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 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|Head Office |CA| rows in set (0.00 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 | row in set (0.17 sec) mysql> SELECT bdesc FROM branches WHERE cid = 104; | bdesc | | Branch Office (East) | | Branch Office (West) | rows in set (0.17 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) | rows in set (0.22 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

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 greater than or equal to 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 customers with exactly two branch offices. First, you need to figure out a way to obtain the number of branch offices per customer SELECT cid, COUNT(bid) FROM branches GROUP BY cid; | cid | count(bid) | | 101 | 3 | | 103 | 3 | | 104 | 2 | | 110 | 1 | rows in set (0.27 sec)

You can nest subqueries to any depth, so long as the basic rules above are followed. List the services used by Sharp Eyes Detective Agency 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 | row in set (0.28 sec)

Then filter out those with just two offices with a HAVING clause SELECT cid, COUNT(bid) FROM branches GROUP BY cid HAVING COUNT(bid) = 2; | cid | count(bid) | | 104 | 2 | row in set (0.16 sec)

Then hand the client ID over to the "clients" table in order to get the client name SELECT cname FROM clients WHERE cid = 104; | cname | | Rabbit Foods Inc | row in set (0.22 sec)

This subquery will take care of the three steps: SELECT cname FROM clients WHERE cid = (SELECT cid FROM branches GROUP BY cid HAVING COUNT(bid) = 2); | cname | | Rabbit Foods Inc | row in set (0.28 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 customers using the service with the maximum service fee 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 | row in set (0.01 sec)

HAVING CLAUSE Which sites are using more than 50% of all available services SELECT bid FROM branches_services GROUP BY bid HAVING COUNT(sid) > (SELECT COUNT(*) FROM services)/2; | bid | | 1011 | row in set (0.22 sec)

Which sites are using more than 50% of all available services and get the branch name and customer name as well 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 | row in set (0.28 sec)

List all clients using all available services across their branch offices 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 | row in set (0.01 sec)

IN OPERATOR A subquery can also return a single column of multiple values HOW? – By using the IN operator makes it possible to test if a particular value exists in the result set, and perform the outer query if the test is successful

List of all services being used by Branch ID 1031 SELECT sid FROM branches_services WHERE bid = 1031; | sid | | 2 | | 3 | | 4 | rows in set (0.16 sec)

And then: SELECT sname FROM services WHERE sid = 2; | sname | | Recruitment | row in set (0.28 sec) SELECT sname FROM services WHERE sid = 3; | sname | | Data Management | row in set (0.17 sec) SELECT sname FROM services WHERE sid = 4; | sname | | Administration | row in set (0.11 sec)

Using the IN operator in a subquery is more efficient for this: SELECT sname FROM services WHERE sid IN (SELECT sid FROM branches_services WHERE bid = 1031); | sname | | Recruitment | | Data Management | | Administration | rows in set (0.27 sec)

List of all branches using the "Accounting" service (service ID 1) 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 | rows in set (0.17 sec)

Now add the customer name for each branch as well 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 | rows in set (0.16 sec)

Now just show the customer list (DISTINCT) 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 | rows in set (0.17 sec)

logfile for your Activtiy 4-1 submission. Review your logfile as Activity 4-2 will continue with subqueries.