1. 376a. Database Design Dept. of Computer Science Vassar College
Class 7: Domain Relational Calculus and beginning SQL
Prof. Billibon Yoshimi
9/19/2018

2. Find employees with no dependents using Domain Relation Calculus.
EMPLOYEE(FNAME, MINIT, LNAME, SSN, BDATE, ADDRESS, SEX, SALARY, SUPERSSN, DNO)
DEPENDENT(ESSN, DEPENDENT_NAME,SEX,BDATE,RELATIONSHIP)
EMPLOYEE(q-z) and NOT (exists l) (DEPENDENT (lmnop) and t=l)
Prof. Billibon Yoshimi
9/19/2018

3. List names of all managers with at least one dependent.
EMPLOYEE(FNAME, MINIT, LNAME, SSN, BDATE, ADDRESS, SEX, SALARY, SUPERSSN, DNO)
DEPARTMENT(DNAME, DNUMBER, MGRSSN, MGRSTARTDATE)
DEPENDENT(ESSN, DEPENDENT_NAME,SEX,BDATE,RELATIONSHIP)
] t ] j ] l EMPLOYEE(q-z) and DEPARTMENTand dep.mgrssn=employee.ssn and employee.ssn=dependent.essn
Prof. Billibon Yoshimi
9/19/2018

4. SELECT statement No relation to sigma
SQL tables are not sets (they’re multi-sets).
Use DISTINCT to regain set-like quality.
Basically:
SELECT
<attribute list>
FROM
<list of tables>
WHERE
<condition list>
Prof. Billibon Yoshimi
9/19/2018

5. Get the birthdate and address of employees with the name “John B
Get the birthdate and address of employees with the name “John B. Smith”
R.A.
SELECT
BDATE, ADDRESS
FROM
EMPLOYEE
WHERE
FNAME=“JOHN” and MINIT=“B” and LNAME=“SMITH”
Prof. Billibon Yoshimi
9/19/2018

6. Can use SELECT to do join operation too
FNAME, ADDRESS
FROM
EMPLOYEE, DEPARTMENT
WHERE
DEPARTMENT=“RESEARCH” AND DNUMBER=DNO
Print the firstname and address of all employees in the research department.
Prof. Billibon Yoshimi
9/19/2018

7. Resolving ambiguous names
Relationships may have same named attributes.
Use relation.attribute to disambiguate.
When using multiple instances of a relation in a SELECT, use aliases..
SELECT
E.FNAME, E.LNAME, S.FNAME, S.LNAME
FROM
EMPLOYEE as E, EMPLOYEE as S
WHERE
E.SUPERSSN = S.SSN
Prof. Billibon Yoshimi
9/19/2018

8. Can also create attribute aliases,
EMPLOYEE AS E(FN,MI,LN,SSN,BD,ADDR,S,SAL,SSSN,DNO)
Prof. Billibon Yoshimi
9/19/2018

9. SELECT - FROM statements
Unspecified WHERE is *.
If multiple relations are specified in FROM then the tuples are formed from the cross product of all FROM relations.
Prof. Billibon Yoshimi
9/19/2018

10. Other modifiers to SELECT
SELECT * FROM - WHERE
* - Selects all attributes
SELECT ALL x FROM WHERE
- get all values, including duplicates
SELECT DISTINCT * FROM WHERE
- removes duplicates *EXPENSIVE*
Prof. Billibon Yoshimi
9/19/2018

11. UNION, EXCEPT and INTERSECT operations
(SELECT *) UNION (SELECT *)
Sub sets should be union compatible, same attribute tuples, same ordering.
UNION gives the union of all tuples (duplicates eliminated).
EXCEPT gives the set difference.
INTERSECT gives the intersection of the tuples.
Prof. Billibon Yoshimi
9/19/2018

12. Comparing strings In WHERE statements use LIKE
WHERE NAME LIKE “%ITH%”
% - replaces arbitrary numbers of characters
_ - replaces a single character
“_____5_____”
In MySQL, use REGEX too.
^ - match beginning of line
$- match end of line
[bB] - match any one char in bracket
* - zero or one instances of preceding thing
Match anywhere in the input, unlike LIKE
Prof. Billibon Yoshimi
9/19/2018

