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arnold 7490

This document provides an overview of SQL (Structured Query Language) including its history, data definition and manipulation capabilities. Key topics covered include SQL's data types, basic queries using SELECT, FROM and WHERE clauses, joins, aggregation, null values, triggers and indexes. The document also discusses SQL standards over time and commercial database implementations of SQL features.

Technologie
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[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
History ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Data Definition Language ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Allows the specification of:
Create Table Construct ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Domain Types in SQL ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Integrity Constraints on Tables ,[object Object],[object Object],Example: Declare branch_name as the primary key for branch . create table branch ( branch_name char(15) , branch_city char(30) not null , assets integer, primary key ( branch_name )) primary key declaration on an attribute automatically ensures not null in SQL-92 onwards, needs to be explicitly stated in SQL-89
Basic Insertion and Deletion of Tuples ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Drop and Alter Table Constructs ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Basic Query Structure ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
The select Clause ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
The select Clause (Cont.) ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
The select Clause (Cont.) ,[object Object],[object Object],[object Object],[object Object],[object Object]
The where Clause ,[object Object],[object Object],[object Object],[object Object],[object Object]
The from Clause ,[object Object],[object Object],[object Object],[object Object],[object Object],select customer_name, borrower.loan_number, amount from borrower, loan where borrower.loan_number = loan.loan_number and branch_name = 'Perryridge'
The Rename Operation ,[object Object],[object Object],[object Object],select customer_name, borrower.loan_number as loan_id, amount from borrower, loan where borrower.loan_number = loan.loan_number
Tuple Variables ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],select customer_name, T.loan_number, S.amount from borrower as T, loan as S where T.loan_number = S.loan_number
Example Instances ,[object Object],[object Object],R1 S1 S2
Basic SQL Query ,[object Object],[object Object],[object Object],[object Object],SELECT [DISTINCT] target-list FROM relation-list WHERE qualification
Conceptual Evaluation Strategy ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Example of Conceptual Evaluation SELECT S.sname FROM Sailors S, Reserves R WHERE S.sid=R.sid AND R.bid=103
A Note on Range Variables ,[object Object],SELECT S.sname FROM Sailors S, Reserves R WHERE S.sid=R.sid AND bid=103 SELECT sname FROM Sailors, Reserves WHERE Sailors.sid=Reserves.sid AND bid=103 It is good style, however, to use range variables always! OR
Find sailors who’ve reserved at least one boat ,[object Object],[object Object],SELECT S.sid FROM Sailors S, Reserves R WHERE S.sid=R.sid
Expressions and Strings ,[object Object],[object Object],[object Object],SELECT S.age, age1=S.age-5, 2*S.age AS age2 FROM Sailors S WHERE S.sname LIKE ‘B_%B’
String Operations ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Ordering the Display of Tuples ,[object Object],[object Object],[object Object],[object Object]
Duplicates ,[object Object],[object Object],[object Object],[object Object],[object Object]
Duplicates (Cont.) ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Set Operations ,[object Object],[object Object],[object Object],[object Object],[object Object]
Set Operations ,[object Object],( select customer_name from depositor ) except (select customer_name from borrower) ( select customer_name from depositor ) intersect (select customer_name from borrower) ,[object Object],(select customer_name from depositor ) union (select customer_name from borrower) ,[object Object]
Find sid’s of sailors who’ve reserved a red or a green boat ,[object Object],[object Object],[object Object],SELECT S.sid FROM Sailors S, Boats B, Reserves R WHERE S.sid=R.sid AND R.bid=B.bid AND (B.color=‘red’ OR B.color=‘green’) SELECT S.sid FROM Sailors S, Boats B, Reserves R WHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘red’ UNION SELECT S.sid FROM Sailors S, Boats B, Reserves R WHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘green’
Find sid’s of sailors who’ve reserved a red and a green boat ,[object Object],[object Object],[object Object],SELECT S.sid FROM Sailors S, Boats B1, Reserves R1, Boats B2, Reserves R2 WHERE S.sid=R1.sid AND R1.bid=B1.bid AND S.sid=R2.sid AND R2.bid=B2.bid AND (B1.color=‘red’ AND B2.color=‘green’) SELECT S.sid FROM Sailors S, Boats B, Reserves R WHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘red’ INTERSECT SELECT S.sid FROM Sailors S, Boats B, Reserves R WHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘green’ Key field!
