Chapter 5 6e & 8 5e: Complex SQL
CSE
4701
Prof. Steven A. Demurjian, Sr.
Computer Science & Engineering Department
The University of Connecticut
191 Auditorium Road, Box U-155
Storrs, CT 06269-3155
[email protected]
http://www.engr.uconn.edu/~steve
(860) 486 - 4818


About one third of these slides are being used with the permission of Dr. Ling
Lui, Associate Professor, College of Computing, Georgia Tech.
About one-half of these slides have been adapted from the AWL web site for
the textbook.
Chapter 5-1
Variety of Complex SQL Queries

CSE
4701






Nested Queries
Grouping and Aggregation
Order by and Having
Views
Sets
Tuple Variable
Other Complex Queries and Operations
Chapter 5-2
Recall Earlier Query 1

CSE
4701
Query 1: Retrieve Name and Address of all Employees
who work for the 'Research' Department
SELECT FNAME, MINIT, LNAME, ADDRESS, DNAME
FROM
EMPLOYEE, DEPARTMENT
WHERE DNAME='Research' AND DNUMBER=DNO

What Action is Being Performed?
Chapter 5-3
Nested Queries

CSE
4701


SQL SELECT Nested Query is Specified within
WHERE-clause of another Query (the Outer Query)
Query 1A: Retrieve the Name and Address of all
Employees who Work for the 'Research' Department
SELECT FNAME, LNAME, ADDRESS
FROM EMPLOYEE
WHERE DNO IN
(SELECT DNUMBER
FROM
DEPARTMENT
WHERE DNAME='Research' )
Note: This Reformulates Earlier Query 1 (prior slide)
Chapter 5-4
How Does Nested Query Work?

CSE
4701




Nested Query Selects DNUMBER of 'Research' Dept.
Outer Query Selects an EMPLOYEE Tuple If Its DNO
Value Is in the Result of Either Nested Query
IN represents Set Inclusion of Result Set
We Can Have Several Levels of Nested Queries
SELECT FNAME, LNAME, ADDRESS
FROM EMPLOYEE
Inner Query Returns
WHERE DNO IN
Set of DNUMBER
(SELECT DNUMBER
D1, D2, etc.
FROM
DEPARTMENT
WHERE Dname=’Research' )
Chapter 5-5
Nested Query Operates on Department Table
CSE
4701
SELECT
FNAME, LNAME, ADDRESS
FROM
EMPLOYEE
WHERE
DNO IN
(SELECT
DNUMBER
FROM
DEPARTMENT
WHERE
Dname=’Research' )

What Does the Nested Query Return?
Set with one Result:
5
Chapter 5-6
What Does Outer Query Do?
CSE
4701
SELECT
FNAME, LNAME, ADDRESS
FROM
EMPLOYEE
WHERE
DNO IN
(SELECT
DNUMBER
DNO IN:
FROM
DEPARTMENT
5
WHERE
Dname=’Research' )


DNO is IN Set of Values Returned by Inner Query
What Does the Outer Query Return?
 All Employees where DNO = 5
Chapter 5-7
Correlated Nested Queries

CSE
4701

When WHERE-clause of a Nested Query References
an Attribute of a Relation Declared in the Outer Query
Query 16: Retrieve the Name of each Employee who
has a Dependent with the Same First Name as the
Employee
SELECT E.FNAME, E.LNAME
Inner Query Returns
FROM
EMPLOYEE AS E
Relation a set of
WHERE E.SSN IN
ESSNs for All Emps
(SELECT
ESSN
that have Dependents
FROM
DEPENDENT with the same FNAME
WHERE
ESSN=E.SSN AND
E.FNAME=DEPENDENT_NAME)

Note: This Differs Slightly from 16 in book.
Chapter 5-8
How Does it Work?
CSE
4701
SELECT
FROM
WHERE
(SELECT
FROM
WHERE
E.FNAME, E.LNAME
Inner Query Returns
EMPLOYEE AS E
Relation a set of
E.SSN IN
ESSNs for All Emps
ESSN
that have Dependents
DEPENDENT
with the same FNAME
ESSN=E.SSN AND
E.FNAME=DEPENDENT_NAME)
Set with two Result:
123456789
333445555
Alice
Franklin
Joy
Abner
John
Alice
Elizabeth
Chapter 5-9
What Does Outer Query Do?
CSE
4701
SELECT
FROM
WHERE
(SELECT
FROM
WHERE
E.FNAME, E.LNAME
Returns the Employee
EMPLOYEE AS E
Names for all elements
E.SSN IN
In the set:
ESSN
123456789
DEPENDENT
333445555
ESSN=E.SSN AND
E.FNAME=DEPENDENT_NAME)
Returns
FNAME
John
Franklin
LNAME
Smith
Wong
Chapter 5-10
Query Equivalence

