
branches GROUP BY cid HAVING COUNT(bid) = 2);
+------------------+
| cname
|
+------------------+
| Rabbit Foods Inc |
+------------------+
1 row in set (0.00 sec)
• The inner query is executed first - this query
takes care of grouping the branches by
customer ID and counting the number of
records (branch offices) in each group.
• Those customers which have exactly two
branch offices can easily be filtered out with a
HAVING clause
• The corresponding customer IDs are returned
to the main query, which then maps the IDs
into the customers table and returns the
corresponding customer name.
Select all those clients using the service with the maximum
service fee
mysql> SELECT cname, bdesc FROM clients, branches,
branches_services, services WHERE services.sid =
branches_services.sid AND clients.cid = branches.cid AND
branches.bid = branches_services.bid AND sfee = (SELECT
MAX(sfee) FROM services);
+----------------+--------------------------------+
| cname
| bdesc
|
+----------------+--------------------------------+
| JV Real Estate | Customer Grievances Department |
+----------------+--------------------------------+
1 row in set (0.00 sec)
Which sites are using more than 50% of all available services
(use the HAVING clause)
mysql> SELECT bid FROM branches_services GROUP BY bid HAVING
COUNT(sid) > (SELECT COUNT(*) FROM services)/2;
+------+
| bid |
+------+
| 1011 |
+------+
1 row in set (0.01 sec)
Which sites are using more than 50% of all available
services and get the branch name and client name as well
mysql> SELECT c.cid, c.cname, b.bid, b.bdesc FROM clients
AS c, branches AS b, branches_services AS bs WHERE c.cid =
b.cid AND b.bid = bs.bid GROUP BY bs.bid HAVING
COUNT(bs.sid) > (SELECT COUNT(*) FROM services)/2;
+-----+----------------+------+--------------+
| cid | cname
| bid | bdesc
|
+-----+----------------+------+--------------+
| 101 | JV Real Estate | 1011 | Corporate HQ |
+-----+----------------+------+--------------+
1 row in set (0.03 sec)
List all clients using all available services across their branch
offices
mysql> SELECT clients.cname FROM clients, branches,
branches_services WHERE branches.bid = branches_services.bid
AND branches.cid = clients.cid GROUP BY clients.cid HAVING
COUNT(sid) = (SELECT COUNT(*) FROM services);
+----------------+
| cname
|
+----------------+
| JV Real Estate |
+----------------+
1 row in set (0.00 sec)
List of all services being used by Branch ID 1031
(use the IN operator)
mysql> SELECT sname FROM services WHERE sid IN (SELECT sid
FROM branches_services WHERE bid = 1031);
+-----------------+
| sname
|
+-----------------+
| Recruitment
|
| Data Management |
| Administration |
+-----------------+
3 rows in set (0.00 sec)
List of all branches using the "Accounting" service (service ID 1)
mysql> SELECT bdesc FROM branches WHERE bid IN (SELECT bid
FROM branches_services WHERE sid = 1);
+-----------------------+
| bdesc
|
+-----------------------+
| Corporate HQ
|
| Accounting Department |
| Branch Office (East) |
| Branch Office (West) |
| Head Office
|
+-----------------------+
5 rows in set (0.00 sec
Now add the client name for each branch as well
mysql> SELECT cname, bdesc FROM branches, clients WHERE
branches.bid IN (SELECT bid FROM branches_services WHERE
sid = 1) AND clients.cid = branches.cid;
+-----------------------------+-----------------------+
| cname
| bdesc
|
+-----------------------------+-----------------------+
| JV Real Estate
| Corporate HQ
|
| JV Real Estate
| Accounting Department |
| Rabbit Foods Inc
| Branch Office (East) |
| Rabbit Foods Inc
| Branch Office (West) |
| Sharp Eyes Detective Agency | Head Office
|
+-----------------------------+-----------------------+
5 rows in set (0.00 sec)
Now just show the customer list
(use the DISTINCT function)
mysql> SELECT DISTINCT cname FROM branches, clients WHERE
branches.bid IN (SELECT bid FROM branches_services WHERE
sid = 1) AND clients.cid = branches.cid;
+-----------------------------+
| cname
|
+-----------------------------+
| JV Real Estate
|
| Rabbit Foods Inc
|
| Sharp Eyes Detective Agency |
+-----------------------------+
3 rows in set (0.00 sec)
Email your logfile for as Activtiy 4-1 submission.
Review your logfile as Activity 4-2 will continue with more
subqueries using the same database.