1. Chapter 5 6e & 8 5e: Complex SQL
Prof. Steven A. Demurjian, Sr.
Computer Science & Engineering Department
The University of Connecticut
191 Auditorium Road, Box U-155
Storrs, CT
(860)
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.

2. Variety of Complex SQL Queries
Nested Queries
Grouping and Aggregation
Order by and Having
Views
Sets
Tuple Variable
Other Complex Queries and Operations

3. Recall Earlier Query 1
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?

4. Nested Queries
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)

5. How Does Nested Query Work?
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 WHERE DNO IN (SELECT DNUMBER FROM DEPARTMENT WHERE Dname=’Research' )
Inner Query Returns
Set of DNUMBER
D1, D2, etc.

6. Nested Query Operates on Department Table
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

7. What Does Outer Query Do?
SELECT FNAME, LNAME, ADDRESS FROM EMPLOYEE WHERE DNO IN (SELECT DNUMBER FROM DEPARTMENT WHERE Dname=’Research' )
DNO is IN Set of Values Returned by Inner Query
What Does the Outer Query Return?
All Employees where DNO = 5
DNO IN: 5

8. Correlated Nested Queries
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 FROM EMPLOYEE AS E WHERE E.SSN IN (SELECT ESSN FROM DEPENDENT WHERE ESSN=E.SSN AND E.FNAME=DEPENDENT_NAME)
Note: This Differs Slightly from 16 in book.
Inner Query Returns
Relation a set of
ESSNs for All Emps
that have Dependents
with the same FNAME

9. How Does it Work?
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)
Inner Query Returns
Relation a set of
ESSNs for All Emps
that have Dependents
with the same FNAME
Alice
Franklin
Joy
Abner
John
Elizabeth
Set with two Result:

10. What Does Outer Query Do?
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)
Returns the Employee
Names for all elements
In the set:
Returns
FNAME LNAME
John Smith
Franklin Wong

11. Query Equivalence
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

12. EXISTS Nested Queries
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 SELECT FNAME, LNAME FROM EMPLOYEE WHERE EXISTS (SELECT * FROM DEPENDENT WHERE SSN=ESSN AND FNAME=DEPENDENT_NAME)
There is a analogous NOT EXISTS
Inner Query Returns
Is True if there is
At least one match.
Can there be 2 matches?

13. NULLS in SQL Queries
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

14. Aggregate Functions in SQL Queries
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

15. Grouping in SQL Queries
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

16. Grouping in SQL Queries
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

17. Results of Query 24:
SELECT DNO, COUNT (*), AVG (SALARY) FROM EMPLOYEE GROUP BY DNO

18. Joins and Grouping in SQL Queries
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

19. The HAVING Clause in SQL Queries
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

20. Two Groups Not Selected Based on Having Constraint
Results of Query 26:
After Applying the WHERE/GROUP BY Clauses
Two Groups Not Selected Based on Having Constraint

21. Results of Query 26:
After Applying the HAVING Clause Condition 3

22. Substring Comparison in SQL Queries
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

23. Substring Comparison in SQL Queries
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

24. Arithmetic Operations in SQL Queries
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'

25. ORDER BY Clause in SQL Queries
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 DNAME, LNAME, FNAME, PNAME FROM DEPARTMENT, EMPLOYEE, WORKS_ON, PROJECT WHERE DNUMBER=DNO AND SSN=ESSN AND PNO=PNUMBER ORDER BY DNAME, LNAME
Default is Ascending - Can be ASC/DESC as we’ll see in a Later Example

26. ASC DESC Example
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 DNAME, LNAME, FNAME, PNAME FROM DEPARTMENT, EMPLOYEE, WORKS_ON, PROJECT WHERE DNUMBER=DNO AND SSN=ESSN AND PNO=PNUMBER ORDER BY ASC DNAME, DESC LNAME

27. SQL Support for Views 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”

28. SQL View Definition 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
Reasons for Views
Security
Increasing Application-Data Independence
CREATE VIEW PQ(P#, SUMQTY)
AS SELECT P#, SUM(QTY)
FROM SP
GROUP BY P#;

29. View Definition in Ongoing Example
First View: Attribute Names are Inherited
CREATE VIEW WORKS_ON1 AS
SELECT FNAME, LNAME, PNAME, HOURS
FROM EMPLOYEE, PROJECT, WORKS_ON
WHERE SSN=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 ;

30. Queries on Views
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

31. What is View Update Problem?
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:
Try to Change the Total Quantity SUMQTY of P1 in PQ from “30” to “40”
Why Does a view Update Problem occur?
CREATE VIEW PQ(P#, SUMQTY)
AS SELECT P#, SUM(QTY)
FROM SP
GROUP BY P#;
If we want to change the total quantity SUMQTY of a part P1 in PQ from “30” to “40”, a view update problem incurs, because this update over view PQ does not give enough information to enable us to map the view update to the base relations without ambiguity.