Nested Queries ,[object Object],[object Object],[object Object],SELECT S.sname FROM Sailors S WHERE S.sid IN ( SELECT R.sid FROM Reserves R WHERE R.bid=103) Find names of sailors who’ve reserved boat #103:
Nested Queries with Correlation ,[object Object],[object Object],[object Object],SELECT S.sname FROM Sailors S WHERE EXISTS ( SELECT * FROM Reserves R WHERE R.bid=103 AND S.sid =R.sid) Find names of sailors who’ve reserved boat #103:
Aggregate Functions ,[object Object],[object Object]
Aggregate Functions (Cont.) ,[object Object],[object Object],[object Object],select avg (balance) from account where branch_name = 'Perryridge' select count (*) from customer select count (distinct customer_name) from depositor
Aggregate Functions – Group By ,[object Object],Note: Attributes in select clause outside of aggregate functions must appear in group by list select branch_name, count (distinct customer_name) from depositor, account where depositor.account_number = account.account_number group by branch_name
Aggregate Functions – Having Clause ,[object Object],Note: predicates in the having clause are applied after the formation of groups whereas predicates in the where clause are applied before forming groups select branch_name, avg ( balance ) from account group by branch_name having avg ( balance ) > 1200
Nested Subqueries ,[object Object],[object Object],[object Object]
“In” Construct ,[object Object],[object Object],select distinct customer_name from borrower where customer_name not in ( select customer_name from depositor ) select distinct customer_name from borrower where customer_name in ( select customer_name from depositor )
Example Query ,[object Object],[object Object],select distinct customer_name from borrower, loan where borrower.loan_number = loan.loan_number and branch_name = 'Perryridge' and ( branch_name, customer_name ) in (select branch_name, customer_name from depositor, account where depositor.account_number = account.account_number )
“Some” Construct ,[object Object],[object Object],select branch_name from branch where assets > some (select assets from branch where branch_city = ' Brooklyn ' ) select distinct T.branch_name from branch as T, branch as S where T.assets > S.assets and S.branch_city = ' Brooklyn '
“All” Construct ,[object Object],select branch_name from branch where assets > all (select assets from branch where branch_city = 'Brooklyn')
“Exists” Construct ,[object Object],select distinct S.customer_name from depositor as S where not exists ( ( select branch_name from branch where branch_city = 'Brooklyn') except ( select R.branch_name from depositor as T, account as R where T.account_number = R.account_number and S.customer_name = T.customer_name )) ,[object Object],[object Object]
Absence of Duplicate Tuples ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Example Query ,[object Object],select distinct T.customer_name from depositor as T where not unique ( select R.customer_name from account, depositor as R where T.customer_name = R.customer_name and R.account_number = account.account_number and account.branch_name = ' Perryridge ' ) ,[object Object]
Modification of the Database – Deletion ,[object Object],[object Object],[object Object],[object Object]