13. Operations on return values
+,-,*,/, BETWEEN (SALARY BETWEEN X AND Y)
|| is string append
Like this
SELECT
name||address, salary*1.3
You can also call functions using the select statement to do evaluations
NOW( ), SIN(3.1415), 3*5
Prof. Billibon Yoshimi
9/19/2018

14. SELECT FROM WHERE ORDER BY
ORDER BY attribute ASC|DESC, attribute ASC|DESC
By default it is in ascending order.
Order on first attribute, then second,then third.
Prof. Billibon Yoshimi
9/19/2018

15. Nested queries
SELECT
FROM
WHERE X IN (SELECT as argument);
X is an attribute name to compare or is a tuple (attribute, attribute, etc.)
In the case of ambiguity, attribute is associated with innermost nested SELECT.
Prof. Billibon Yoshimi
9/19/2018

16. Example SELECT DISTINCT PNUMBER FROM PROJECT WHERE PNUMBER IN
(SELECT PNUMBER
FROM PROJECT, DEPARTMENT, EMPLOYEE
WHERE DNUM=DNUMBER AND MGRSSN=SSN AND LNAME=‘Smith’
) OR
FROM WORKS_ON_EMPLOYEE
WHERE ESSN=SSN AND LNAME=‘Smith’);
Prof. Billibon Yoshimi
9/19/2018

17. Other operators used like IN
= ANY or = SOME
> ANY
< ANY, <= ANY, >= ANY, <> ANY
(similar for SOME but only needs to find one case)
Why isn’t there a NONE?
Prof. Billibon Yoshimi
9/19/2018

18. When nested query is correlated to outer query.
For each tuple satisfying the outside query,
Apply the nested query.
Select all employees with the same first name as one of their progeny.
E.g. SELECT *
FROM EMPLOYEE AS E
WHERE E.SSN IN (SELECT ESSN
FROM DEPENDENT
WHERE E.FNAME=DEPEND_NAME)
Prof. Billibon Yoshimi
9/19/2018

19. All nested queries can ALWAYS be expressed as single block queries
SELECT *
FROM EMPLOYEE AS E
WHERE E.SSN IN (SELECT ESSN
FROM DEPENDENT
WHERE E.FNAME=DEPEND_NAME);
We explicitly indicate the JOIN.
FROM EMPLOYEE AS E, DEPENDENT AS D
WHERE E.SSN=D.ESSN AND E.FNAME=D.DEPEND_NAME;
Prof. Billibon Yoshimi
9/19/2018

20. EXISTS Also used in WHERE clause SELECT * FROM EMPLOYEE AS E
WHERE EXISTS (SELECT * FROM DEPENDENT
WHERE E.SSN=ESSN AND E.FNAME=DEPEND_NAME);
WHERE NOT EXISTS (SELECT * FROM DEPENDENT
WHERE E.SSN=ESSN);
Prof. Billibon Yoshimi
9/19/2018

21. General form of EXISTS EXISTS (Q) where Q is the query.
Returns TRUE if the query returns at least one tuple.
Returns FALSE if query returns no tuples.
Can use multiple EXISTS in a WHERE statement
e.g. WHERE EXISTS (employee who is a manager) AND EXISTS (EMPLOYEE HAS DEPENDENTS)
What is same query without EXISTS (use one nested select)
What is same query without any nesting?
Select
FNAME, LNAME
FROM EMPLOYEE AS E, DEPARTMENT AS D, DEPENDENT AS CHILD
WHERE
E.SSN=D.MGRSSN AND E.SSN=CHILD.ESSN
Prof. Billibon Yoshimi
9/19/2018

22. Explicit sets
Another way to write the argument for IN or similar functions
Normally, WHERE X IN (SELECT …)
Can also be written WHERE NAME IN (‘Bill’, ‘Bob’, ‘Jack’)
Prof. Billibon Yoshimi
9/19/2018

