Vocabulary
Identify the vocabulary word for each definition below.
An entry in a table, consisting of values for each appropriate col-umn. [Row]
The set of mandatory columns within a table that is used to en-force uniqueness of rows, and that is
normally the most frequent means by which rows are accessed. [Premary Key]
An arrangement of data in rows and columns. [Table]
A column or set of columns that refers to a primary key in the same table or another table. [Foreign Key]
Collections of objects or relations, set of operators to act on those relations, and data integrity for
accuracy and consistency. [Relational dB]
Intersection of a row and column. [Field]
Used to modify the table data by entering, changing, or removing
Rows. [Data Manipulation Language]
Creates, changes, and removes data structures from the data-base. [Data Manipulation Language]
Used to manage the changes made by DML statements. [Transaction Control Language]
Used to give or remove access rights to the database and the structures within it. [Data Control
Language]
1. The Global Fast Foods database consists of how many tables? ____ tables
SELECT DISTINCT owner, count(table_name)
FROM all_tables
GROUP BY owner;
OWNER
MDSYS
CTXSYS
OLAPSYS
SYSTEM
COUNT(TABLE_NAME)
48
5
2
7
XDB
HKUMAR
SYS
But this contains tables from all the ERD’s in:
2
35
37
And I need for:
Specific to Global Fast Foods ERD [9] are:

F_ORDER_LINES

F_FOOD_ITEMS

F_PROMOTIONAL_MENUS

F_REGULAR_MENUS

F_SHIFTS

F_SHIFT_ASSIGNMENTS

F_STAFFS

F_CUSTOMERS

F_ORDERS
2. How is the F_SHIFTS table related to the F_STAFFS table?
SELECT DISTINCT table_name from user_constraints
WHERE r_constraint_name in
(
SELECT constraint_name
FROM user_constraints
WHERE table_name in (upper('f_shifts'), upper('F_STAFFS'))
);
Above query gives an idea that, go look study tables F_ORDERS, F_SHIFT_ASSIGNMENTS
Now finally I found that, F_SHIFTS table related to the F_STAFFS – both having foreign key
reference in F_SHIFT_ASSIGNMENTS.
3. What are the names of the columns in the F_CUSTOMERS table?
SELECT column_name
FROM all_tab_cols
WHERE table_name = 'F_CUSTOMERS'
COLUMN_NAME:

ID

FIRST_NAME

LAST_NAME

ADDRESS

CITY

STATE

ZIP

PHONE_NUMBER
4. How many rows of data have been entered in the F_PROMOTIONAL_MENUS table?
select count(*) from F_PROMOTIONAL_MENUS
Result is 2.
5. In the F_FOOD_ITEMS table, column _________ is a foreign-key column. What table
and column is this key referencing?
select *
from user_constraints
where table_name = 'F_FOOD_ITEMS' and CONSTRAINT_TYPE = 'R';
select table_name from user_constraints
where constraint_name in
(
select r_constraint_name
from user_constraints
where table_name = 'F_FOOD_ITEMS' and CONSTRAINT_TYPE = 'R'
);
"REGULAR_CODE" to "CODE" in "F_REGULAR_MENUS"
"PROMO_CODE" to "CODE" in "F_PROMOTIONAL_MENUS"
6. List the primary key to foreign key relationships required to go from the F_SHIFTS
table to the F_REGULAR_MENUS table.

This must be there to identify hours served.

Now, ideally it should have been like this:
But seems to be, our fellow restaurant owner have separate regular menu for each shift.
So ignore suggestion above.