CSE
4701
Query 16:
SELECT E.FNAME, E.LNAME
FROM
EMPLOYEE AS E
WHERE E.SSN IN
(SELECT
ESSN
FROM
DEPENDENT
WHERE
ESSN=E.SSN AND
E.FNAME=DEPENDENT_NAME)

Query 16A:
SELECT E.FNAME, E.LNAME
FROM
EMPLOYEE E, DEPENDENT D
WHERE E.SSN=D.SSN AND
E.FNAME=D.DEPENDENT_NAME
Chapter 5-11
EXISTS Nested Queries

CSE
4701

EXISTS checks Whether the Result of a Correlated
Nested Query is Empty (contains no tuples) or not
Query 16B: Retrieve the Name of each Employee who
has a Dependent with the Same First Name as the
Employee
Inner Query Returns
SELECT FNAME, LNAME
Is True if there is
FROM
EMPLOYEE
At least one match.
WHERE EXISTS
(SELECT
*
Can there be 2 matches?
FROM
DEPENDENT
WHERE
SSN=ESSN AND
FNAME=DEPENDENT_NAME)

There is a analogous NOT EXISTS
Chapter 5-12
NULLS in SQL Queries

CSE
4701


SQL Allows Queries that Check if a value is NULL
(Missing or Undefined or not Applicable)
SQL uses IS or IS NOT to compare NULLs since it
Considers each NULL value Distinct from other NULL
Values, so Equality Comparison is not Appropriate
Query 18: Retrieve the names of all employees who do
not have supervisors.
SELECT FNAME, LNAME
FROM
EMPLOYEE
WHERE SUPERSSN IS NULL

Why Would Such a Capability be Useful?
 Downloading/Crossloading a Database
 Promoting a Attribute to PK/FK
Chapter 5-13
Aggregate Functions in SQL Queries

CSE
4701
Query 19: Find Maximum Salary, Minimum Salary,
and Average Salary among all Employees
SELECT MAX(SALARY), MIN(SALARY),
AVG(SALARY)
FROM
EMPLOYEE

Query 20: Find maximum and Minimum Salaries
among 'Research' Department Employees
SELECT MAX(SALARY), MIN(SALARY)
FROM
EMPLOYEE, DEPARTMENT
WHERE DNAME='Research' AND DNUMBER=DNO

What Does Query 22 Do?
SELECT COUNT(*)
FROM
EMPLOYEE, DEPARTMENT
WHERE DNAME='Research' AND DNUMBER=DNO
Chapter 5-14
Grouping in SQL Queries

CSE
4701



In Many Cases, We Want to Apply the Aggregate
Functions to Subgroups of Tuples in a Relation
Each Subgroup of Tuples is Set of Tuples that Have the
Same Value for the Grouping Attribute(s)
Function is Applied to Each Subgroup Independently
Query 24: For Each Department, Retrieve the DNO,
Number of Employees, and Their Average Salary
SELECT DNO, COUNT (*), AVG (SALARY)
FROM EMPLOYEE
GROUP BY
DNO
Chapter 5-15
Grouping in SQL Queries

CSE
4701



Query 24: For Each Department, Retrieve the DNO,
Number of Employees, and Their Average Salary
SELECT DNO, COUNT (*), AVG (SALARY)
FROM EMPLOYEE
GROUP BY
DNO
EMPLOYEE tuples are Divided into Groups; each
group has the Same Value for Grouping Attribute DNO
COUNT and AVG functions are applied to each Group
of Tuples Aeparately
SELECT-clause Includes only the Grouping Attribute
and the Functions to be Applied on each Tuple Group
Chapter 5-16
Results of Query 24:

CSE
4701
SELECT DNO, COUNT (*), AVG (SALARY)
FROM EMPLOYEE
GROUP BY
DNO
Chapter 5-17
Joins and Grouping in SQL Queries

CSE
4701


A Join Condition can be used in Conjunction with
Grouping
Query 25: For each Project, Retrieve its Number,
Name, and Number of Employees working on Project
SELECT
PNUMBER, PNAME, COUNT (*)
FROM
PROJECT, WORKS_ON
WHERE
PNUMBER=PNO
GROUP BY
PNUMBER, PNAME
In this case, the Grouping and Functions are Applied
after the Joining of the two Relations
Chapter 5-18
The HAVING Clause in SQL Queries