32. The Book Example - DDL - Create Tables
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);

33. DDL - Create Tables (continued)
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));

34. DDL - Change Table Structure
Add a Column to a Table:
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

35. Views Basis of APIs 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

36. Views in Medical Domain
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

37. Other SQL Query Examples
Homework 2 Spring 2015
Book Database
Employee/Project/Department Database
Northwind

38. 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
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)
WHERE AUTHORS LIKE “%Maier”;

39. Other SQL Search: String Matching
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

40. Other SQL Queries 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
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)

41. Using Tuple Variables
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 = “ ”;

42. Ordering Results
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?

43. Set Operations Find books written by Mary or Lisa
SELECT *
FROM BOOK
WHERE AUTHORS LIKE “LISA%”
UNION
WHERE AUTHORS LIKE “MARY%”;
UNION, INTERSECT, EXCEPT
UNION ALL, INTERSECT ALL, and EXCEPT ALL Preserve Duplicates
SELECT *
FROM BOOK
WHERE AUTHORS LIKE “LISA%”
OR AUTHORS LIKE “MARY%”;

44. Built-in Aggregate Functions
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;

45. 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%”;

46. Nested Subqueries
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…

47. Set Membership: IN, NOT IN
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
WHERE WEEKDAY NOT IN (“MON”, “TUE”);

48. Set Comparison 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);

49. Delete and Update Cancel all orders by Mary on 2/17/2000
DELETE FROM BOOK
WHERE DATE=
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= ;

50. View Concepts/Examples
REM updatable view
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%”;

51. View Concepts/Examples
REMcreate row outside view
INSERT INTO LOW-PRICE-BOOKS
VALUES (“ ”, ”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;

52. Homework 3 from Spr 2015
Problem 4.10 in 6th edition

53. Problem 4.10 in 6th edition

54. Problem 4.10 in 6th edition LNAME FNAME Smith John English Joyce
(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
SELECT LNAME, FNAME
FROM EMPLOYEE
WHERE DNO=5 AND SSN IN
( SELECT ESSN
FROM WORKS_ON
WHERE HOURS>10 AND PNO IN
( SELECT PNUMBER
FROM PROJECT
WHERE PNAME='ProductX' ) )
LNAME FNAME
Smith John
English Joyce

55. Problem 4.10 in 6th edition
(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:
FROM EMPLOYEE
WHERE EXISTS
( SELECT *
FROM DEPENDENT
WHERE FNAME=DEPENDENT_NAME
AND SSN=ESSN )
Result (empty):

56. Problem 4.10 in 6th edition LNAME FNAME Smith John Narayan Ramesh
English Joyce
(c) Find the names of employees that are
directly supervised by 'Franklin Wong'.
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
WHERE FNAME='Franklin' AND LNAME='Wong' )

57. Problem 4.10 in 6th edition
(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
ProductY
ProductZ
Computerization55.0
Reorganization 25.0
Newbenefits 55.0

58. Problem 4.10 in 6th edition
(e) Retrieve the names of employees who work on every project.
SELECT LNAME, FNAME
FROM EMPLOYEE
WHERE NOT EXISTS
( SELECT PNUMBER
FROM PROJECT
( SELECT *
FROM WORKS_ON
WHERE PNUMBER=PNO AND ESSN=SSN ) )
Result (empty):

59. Problem 4.10 in 6th edition
(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):

60. Problem 4.10 in 6th edition
(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
Administration 31000
Headquarters

61. Problem 4.10 in 6th edition
(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
FROM DEPT_LOCATIONS
WHERE DNO=DNUMBER
AND DLOCATION='Houston' )
Result:
LNAME FNAME ADDRESS
Wallace Jennifer 291 Berry, Bellaire, TX

62. Problem 4.10 in 6th edition
(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
FROM DEPENDENT
WHERE SSN=ESSN )
Result:
LNAME FNAME
Borg James

63. Other SQL Queries Hmwk 4 Spring 2015
Problem 3.4 for the Northwind Database Schema
Find and print the company names and company addresses of all Suppliers that supply the category name Seafood.
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.
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

64. Explaining Northwind Schema
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.

65. Northwind Schema

66. Find and print the company names and company addresses of all Suppliers that supply the category name Seafood.
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';

67. 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.
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;

68. For each country (ShipCountry in Orders), total the Freight charges.
SELECT orders.ShipCountry, SUM(orders.Freight)
FROM northwind.orders
GROUP BY orders.ShipCountry;

69. Other SQL Queries Hmwk 2 Fall 2015
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”.

70. Find list of all customers from United Kingdom
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.

71. Find number of customers in each country.
Argentina 3
Austria 2
Belgium
Brazil 9
Canada
Denmark
Finland
France 11
Germany
Ireland 1
Italy
Mexico 5
Norway
Poland
Portugal
Spain
Sweden
Switzerland UK 7
USA 13
Venezuela 4
SELECT northwind.customers.country, COUNT(*)
FROM northwind.customers
WHERE northwind.customers.country IS NOT NULL
GROUP BY northwind.customers.country

72. Find the Product Name …starts with “Ch”.
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
Chartreuse verte 69 4
Chef Anton's Cajun Seasoning 53 5
Chef Anton's Gumbo Mix 48
Chocolade 15

73. Concluding Remarks 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