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IT420: Database Management and Organization

SQL part 37 February 2006

Adina Crăiniceanuwww.cs.usna.edu/~adina

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Announcements

Exam next Tuesday 2 hours Closed book/closed notes No computers Covers all material

Labs returned on Friday

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Previously

SQL DDL: Data Definition Language CREATE, DROP, ALTER

DML: Data Manipulation Language INSERT DELETE UPDATE SELECT

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Today

More about SELECT

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The SQL SELECT Statement

Basic SQL Query: SELECT [DISTINCT] column_name(s)

FROM table_name(s)

[WHERE conditions]

[ORDER BY some_column_names [ASC/DESC] ]

*Terms between [ ] are optional

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WHERE Clause Options

AND, OR IN, NOT IN, BETWEEN =, >, <, <>, >=

SELECT SNbFROM Students S, Enrolled EWHERE S.SNb = E.Nb AND

E.Cid NOT IN (‘ComSci’, ‘Math’)

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Calculations in SQL

Simple arithmetic Five SQL aggregate operators:

COUNT SUM AVG MIN MAX

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Find the age of the youngest student Students(Alpha, LName, FName, Class,

Age)

SELECT MIN(Age)

FROM Students

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Aggregate Operators

So far, aggregate operations applied to all (qualifying) rows

We want to apply them to each of several groups of rows

Students(Alpha, LName, SName, Class, Age) Find the age of the youngest student for each

class

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Example

Students(Alpha, LName, FName, Class, Age) Find the age of the youngest student for each

class

If class values go from 1 to 4 we can write 4 queries that look like this:

SELECT MIN (S.Age)

FROM Students S

WHERE S.Class = i

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GROUP-BY Clause

SELECT grouping_columns(s), aggregates

FROM table_name(s)

[WHERE conditions]

GROUP BY grouping_columns

SELECT Class, MIN(Age)

FROM Students

GROUP BY Class

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Conceptual Evaluation

SQL query semantics: Compute the cross-product of table_names Discard resulting rows if they fail conditions Delete columns that are not specified in SELECT Remaining rows are partitioned into groups by the

value of the columns in grouping-columns One answer row is generated per group

Note: Does not imply query will actually be evaluated this way!

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GROUP BY Exercise

Students(Alpha, LName, FName, Class, Age)

For each last name, find the number of students with same last name

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HAVING Clause

SELECT [grouping_columns(s), aggregates

FROM table_name(s)

[WHERE conditions]

GROUP BY grouping_columns

HAVING group_conditions

HAVING restricts the groups presented in the result

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Example

SELECT Class, MIN(Age)

FROM Students

WHERE MajDeptName = ‘ComSci’

GROUP BY Class

HAVING Class > 2

What does the query compute?

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Another GROUP BY Example

SKU_DATA(SKU, SKU_description, Buyer, Department)

SELECT Department, COUNT(*) AS

Dept_SKU_Count

FROM SKU_DATA

WHERE SKU <> 302000

GROUP BY Department

HAVING COUNT (*) > 1

ORDER BY Dept_SKU_Count;

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Select students with age higher than average SELECT *

FROM Students

WHERE Age > AVG(Age)

Illegal!

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Subqueries

SELECT *

FROM Students

WHERE Age > (SELECT AVG(Age)

FROM Students)

Second select is a subquery (or nested query) You can have subqueries in FROM or HAVING

clause also

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Subqueries Exercise

Students(Alpha, LName, FName, Class, Age) Enroll(Alpha, CourseID, Semester, Grade)

1. Find alpha for students enrolled in both ‘IT420’ and ‘IT334’

2. Find name of students enrolled in both ‘IT420’ and ‘IT334’

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Class Exercise

Students(Alpha, LName, FName, Class, Age)

Enroll(Alpha, CourseID, Semester, Grade)

Find the name of students enrolled in ‘IT420’ Usual way Use subqueries

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SELECT FName, LName

FROM Students S

WHERE S.Alpha IN

(SELECT Alpha

FROM Enroll E

WHERE E.CourseID = ‘IT420’)

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Correlated Subqueries

SELECT FName, LName

FROM Students S

WHERE EXISTS

(SELECT *

FROM Enroll E

WHERE E.CourseID = ‘IT420’

AND E.Alpha = S.Alpha)

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Subqueries versus Joins

Subqueries and joins both process multiple tables.

Subquery can only be used to retrieve data from the top table.

Join can be used to obtain data from any number of tables

Correlated subquery can do work that is not possible with joins.

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Class Exercise

What does this query compute: SELECT FName, LName

FROM Students S, Enroll E1, Enroll E2WHERE S.Alpha = E1.Alpha

AND S.Alpha = E2.AlphaAND E1.CourseID = ‘IT420’AND E2.CourseID = ‘IT344’

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JOIN ON Syntax Students(Alpha, LName, FName, Class, Age) Courses(CourseID, Description, Textbook) Enroll(Alpha, CourseID, Semester, Grade)

Find the names of students enrolled in ‘IT420’

SELECT LName, FNameFROM Students S JOIN Enroll C

ON S.Alpha = C.AlphaWHERE CourseID = ‘IT420’

Find the names of students enrolled in ‘Database Management’

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Multiple JOIN ON

Find the names of students enrolled in ‘Database Management’

SELECT LName, FNameFROMEnroll E JOIN Courses C

ON E.CourseID = C.CourseIDJOIN Students ON E.Alpha = S.Alpha

WHERE C.Description = ‘Database Management’

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Outer Joins

Find all students and courses in which they are enrolled

SELECT S.Alpha, S.LName, S.FName, E.CourseID

FROM Students S LEFT JOIN Enrolled E

ON S.Alpha = E.Alpha

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Class Exercise

Students(Alpha, LName, FName, Class, Age)

Courses(CourseID, Description, Textbook) Enroll(Alpha, CourseID, Semester, Grade)

Find the age of youngest student older than 18, for each course with at least one such student enrolled in it