CSE
4701


In Some Cases, we want to retrieve values of Functions
for only those Groups that Satisfy Certain Condition(s)
The HAVING-clause is used for Specifying a Selection
Condition on Groups (rather than Individual Tuples)
Query 26: For each Project on which more than two
employees work, Retrieve its Number, Name, and
Number of Employees working on Project project
SELECT
PNUMBER, PNAME, COUNT (*)
FROM
PROJECT, WORKS_ON
WHERE
PNUMBER=PNO
GROUP BY
PNUMBER, PNAME
HAVING
COUNT (*) > 2
Chapter 5-19
Results of Query 26:

After Applying the WHERE/GROUP BY Clauses
Two Groups Not Selected Based
on Having Constraint
CSE
4701
Chapter 5-20
Results of Query 26:

After Applying the HAVING Clause Condition
CSE
4701
3
3
3
3
Chapter 5-21
Substring Comparison in SQL Queries

CSE
4701




In Regard to Strings, Most DBMSs Support SQL
Queries for Exact, Near, and Starts with Matching
LIKE is Used to Compare Partial Strings
'%' (or '*') Replaces an Arbitrary # of characters
'_' replaces a single arbitrary character
Query 12: Retrieve all Employees whose Address is in
Houston, Texas.
SELECT FNAME, LNAME
FROM
EMPLOYEE
WHERE ADDRESS LIKE '%Houston,TX% '
Houston, TX can be anywhere within the ADDRESS
VAR CHAR String
Chapter 5-22
Substring Comparison in SQL Queries

CSE
4701



The LIKE Operator Allows us to get Around the Fact
that each Value is Considered Atomic and Indivisible
SQL: Character String Attribute values are not Atomic
Query 12A: Retrieve all employees who were born
during the 1950s.
SELECT
FNAME, LNAME
FROM
EMPLOYEE
WHERE
BDATE LIKE
' __5_______'
There are two “_” before 5 and seven “_” after 5
Chapter 5-23
Arithmetic Operations in SQL Queries

CSE
4701

Standard Arithmetic Operators '+', '-'. '*', and '/' can be
Applied to Numeric Values in an SQL Query Result
Query 13: Show the Effect of Giving all Employees
who work on the 'ProductX' project a 10% raise.
SELECT FNAME, LNAME, 1.1*SALARY
FROM
EMPLOYEE, WORKS_ON, PROJECT
WHERE SSN=ESSN AND PNO=PNUMBER
AND PNAME='ProductX'
Chapter 5-24
ORDER BY Clause in SQL Queries

CSE
4701

ORDER BY used to Sort the Tuples in a Query Result
based on the Values of one or More Attribute(s)
Query 15: Retrieve a list of Employees and the Projects
each works in, ordered by Dept., and within each Dept.,
alphabetically by Employee last name
SELECT
FROM
WHERE
ORDER BY

DNAME, LNAME, FNAME, PNAME
DEPARTMENT, EMPLOYEE,
WORKS_ON, PROJECT
DNUMBER=DNO AND SSN=ESSN
AND PNO=PNUMBER
DNAME, LNAME
Default is Ascending - Can be ASC/DESC as we’ll see
in a Later Example
Chapter 5-25
ASC DESC Example

CSE
4701
Query 15: Retrieve a list of Employees and the Projects
each works in, ordered by Dept in alphabetical order,
and within each Dept., alphabetically in reverse by
Employee last name
SELECT
FROM
WHERE
ORDER BY
DNAME, LNAME, FNAME, PNAME
DEPARTMENT, EMPLOYEE,
WORKS_ON, PROJECT
DNUMBER=DNO AND SSN=ESSN
AND PNO=PNUMBER
ASC DNAME, DESC LNAME
Chapter 5-26
SQL Support for Views

CSE
4701

Views are Part of the SQL DDL
Abstracting from Conceptual to External Schema
 View Hides the Details
 One or More Tables in Conceptual Schema May be
Combined (in Part) to Form a View
 Don’t Include FKs and Other Internal Attributes
 Typically, View is Formed by Join of Two or More
Relations Utilizing FKs, PKs, etc.
 As a Result - View is Independent
Once
Formed - View Static/Unchangeable to Insulate
User Applications from Conceptual Schema
Similar in Concept/Intent to “Public Interface”
Chapter 5-27
SQL View Definition