23. Comparison to NULL WHERE X IS NULL
Compares the value of attribute X with NULL, returns TRUE when X is NULL.
e.g. employees with no manager
SELECT * FROM EMPLOYEE WHERE SUPERSSN IS NULL
Prof. Billibon Yoshimi
9/19/2018

24. Use AS to rename any attribute or relation
Remember
SELECT
FROM EMPLOYEE AS E, PROJECT AS P;
Can rename columns returned in relation too.
SELECT FNAME AS FIRST_NAME, LNAME AS LAST_NAME
Will output a relation with new attribute names.
Prof. Billibon Yoshimi
9/19/2018

25. Explicit JOINs
In the select statement there is an implict join whenever there is more than 1 relation in the FROM section.
Control the join explicitly by writing it in.
SELECT *
FROM EMPLOYEE JOIN DEPARTMENT AS D ON DNO=DNUMBER
WHERE D.DNAME=‘RESEARCH’;
By default any join is an INNER JOIN (added only if matching tuple is found in other relation.)
Prof. Billibon Yoshimi
9/19/2018

26. JOIN types Remember different types again.
NATURAL JOIN - removes duplicate attribute (by name). Joins on all same named attributes.
EMPLOYEE(FNAME, MINIT, LNAME, SSN, BDATE, ADDRESS, SEX, SALARY, SUPERSSN, DNO)
DEPARTMENT(DNAME, DNUMBER, MGRSSN, MGRSTARTDATE)
SELECT * FROM EMPLOYEE NATURAL JOIN (DEPARTMENT AS DEPT(DNAME,DNO,MSSN,MSDATE))
WHERE …
Prof. Billibon Yoshimi
9/19/2018

27. OUTER JOIN Show employee name and supervisor name for all employees.
(some employees have null for superssn field!)
SELECT E.FNAME| ‘ ‘ | E.LNAME, S.FNAME| ‘ ‘ | S.LNAME
FROM EMPLOYEE AS E LEFT OUTER JOIN (EMPLOYEE AS S ON E.SUPERSSN = S.SSN);
Why no WHERE clause?
Prof. Billibon Yoshimi
9/19/2018

28. Rest of joins. INNER JOIN (or just JOIN), LEFT OUTER JOIN,
FULL OUTER JOIN,
RIGHT OUTER JOIN
Note: in last 3 cases the OUTER can be left out.
Prof. Billibon Yoshimi
9/19/2018

29. Aggregate operations A few of these (many more)
COUNT() - counts the number of tuples or values in the query
SUM() - applied to set of multiset of attributes.
MAX() - may be applied to any totally orderable domain.
MIN()
AVG()
e.g. SELECT SUM(SALARY), MAX (SALARY), MIN(SALARY), AVG(SALARY
FROM EMPLOYEE;
Prof. Billibon Yoshimi
9/19/2018

30. COUNT() SELECT COUNT (*) FROM EMPLOYEE; SELECT COUNT (SALARY)
SELECT COUNT (DISTINCT SALARY)
First 2 results are different from 3rd.
Prof. Billibon Yoshimi
9/19/2018

31. Use COUNT in nested SELECT
Select all employees with 2 or more dependents
SELECT *
FROM EMPLOYEE
WHERE ( SELECT COUNT (*)
FROM DEPENDENT
WHERE SSN=ESSN) >= 2;
For each employee, if the number of dependents for that employee is > 2, include the tuple.
Prof. Billibon Yoshimi
9/19/2018

32. Use GROUP BY to group statistics
SELECT DNO, COUNT(*), AVG(SALARY)
FROM EMPLOYEE
GROUP BY DNO;
GROUP BY specifies the subgroups the statistics are applied to.
What if GROUP BY is given a primary key?
Prof. Billibon Yoshimi
9/19/2018