Now again, values of hours served in regular menu table ['6-11am', '11-9pm'] don’t
match any shift timings [8am to 12pm, 6pm to 10pm], seems to be these values on
menu side are meant to be lost while this exercise 
--remove HOURS_SERVED
alter table "F_REGULAR_MENUS" drop column "HOURS_SERVED" ;
--add SHIFT_CODE
alter table "F_REGULAR_MENUS" add "SHIFT_CODE" NUMBER(5, 0) ;
--give some value so that it may be set to not null later
UPDATE "F_REGULAR_MENUS" SET "SHIFT_CODE" = 1;
--set SHIFT_CODE to not null
ALTER TABLE "F_REGULAR_MENUS" MODIFY ("SHIFT_CODE" NUMBER(5,0)
CONSTRAINT "F_RMU_SHIFT_CODE_NN" NOT NULL ENABLE);
--now create the foreign key constraint
ALTER TABLE "F_REGULAR_MENUS" ADD CONSTRAINT
"F_RMU_SHIFT_CODE_FK" FOREIGN KEY ("SHIFT_CODE") REFERENCES
"F_SHIFTS" ("CODE") ENABLE;
7. Which table(s) contains null values?
As per definition available at
https://docs.oracle.com/cd/B19306_01/server.102/b14237/statviews_2094.htm, below
mentioned query would have been sufficient to give this result:
SELECT distinct "TABLE_NAME" FROM "USER_TAB_COLUMNS" WHERE
"NULLABLE" = 'Y' and "NUM_NULLS" > 0
Or
SELECT count(distinct "TABLE_NAME") FROM "USER_TAB_COLUMNS" WHERE
"NULLABLE" = 'Y' and "NUM_NULLS" > 0
But, It didn’t work as per design, due to some index issues.
So I removed the last condition in above query and manually verified each table: [10]
Table
EMPLOYEES
LOCATIONS
EMP_DETAILS_VIEW
JOBS
JOB_GRADES
REGIONS
COUNTRIES
D_CLIENTS
D_JOB_ASSIGNMENTS
D_SONGS
D_VENUES
F_ORDERS
F_STAFFS
Has some field null
yes
no
n/a
no
no
no
no
no
no
no
no
no
yes
it's a view
JOB_HISTORY
WF_COUNTRIES
D_EVENTS
HTMLDB_PLAN_TABLE
WF_CURRENCIES
WF_SPOKEN_LANGUAGES
D_PLAY_LIST_ITEMS
F_PROMOTIONAL_MENUS
DEPARTMENTS
F_FOOD_ITEMS
D_PARTNERS
no
yes
no
yes
yes
yes
yes
no
yes
yes
yes
But to verify my manual observation above, I needed something, for which I created below
mentioned query:
SELECT 'SELECT count(*) from "' || "TABLE_NAME" || '" where "' ||
"COLUMN_NAME" ||'" is null;' as query FROM "USER_TAB_COLUMNS" WHERE
"NULLABLE" = 'Y';
This gives output like:
SELECT count(*) from "D_PARTNERS" where "EXPERTISE" is null;
(112 count)
I executed each query generated above to verify above results.
Sample queries to verify above results (off the 112 generated):
SELECT count(*) from "DEPARTMENTS" where "MANAGER_ID" is null;
SELECT count(*) from "D_PARTNERS" where "EXPERTISE" is null;
SELECT count(*) from "D_PLAY_LIST_ITEMS" where "COMMENTS" is null;
SELECT count(*) from "EMPLOYEES" where "COMMISSION_PCT" is null;
SELECT count(*) from "F_FOOD_ITEMS" where "REGULAR_CODE" is null;
SELECT count(*) from "F_STAFFS" where "OVERTIME_RATE" is null;
SELECT count(*) from "HTMLDB_PLAN_TABLE" where "REMARKS" is null;
SELECT count(*) from "WF_COUNTRIES" where "COUNTRY_TRANSLATED_NAME"
is null;
SELECT count(*) from "WF_CURRENCIES" where "COMMENTS" is null;
SELECT count(*) from "WF_SPOKEN_LANGUAGES" where "COMMENTS" is null;
Vocabulary
Identify the vocabulary word for each definition below.
Display data from two or more related tables. [Join]
A symbol used to perform an operation on some values. [Arithmetic Operator]
An implementation of an attribute or relationship in a table. [Column]
The capability in SQL to choose the columns in a table that you want returned from a query. [Projection]
A value that is unavailable, unassigned, unknown, or inapplica-ble. [NULL]
Renames a column heading. [Column Alias]
A mathematical equation. [Arithmetic expression]
The capability in SQL to choose the rows in a table returned from a query. [Selection]
Retrieves information from the database. [Select statement]
Specifies the columns to be displayed. [Select clause]
Specifies the table containing the column listed in the select clause. [From clause]
An individual SQL statement.[Keyword]
Part of a SQL statement. [Clause]
A combination of the two clauses. [Statement]
Try It / Solve It
Now you know the basics of a SELECT statement, It's time to practice what you've learned.
1. Write a SQL statement that demonstrates projection and selection.
SELECT constraint_name
FROM user_constraints
WHERE table_name in (upper('f_shifts'), upper('F_STAFFS'));
2. Write a query that displays the last_name and email addresses for all the people in the
DJs on Demand d_client table. The column headings should appear as “Client” and
“Email Address.”
SELECT last_name AS "Client", email AS "Email Address"
FROM d_clients;
3. The manager of Global Fast Foods decided to give all employees at 5%/hour raise + a
$.50 bonus/hour. However, when he looked at the results, he couldn't figure out why the
new raises were not as he predicted. Ms. Doe should have a new salary of $7.59, Mr.
Miller's salary should be $11.00, and Monique Tuttle should be $63.50. He used the
following query. What should he have done?
SELECT last_name, salary *.05 +.50
FROM f_staffs;
He assumed wrong that if he has to give 5% raise, he should multiply by ‘.05’. It
should be multiplied by ‘1.05’ and rounded to two digits after decimal. Precedence
he knows about correctly. (*/+- grouped, */ being higher precedence than +- and
then left to right in statement for the two in same group)
SELECT last_name, ROUND(salary*1.05 +.50, 2) as "Salary"
FROM f_staffs;
4. Which of the following would be the easiest way to see all rows in the d_songs table?
a. SELECT id, title, duration, artist, type_code
b. SELECT columns
c. SELECT *
d. SELECT all
5. If tax = 8.5% * car_cost and license = car_cost * .01%, which value will produce the
largest car payment?
a. Payment = (car_cost * 1.25) + 5.00 - (tax) - (license)
b. Payment = car_cost * 1.25 + 5.00 - (tax - license)