CSE
4701

Features of Views
 View Represents a Restricted Portion (Rows,
Columns) of a Relation - External Schema in SQL
 View is Virtual Table View (Not Stored) and Must
be Re-evaluated Every Time - Dynamic
 Like Relation, a View Can Be Deleted at Any Time
 Attributes Can Be Renamed in View
CREATE VIEW PQ(P#, SUMQTY)
AS SELECT P#, SUM(QTY)
FROM SP
GROUP BY P#;
Reasons for Views
 Security
 Increasing Application-Data Independence
Chapter 5-28
View Definition in Ongoing Example

CSE
4701

First View: Attribute Names are Inherited
CREATE VIEW WORKS_ON1 AS
SELECT FNAME, LNAME, PNAME, HOURS
FROM EMPLOYEE, PROJECT, WORKS_ON
WHERESSN=ESSN AND PNO=PNUMBER ;
Second View: View attribute names are Aliased via a
one-to-one Correspondence with the SELECT-clause
CREATE VIEW DEPT_INFO
(DEPT_NAME, NO_OF_EMPS, TOTAL_SAL) AS
SELECT
DNAME, COUNT (*), SUM (SALARY)
FROM
DEPARTMENT, EMPLOYEE
WHERE
DNUMBER=DNO
GROUP BY
DNAME ;
Chapter 5-29
Queries on Views

CSE
4701
Retrieve the Last Name and First Name of All
Employees Who Work on 'ProjectX'.
SELECT PNAME, FNAME, LNAME
FROM
WORKS_ON1
WHERE PNAME='ProjectX' ;




Without the View WORKS_ON1, this Query
Specification Would Require Two Join Conditions
A View Can Be Defined to Simplify Frequently
Occurring Queries
DBMS Keeps the View Up-to-date if the Base Tables
on Which the View is Defined are Modified
Hence, the View is Realized at the Time we Specify a
Query on the View
Chapter 5-30
What is View Update Problem?

CSE
4701



Retrieval over View Mirrors a Retrieval over Relation
However, Update over View may cause Problems!
In general, a View Update may Introduce Ambiguity
when there is more than one way to Update Underlying
Relations
Consider the view PQ Created Below:
CREATE VIEW PQ(P#, SUMQTY)
AS SELECT P#, SUM(QTY)
FROM SP
GROUP BY P#;


Try to Change the Total Quantity SUMQTY of P1 in
PQ from “30” to “40”
Why Does a view Update Problem occur?
Chapter 5-31
The Book Example - DDL - Create Tables

CSE
4701

Create a Table in the MY_BOOK_DB Schema:
What do ISBNNUMBER and PUBLISHERID
Represent?
CREATE TABLE BOOK_CATALOG
(ISBN
ISBNNUMBER NOT NULL,
TITLE
VARCHAR(25),
AUTHORS VARCH(100),
...
PUBLISHER PUBLISHERID,
PRIMARY KEY (ISBN)
FOREIGN KEY (PUBLISHER)
REFERENCES PUBLISHING_HOUSE(NAME);
Chapter 5-32
DDL - Create Tables (continued)
CSE
4701
CREATE TABLE PUBLISHING_HOUSE
(PUB_ID
PUBLISHERID
NOT NULL,
PUB_NAME
VARCHAR(50)
LOC
CITYNAME
CONTACT
ADDRESS,
UNIQUE(PUB_NAME, LOC);
CREATE TABLE ORDER
(ISBN
ISBNNUMBER NOT NULL,
PUBLISHER
PUBLISHERID
DATE
DATE
NOT NULL,
PRIMARY KEY(ISBN, PUBLISHER),
FOREIGN KEY ISBN
REFERENCES BOOK_CATALOG(ISBN),
FOREIGN KEY PUBLISHER
REFERENCES PUBLISHING_HOUSE(PUB_ID));
Chapter 5-33
DDL - Change Table Structure

Add a Column to a Table:
CSE
4701
ALTER TABLE BOOK
ADD PRICE DECIMAL(7,2),
ADD MEMBER_DISCOUNT, DEFAULT 5;
No DEFAULT Implies NULL Values for all Tuples
Drop a Column from a table


ALTER TABLE BOOK
DROP PRICE RESTRICT (or CASCADE);


Restrict: Drop Operation Fails if Column is
Referenced
Cascade: Drop Operation Removes Referencing View
and Constraint Definitions
Chapter 5-34
Views Basis of APIs