33. What does the following do?
SELECT PNUMBER, PNAME, COUNT (*)
FROM PROJECT, WORKS_ON
WHERE PNUMBER=PNO
GROUP BY PNUMBER, PNAME;
In this case, the JOIN happens first, then the GROUP BY statistics are applied.
Remember there is one WORKS_ON tuple for every project every person works on.
Prof. Billibon Yoshimi
9/19/2018

34. Use HAVING to filter GROUP BY results
SELECT PNUMBER, PNAME, COUNT (*)
FROM PROJECT, WORKS_ON
WHERE PNUMBER=PNO
GROUP BY PNUMBER, PNAME
HAVING COUNT(*)>2;
Same as before except, after the group statistics are calculated, the troups are passed through the HAVING clause.
Prof. Billibon Yoshimi
9/19/2018

35. Be careful with the order though
What does this do?
SELECT DNAME, COUNT(*)
FROM DEPARTMENT, EMPLOYEE
WHERE DNUMBER=DNO AND SALARY>4000
GROUP BY DNAME
HAVING COUNT(*) > 5;
Prof. Billibon Yoshimi
9/19/2018

36. What if we really wanted
Number of employees making over $40,000 in departments with greater than 5 employees….
Prof. Billibon Yoshimi
9/19/2018

37. Result Query SELECT DNAME, COUNT(*) FROM EMPLOYEE, DEPARTMENT
WHERE DNO IN (
SELECT DNO
FROM EMPLOYEE
GROUP BY DNO
HAVING COUNT(*) >5 )
GROUP BY DNAME;
Prof. Billibon Yoshimi
9/19/2018

38. How to think about SELECT queries
1. FROM first
2. WHERE
3. GROUP BY
4. HAVING
5. ORDER BY
6. SELECT
Prof. Billibon Yoshimi
9/19/2018

39. INSERT command INSERT INTO EMPLOYEE
VALUES (‘john’,’q’,’public’,’ ’,’ ’,’address’,’m’,20000,NULL,3)
Or a list of values separated by commas.
Can also specify attribute mapping
INSERT INTO EMPLOYEE(LNAME,FNAME,SSN)
VALUES (same order).
Prof. Billibon Yoshimi
9/19/2018

40. INSERT constraints
Values not specified map to NULL or default value if specified.
All integrity checks are checked.
- Entity integrity (no NULL primary key)
- Referential integrity (Foreign key must be valid)
Prof. Billibon Yoshimi
9/19/2018

41. How to implement nested queries in MySQL.
CREATE TABLE TEMPTBL
(DNAME VARCHAR(15),
NO_EMPL INT,
TOTAL_SAL INT);
INSERT INTO TEMPTBL (DEPT_NAME, NO_EMPL, TOTAL_SAL)
SELECT DNAME, COUNT(*),SUM(SALARY)
FROM EMPLOYEE JOIN DEPARTMENT ON DNO=DNUMBER
GROUP BY DNAME;
Then use TEMPTBL where you’d normally use the nested SELECT.
Prof. Billibon Yoshimi
9/19/2018

42. DELETE command DELETE FROM <relation> WHERE <condition>;
DELETE FROM EMPLOYEE WHERE SSN=‘ ’;
DELETE FROM EMPLOYEE;
Beware of the last one, it deletes all tuples from the EMPLOYEE relations.
Depending on referential triggers, references to deleted tuples can cascase or delete referers.
Prof. Billibon Yoshimi
9/19/2018

43. UPDATE command UPDATE <relation>
SET <attribute=value pairs>
WHERE <condition>
Prof. Billibon Yoshimi
9/19/2018

44. Views
Virtual tables.
Prof. Billibon Yoshimi
9/19/2018

45. Project
Groups of 2.
Demonstrate projects instead of final exam.( Monday Dec AM in class)
Deadlines:
Summary of system requirements
Entity-relation diagram
Relational database design
Application program design
Prof. Billibon Yoshimi
9/19/2018

46. Project
Document all design decisions and justifications for those decisions.
Include all instances used for the project in Appendix.
Oct 21 -
Prof. Billibon Yoshimi
9/19/2018

47. Prof. Billibon Yoshimi
9/19/2018