Example Query ,[object Object],delete from account where balance < ( select avg ( balance ) from account ) ,[object Object],[object Object],[object Object],[object Object]
Modification of the Database – Insertion ,[object Object],[object Object],[object Object],[object Object],[object Object]
Modification of the Database – Insertion ,[object Object],[object Object],[object Object],[object Object]
Modification of the Database – Updates ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Case Statement for Conditional Updates ,[object Object],[object Object]
More on Set-Comparison Operators ,[object Object],[object Object],[object Object],SELECT * FROM Sailors S WHERE S.rating > ANY ( SELECT S2.rating FROM Sailors S2 WHERE S2.sname=‘Horatio’)
Rewriting INTERSECT Queries Using IN ,[object Object],[object Object],Find sid’s of sailors who’ve reserved both a red and a green boat: SELECT S.sid FROM Sailors S, Boats B, Reserves R WHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘red’ AND S.sid IN ( SELECT S2.sid FROM Sailors S2, Boats B2, Reserves R2 WHERE S2.sid=R2.sid AND R2.bid=B2.bid AND B2.color=‘green’)
Division in SQL ,[object Object],SELECT S.sname FROM Sailors S WHERE NOT EXISTS (( SELECT B.bid FROM Boats B) EXCEPT ( SELECT R.bid FROM Reserves R WHERE R.sid=S.sid)) SELECT S.sname FROM Sailors S WHERE NOT EXISTS ( SELECT B.bid FROM Boats B WHERE NOT EXISTS ( SELECT R.bid FROM Reserves R WHERE R.bid=B.bid AND R.sid=S.sid)) Sailors S such that ... there is no boat B without ... a Reserves tuple showing S reserved B Find sailors who’ve reserved all boats. (1) (2)
Aggregate Operators ,[object Object],COUNT (*) COUNT ( [ DISTINCT ] A) SUM ( [ DISTINCT ] A) AVG ( [ DISTINCT ] A) MAX (A) MIN (A) SELECT AVG (S.age) FROM Sailors S WHERE S.rating=10 SELECT COUNT (*) FROM Sailors S SELECT AVG ( DISTINCT S.age) FROM Sailors S WHERE S.rating=10 SELECT S.sname FROM Sailors S WHERE S.rating= ( SELECT MAX (S2.rating) FROM Sailors S2) single column SELECT COUNT ( DISTINCT S.rating) FROM Sailors S WHERE S.sname=‘Bob’
Find name and age of the oldest sailor(s) ,[object Object],[object Object],SELECT S.sname, MAX (S.age) FROM Sailors S SELECT S.sname, S.age FROM Sailors S WHERE S.age = ( SELECT MAX (S2.age) FROM Sailors S2) SELECT S.sname, S.age FROM Sailors S WHERE ( SELECT MAX (S2.age) FROM Sailors S2) = S.age
Motivation for Grouping ,[object Object],[object Object],[object Object],[object Object],SELECT MIN (S.age) FROM Sailors S WHERE S.rating = i For i = 1, 2, ... , 10:
Queries With GROUP BY and HAVING ,[object Object],[object Object],SELECT [DISTINCT] target-list FROM relation-list WHERE qualification GROUP BY grouping-list HAVING group-qualification
Conceptual Evaluation ,[object Object],[object Object],[object Object],[object Object]
Find age of the youngest sailor with age 18, for each rating with at least 2 such sailors SELECT S.rating, MIN (S.age) AS minage FROM Sailors S WHERE S.age >= 18 GROUP BY S.rating HAVING COUNT (*) > 1 Answer relation: Sailors instance:
Find age of the youngest sailor with age 18, for each rating with at least 2 such sailors.
Find age of the youngest sailor with age 18, for each rating with at least 2 such sailors and with every sailor under 60. HAVING COUNT (*) > 1 AND EVERY (S.age <=60) What is the result of changing EVERY to ANY?