CSE
4701

WSDL, SOAP, REST, etc.
View Presents a Limited Picture of the DB
 Defines Data Available to Different
User
Groups
Applications (web or mobile)
Other Systems (for Interoperability/Sharing)
Most Typically Read-Based Views
Update-Based Views Carefully Define
 Insure that Updates Don’t Alter Consistency
 May Limit What can be Modified


Chapter 5-35
Views in Medical Domain

CSE
4701




Prescription Object has Multiple Views
MD/Prescriber
 Ability to Write the Script with Drug, Dosage,
Instructions, etc.
Pharmacist -Fill Prescription (Drug dispensed)
 Ability to Substitute Generic for Brand
 A Refill Reduces Future Availability
Patient
 Submit Script to Pharmacy/Insurance
Insurer
 See Drug Dispensed, Script, Approve Payment
Chapter 5-36
Other SQL Query Examples

CSE
4701



Homework 2 Spring 2015
Book Database
Employee/Project/Department Database
Northwind
Chapter 5-37
Another Set of Examples- Book DB

BOOK(ISBN, TITLE,AUTHORS,PRICE,PUBLISHER,YEAR)
ORDER(ISBN, CUST_NAME, LOC, DATE,WEEKDAY)

Find the books Written by Maier

CSE
4701
SELECT TITLE, ISBN
FROM BOOK;
WHERE AUTHOR LIKE ‘%MAIER’ ;

Find ISBN of the Books Whose Price is at Least 5%
less than the Average Price of the Books by Maier
SELECT ISBN, PRICE
FROM BOOK
WHERE PRICE*(1-0.05) <
SELECT AVG(PRICE)
FROM BOOK
WHERE AUTHORS LIKE “%Maier”;
Chapter 5-38
Other SQL Search: String Matching

CSE
4701
Wildcard:
%: Matches any Substring
_: Matches any Character
SELECT *
FROM BOOK
WHERE PUBLISHER LIKE “%IEEE%”;

“%can%” Matches American, Canada, Scandinavian, ...

“Ca%” Matches Canada, Canadian, ...

“ %” Matches any string with at least two characters
Chapter 5-39
Other SQL Queries

CSE
4701


Find ISBN and Price of Books Published by ACM
SELECT ISBN, PRICE
FROM BOOK
WHERE PUBLISHER=“ACM”;
Find ISBN and price for all books ordered from Atlanta
with a price over $50
SELECT ISBN, PRICE
FROM BOOK, ORDER
WHERE BOOK.ISBN = ORDER.ISBN
AND ORDER.LOC=‘ATL’
AND PRICE > 50.00;
Note the Distinguishing Between Attributes with Same
Name in Different Tables (TableName.AttributeName)
Chapter 5-40
Using Tuple Variables

CSE
4701


Tuple Variables Simplify Query Since Don’t Need to
Repeat the Entire Table Name
Find ISBN and Price for all Books ordered from
Atlanta with a price over $50
SELECT B.ISBN, B.PRICE
FROM BOOK B, ORDER O
WHERE B.ISBN = O.ISBN AND O.LOC=‘ATL’
AND B.PRICE > 50.00;
Also Useful if Relation is Used “twice” in a Query:
SELECT B1.ISBN, B1.TITLE, B1.AUTHORS
FROM BOOK B1, BOOK B2
WHERE B1.PRICE > B2.PRICE
AND B2.ISBN = “111001100”;
Chapter 5-41
Ordering Results

CSE
4701


Order by Clause Sorts the rows in a Query Result in
Ascending (asc) or Descending (desc) Order
Find all books Published by ACM in the Ascending
order of Price and Descending order of year
SELECT *
FROM BOOK
WHERE PUBLISHER LIKE “ACM%”
ORDER BY PRICE ASC, YEAR DESC;
Questions:
 What Does “*” Indicate?
 What Does ACM% Retrieve?
Chapter 5-42
Set Operations

Find books written by Mary or Lisa
SELECT *
FROM BOOK
WHERE AUTHORS LIKE “LISA%”
UNION
SELECT *
FROM BOOK
WHERE AUTHORS LIKE “MARY%”;
CSE
4701
SELECT *
FROM BOOK
WHERE AUTHORS LIKE “LISA%”
OR AUTHORS LIKE “MARY%”;


UNION, INTERSECT, EXCEPT
UNION ALL, INTERSECT ALL, and EXCEPT
ALL Preserve Duplicates
Chapter 5-43
Built-in Aggregate Functions

CSE
4701


Count (COUNT), Sum (SUM), Average (AVG),
Minimum (MIN), Maximum (MAX)
Count Books Ordered on 2/16
SELECT COUNT( *)
FROM ORDER
WHERE ORDER.DATE = “2/16/2000”;
Find the Average Price of Books by each Publisher
SELECT PUBLISHER, AVG(PRICE)
FROM BOOK
GROUP BY PUBLISHER;
Chapter 5-44
Built-in Aggregate Functions

Find the Average Book Price of all Publishers that have
Published more than 1000 Books
SELECT PUBLISHER, AVG(PRICE)
FROM BOOK
GROUP BY PUBLISHER
HAVING COUNT (ISBN) >= 1000;

Find the Highest Priced book(s) by Maier
SELECT ISBN, MAX(PRICE)
FROM BOOK
WHERE AUTHORS LIKE “%Maier%”;
CSE
4701
Chapter 5-45
Nested Subqueries

CSE
4701



Nested Subqueries Allow us to Ask More Complex
Questions Regarding the Database Content
Queries are Nested and Involve Set Relationships
Relationships Supported Include:
 Set Membership: IN, NOT IN
 Set Comparison
(=, <, <=, >, >=, <>) ALL
(=, <, <=, >, >=, <>) SOME
 Test Empty Relation: EXISTS, NOT EXISTS
Let’s see Some Examples…
Chapter 5-46
Set Membership: IN, NOT IN

CSE
4701

Find Title of the Books Ordered on Mondays
SELECT DISTINCT TITLE
FROM BOOK
WHERE ISBN IN
(SELECT ISBN
FROM ORDER
WHERE WEEKDAY = “MON”);
Find Titles of Books ordered on Wednesday to Friday
SELECT DISTINCT ISBN
FROM BOOK
WHERE WEEKDAY NOT IN (“MON”, “TUE”);
Chapter 5-47
Set Comparison

CSE
4701

Operators
(=, <, <=, >, >=, <>) ALL
(=, <, <=, >, >=, <>) SOME
Find ISBN of Books Published by ACM, which are
Cheaper than all Books Ordered by Smith
SELECT ISBN
FROM BOOK
WHERE PUBLISHER LIKE “%ACM%”
AND PRICE <
ALL (SELECT B.PRICE
FROM BOOK B, ORDER O
WHERE CUST_NAME LIKE “%SMITH%”
AND B.ISBN = O.ISBN);
Chapter 5-48
Delete and Update

CSE
4701

Cancel all orders by Mary on 2/17/2000
DELETE FROM BOOK
WHERE DATE=2000-02-17
AND ISBN IN
(SELECT ISBN
FROM ORDER
WHERE CUST_NAME LIKE “%Mary%”);
Update all Orders for Customers on 2/15/2002 by
giving a discount of 5%
UPDATE ORDER
SET PRICE = PRICE * (1-0.05)
WHERE DATE=2002-02-50;
Chapter 5-49
View Concepts/Examples
REM updatable view
CSE
4701
CREATE VIEW LOW-PRICE-BOOKS
AS SELECT *
FROM BOOKS
WHERE PRICE<50.00;
REM moves row outside view
UPDATE LOW-PRICE-BOOKS
SET PRICE = 60.00
WHERE PUBLISHER LIKE “%ACM%”;
Chapter 5-50
View Concepts/Examples
REMcreate row outside view
CSE
4701
INSERT INTO LOW-PRICE-BOOKS
VALUES (“1010110022”, ”Java Beans”,
“Smith”, 45, ”ACM”);
REM prevents updates outside the view
CREATE VIEW LOW-PRICE-BOOKS
AS SELECT *
FROM BOOK
WHERE PRICE<50.00
WITH CHECK OPTION;
Chapter 5-51
Homework 3 from Spr 2015