Find age of the youngest sailor with age 18, for each rating with at least 2 sailors between 18 and 60. SELECT S.rating, MIN (S.age) AS minage FROM Sailors S WHERE S.age >= 18 AND S.age <= 60 GROUP BY S.rating HAVING COUNT (*) > 1 Answer relation: Sailors instance:
For each red boat, find the number of reservations for this boat ,[object Object],[object Object],[object Object],SELECT B.bid, COUNT (*) AS scount FROM Sailors S, Boats B, Reserves R WHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘red’ GROUP BY B.bid
Find age of the youngest sailor with age > 18, for each rating with at least 2 sailors (of any age) ,[object Object],[object Object],[object Object],[object Object],SELECT S.rating , MIN (S.age) FROM Sailors S WHERE S.age > 18 GROUP BY S.rating HAVING 1 < ( SELECT COUNT (*) FROM Sailors S2 WHERE S.rating=S2.rating )
Find those ratings for which the average age is the minimum over all ratings ,[object Object],SELECT S.rating FROM Sailors S WHERE S.age = ( SELECT MIN ( AVG (S2.age)) FROM Sailors S2) SELECT Temp.rating, Temp.avgage FROM ( SELECT S.rating, AVG (S.age) AS avgage FROM Sailors S GROUP BY S.rating) AS Temp WHERE Temp.avgage = ( SELECT MIN (Temp.avgage) FROM Temp) ,[object Object]
Null Values ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Null Values ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Null Values and Three Valued Logic ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Null Values and Aggregates ,[object Object],[object Object],[object Object],[object Object],[object Object]
Joined Relations** ,[object Object],[object Object],[object Object],[object Object]
Joined Relations – Datasets for Examples ,[object Object],[object Object],[object Object]
Joined Relations – Examples ,[object Object],[object Object]
Joined Relations – Examples ,[object Object],[object Object],[object Object],select customer_name from ( depositor natural full outer join borrower ) where account_number is null or loan_number is null
Joined Relations – Examples ,[object Object],[object Object],[object Object],[object Object]
Derived Relations ,[object Object],[object Object],[object Object],[object Object]
Integrity Constraints (Review) ,[object Object],[object Object],[object Object],[object Object],[object Object]
General Constraints ,[object Object],[object Object],[object Object],CREATE TABLE Sailors ( sid INTEGER, sname CHAR(10), rating INTEGER, age REAL, PRIMARY KEY (sid), CHECK ( rating >= 1 AND rating <= 10 ) CREATE TABLE Reserves ( sname CHAR(10), bid INTEGER, day DATE, PRIMARY KEY (bid,day), CONSTRAINT noInterlakeRes CHECK (`Interlake’ <> ( SELECT B.bname FROM Boats B WHERE B.bid=bid)))
Constraints Over Multiple Relations ,[object Object],[object Object],[object Object],CREATE TABLE Sailors ( sid INTEGER, sname CHAR(10), rating INTEGER, age REAL, PRIMARY KEY (sid), CHECK ( (SELECT COUNT (S.sid) FROM Sailors S) + (SELECT COUNT (B.bid) FROM Boats B) < 100 ) CREATE ASSERTION smallClub CHECK ( (SELECT COUNT (S.sid) FROM Sailors S) + (SELECT COUNT (B.bid) FROM Boats B) < 100 ) Number of boats plus number of sailors is < 100
Triggers ,[object Object],[object Object],[object Object],[object Object],[object Object]
Triggers: Example (SQL:1999) ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

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  • 21. Example of Conceptual Evaluation SELECT S.sname FROM Sailors S, Reserves R WHERE S.sid=R.sid AND R.bid=103
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  • 61. Find age of the youngest sailor with age 18, for each rating with at least 2 such sailors SELECT S.rating, MIN (S.age) AS minage FROM Sailors S WHERE S.age >= 18 GROUP BY S.rating HAVING COUNT (*) > 1 Answer relation: Sailors instance:
  • 62. Find age of the youngest sailor with age 18, for each rating with at least 2 such sailors.
  • 63. Find age of the youngest sailor with age 18, for each rating with at least 2 such sailors and with every sailor under 60. HAVING COUNT (*) > 1 AND EVERY (S.age <=60) What is the result of changing EVERY to ANY?
  • 64. Find age of the youngest sailor with age 18, for each rating with at least 2 sailors between 18 and 60. SELECT S.rating, MIN (S.age) AS minage FROM Sailors S WHERE S.age >= 18 AND S.age <= 60 GROUP BY S.rating HAVING COUNT (*) > 1 Answer relation: Sailors instance:
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