Problem 4.10 in 6th edition
CSE
4701
Chapter 5-52
Problem 4.10 in 6th edition
CSE
4701
Chapter 5-53
Problem 4.10 in 6th edition
CSE
4701
(a) Retrieve the names of employees in department 5 who
work more than 10 hours per week on the 'ProductX' project.
SELECT LNAME, FNAME
FROM EMPLOYEE, WORKS_ON, PROJECT
WHERE DNO=5 AND SSN=ESSN AND
PNO=PNUMBER AND PNAME='ProductX' AND HOURS>10
LNAME
FNAME
SELECT LNAME, FNAME
FROM EMPLOYEE
Smith
John
WHERE DNO=5 AND SSN IN
English Joyce
( SELECT ESSN
FROM WORKS_ON
WHERE HOURS>10 AND PNO IN
( SELECT PNUMBER
FROM PROJECT
WHERE PNAME='ProductX' ) )
Chapter 5-54
Problem 4.10 in 6th edition
CSE
4701
(b) List the names of employees who have a dependent
with the same first name as themselves.
SELECT LNAME, FNAME
FROM EMPLOYEE, DEPENDENT
WHERE SSN=ESSN AND FNAME=DEPENDENT_NAME
Another possible SQL query uses nesting as follows:
SELECT LNAME, FNAME
FROM EMPLOYEE
WHERE EXISTS
( SELECT *
FROM DEPENDENT
WHERE FNAME=DEPENDENT_NAME
AND SSN=ESSN )
Result (empty):
Chapter 5-55
Problem 4.10 in 6th edition
CSE
4701
(c) Find the names of employees that are
directly supervised by 'Franklin Wong'.
LNAME FNAME
Smith John
Narayan Ramesh
English Joyce
SELECT E.LNAME, E.FNAME
FROM EMPLOYEE E, EMPLOYEE S
WHERE S.FNAME='Franklin' AND
S.LNAME='Wong' AND E.SUPERSSN=S.SSN
Another possible SQL query uses nesting as follows:
SELECT LNAME, FNAME
FROM EMPLOYEE
WHERE SUPERSSN IN
( SELECT SSN
FROM EMPLOYEE
WHERE FNAME='Franklin' AND LNAME='Wong' )
Chapter 5-56
Problem 4.10 in 6th edition
CSE
4701
(d) For each project, list the project name and the total hours per week (by
all employees) spent on that project.
SELECT PNAME, SUM (HOURS)
FROM PROJECT, WORKS_ON
WHERE PNUMBER=PNO
GROUP BY PNAME
Result:
PNAME
SUM(HOURS)
ProductX
52.5
ProductY
37.5
ProductZ
50.0
Computerization55.0
Reorganization
25.0
Newbenefits
55.0
Chapter 5-57
Problem 4.10 in 6th edition
CSE
4701
(e) Retrieve the names of employees who work on every project.
SELECT LNAME, FNAME
FROM EMPLOYEE
WHERE NOT EXISTS
( SELECT PNUMBER
FROM PROJECT
WHERE NOT EXISTS
( SELECT *
FROM WORKS_ON
WHERE PNUMBER=PNO AND ESSN=SSN ) )
Result (empty):
Chapter 5-58
Problem 4.10 in 6th edition
CSE
4701
(f) Retrieve the names of employees who do not work on any project.
SELECT LNAME, FNAME
FROM EMPLOYEE
WHERE NOT EXISTS
( SELECT *
FROM WORKS_ON
WHERE ESSN=SSN )
Result (empty):
Chapter 5-59
Problem 4.10 in 6th edition
CSE
4701
(g) For each department, retrieve the department name,
and the average salary of employees working in that department.
SELECT DNAME, AVG (SALARY)
FROM DEPARTMENT, EMPLOYEE
WHERE DNUMBER=DNO
GROUP BY DNAME
Result:
DNAME
AVG(SALARY)
Research
33250
Administration 31000
Headquarters
55000
Chapter 5-60
Problem 4.10 in 6th edition
CSE
4701
(i) Find the names and addresses of employees who work on at least one
project located in Houston but whose department has no location in Houston.
SELECT LNAME, FNAME, ADDRESS
FROM EMPLOYEE
WHERE EXISTS
( SELECT *
FROM WORKS_ON, PROJECT
WHERE SSN=ESSN AND PNO=PNUMBER
AND PLOCATION='Houston' )
AND NOT EXISTS
( SELECT *
FROM DEPT_LOCATIONS
WHERE DNO=DNUMBER
AND DLOCATION='Houston' )
Result:
LNAME
FNAME
ADDRESS
Wallace Jennifer 291 Berry, Bellaire, TX
Chapter 5-61
Problem 4.10 in 6th edition
CSE
4701
(j) List the last names of department managers who have no dependents.
SELECT LNAME, FNAME
FROM EMPLOYEE
WHERE EXISTS
( SELECT *
FROM DEPARTMENT
WHERE SSN=MGRSSN )
AND NOT EXISTS
( SELECT *
FROM DEPENDENT
WHERE SSN=ESSN )
Result:
LNAME FNAME
Borg James
Chapter 5-62
Other SQL Queries Hmwk 4 Spring 2015
CSE
4701
Problem 3.4 for the Northwind Database Schema
a.
Find and print the company names and company addresses of all
Suppliers that supply the category name Seafood.
b. Count and print the number of suppliers for each of the eight
different categories of food which by name are: Beverages,
Condiments, Confections, Dairy Products, Grains/Cereals,
Meat/Poultry, Produce, Seafood.
c.
For each country (ShipCountry in Orders), total the Freight
charges. The countries are: France, Germany, Brazil, Belgium,
Switzerland, Venezuela, Austria, Mexico, USA, Sweden,
Finland, Italy, Spain, UK, Ireland, Portugal, Canada, Denmark,
Poland, Norway, Argentina
Chapter 4-63
Explaining Northwind Schema

CSE
4701







Suppliers: A Suppliers Contact Info and web link.
Products: Names, suppliers, and Prices
Categories: Categories of Northwind products such as
Beverages, Condiments, Confections, Dairy Products,
Grains/Cereals, Meat/Poultry, Produce, Seafood
Orders: For each Customer with dates &Shipping
Order Details: Products, quantities, and price.
Employees: Typical Info.
Customers: Typical Info.
Shippers: Typical Info.
Chapter 4-64
Northwind Schema
CSE
4701
Chapter 4-65
a.Find and print the company names and company addresses of all
Suppliers that supply the category name Seafood.
CSE
4701
SELECT DISTINCT suppliers.CompanyName, suppliers.Address
FROM northwind.suppliers, northwind.categories, northwind.products
WHERE suppliers.SupplierID = products.SupplierID
AND categories.CategoryID = products.CategoryID
AND categories.CategoryName = 'Seafood';
Chapter 4-66
a.Count/the number of suppliers for each of the eight different categories
of food which by name are: Beverages, Condiments, Confections, Dairy
Products, Grains/Cereals, Meat/Poultry, Produce, Seafood.
CSE
4701
SELECT categories.CategoryName, COUNT(suppliers.SupplierID)
FROM northwind.categories, northwind.products, northwind.suppliers
WHERE suppliers.SupplierID = products.SupplierID
AND products.categoryID = categories.CategoryID
GROUP BY categories.CategoryName;
Chapter 4-67
For each country (ShipCountry in
Orders), total the Freight charges.
CSE
4701
SELECT orders.ShipCountry, SUM(orders.Freight)
FROM northwind.orders
GROUP BY orders.ShipCountry;
Chapter 4-68
Other SQL Queries Hmwk 2 Fall 2015

CSE
4701


Find the list of all customers from United Kingdom sorted by
Company Name in ascending order and return the company
Name and Country.
Find the number of customers that are located in each country.
Find the Product Name, UnitPrice, and UnitsInStock and Sorted
by ProductName for all products where the product name starts
with “Ch”.
Chapter 4-69
Find list of all customers from United Kingdom
CSE
4701
SELECT northwind.suppliers.CompanyName, northwind.suppliers.country
FROM northwind.suppliers
WHERE northwind.suppliers.Country="UK"
ORDER BY northwind.suppliers.CompanyName ASC
Exotic Liquids
UK
Specialty Biscuits, Ltd.
UK
Chapter 4-70
Find number of customers in each country.
CSE
4701
SELECT northwind.customers.country, COUNT(*)
FROM northwind.customers
WHERE northwind.customers.country IS NOT NULL
GROUP BY northwind.customers.country
Argentina
3
Austria
2
Belgium
2
Brazil
9
Canada
3
Denmark
2
Finland
2
France
11
Germany
11
Ireland
1
Italy
3
Mexico
5
Norway
1
Poland
1
Portugal
2
Spain
5
Sweden
2
Switzerland
2
UK
7
USA
13
Venezuela
4
Chapter 4-71
Find the Product Name …starts with “Ch”.
CSE
4701
SELECT
northwind.products.productid, northwind.products.productname,
northwind.products.UnitsinStock
FROM northwind.products
WHERE northwind.products.productname LIKE 'CH%';
1
Chai
39
2
Chang
17
39
Chartreuse verte
69
4
Chef Anton's Cajun
Seasoning
53
5
Chef Anton's Gumbo Mix
0
48
Chocolade
15
Chapter 4-72
Concluding Remarks

CSE
4701
What have we Seen in Chapter 5?
 Complex Data Manipulation in SQL
Nested
Queries
Grouping and Aggregation
Order by and Having
Views
Sets
Tuple Variable
Other Complex Queries and Operations

Strongly Encouraged to Engage in Practice with your
DBMS of Choice for your Project
Chapter 5-73