1. 3.1 Chapter 3: SQL Schema used in examples p. 75-109 (omit 3.8.2, 3.10.5, 3.11)

2. 3.2 Data Definition Language the schema for each relation the domain of values associated with each attribute integrity constraints the set of indices to be maintained for each relations security and authorization information for each relation the physical storage structure of each relation on disk Allows the specification of a set of relations and information about each relation Including:

3. 3.3 Domain Types in SQL char(n). Fixed length character string with user-specified length n varchar(n). Variable length character strings with user-specified maximum length n int. Integer (machine-dependent) smallint. Small integer (machine-dependent) numeric(p,d). Fixed point number with user-specified precision of p digits, with n digits to the right of decimal point real, double precision. Floating point and double-precision floating point numbers with machine-dependent precision float(n). Floating point number with user-specified precision More in Chapter 4

4. 3.4 Create Table A relation is defined using create table: create table r (A 1 D 1, A 2 D 2,..., A n D n, (integrity-constraint 1 ),..., (integrity-constraint k ))  r is the name of the relation  each A i is an attribute name in the schema of relation r  D i is the data type of values in the domain of attribute A i Example (requires that branch_name not be null) create table branch (branch_namechar(15) not null, branch_citychar(30), assetsinteger)

5. 3.5 Integrity Constraints in Create Table not null primary key (A 1,..., A n ) Declare branch_name as the primary key for branch. create table branch (branch_namechar(15), branch_citychar(30), assetsinteger, primary key (branch_name)) primary key ensures not null in SQL-92 onwards A primary key can have multiple attributes : primary key (branch_name, branch_city) (more integrity constraints later)

6. 3.6 Drop and Alter Table drop table deletes a relation from the database alter table to add attributes to an existing relation: alter table r add A D  A is an attribute to be added to relation r  D is the domain of A.  all tuples in the relation are assigned null for the new attribute alter table to drop attributes: alter table r drop A  A is an attribute of relation r

7. 3.7 Basic Query Structure Typical query: select A 1, A 2,..., A n from r 1, r 2,..., r m where P  A i is an attribute  r i s a relation  P is a predicate Equivalent to the relational algebra The result of an SQL query is a relation

8. 3.8 select select corresponds to projection in the relational algebra Find the names of all branches in loan: select branch_name from loan Relational algebra:  branch_name (loan) SQL does not permit '–' (branch_name, not branch_name) SQL names are case-insensitive

9. 3.9 select (cont.) SQL allows duplicates in relations as well as in query results Keyword distinct eliminates duplicates Find the names of all branches in the loan relations and remove duplicates select distinct branch_name from loan Keyword all specifies that duplicates not be removed select all branch_name from loan all is the default

10. 3.10 select (cont.) Asterisk denotes “all attributes” select * from loan select can contain arithmetic expressions: select loan_number, branch_name, amount  100 from loan returns a relation with attribute amount multiplied by 100

11. 3.11 where where corresponds to selection (  in the relational algebra Find loan number for all loans made at the Aurora branch with loan amounts greater than $1200: select loan_number from loan where branch_name = “Aurora” and amount > 1200 (SQL uses the logical connectives and, or, and not) Find the loan number of loans with loan amounts between $90,000 and $100,000: select loan_number from loan where amount  90000 and amount  100000 Using between: select loan_number from loan where amount between 90000 and 100000

12. 3.12 from from corresponds to Cartesian product in the relational algebra. Find borrower x loan: select  from borrower, loan Natural join has to be made explicit E.g. find the name and loan number of all customers with a loan at the Aurora branch: select customer_name, borrower.loan_number from borrower, loan where borrower.loan_number = loan.loan_number and branch_name = “Aurora”

13. 3.13 MySQL MySQL can be run from a command line, Query Browser, or application program interface Free, popular, well-supported Download from mysql.com Access Its own version of SQL Query by Example (QBE)

14. 3.14 The Rename Operation Renaming relations and attributes is accomplished through as: old_name as new_name Find the name and loan number of all customers with a loan at the Aurora branch; replace the column name loan_number with the name loan_id. select customer_name, borrower.loan_number as loan_id, amount from borrower, loan where borrower.loan_number = loan.loan_number and branch_number = “Aurora” (customer_name, loan_number) (loan_number, branch_name, amount) = "Aurora" (customer_name, loan_id, amount)

15. 3.15 Tuple Variables Tuple variables are defined in the from clause via as Find the customer names and their loan numbers for all customers having a loan at some branch select customer_name, T.loan_number from borrower as T, loan as S where T.loan_number = S.loan_number Find the names of all branches that have greater assets than some branch located in Brooklyn. select distinct T.branch_name from branch as T, branch as S where T.assets > S.assets and S.branch_city = “Brooklyn” Try: find the names of all customers with more than one loan select distinct S.customer_name from borrower as S, borrower as T, where S. customer_namer = T. customer_name and S.loan_number ≠ T.loan_number

16. 3.16 String Operations Patterns are described using two special characters:  % matches any substring.  _ matches any character. Find the names of all customers whose street includes the substring “Main”. select customer_name from customer where customer_street like ' % Main% ' What does this do: where customer_street like "M_in"

17. 3.17 Ordering the Tuples List the names of all customers having a loan in Perryridge branch alphabetically select distinct customer_name from borrower, loan where borrower loan_number = loan.loan_number and branch_name = ' Perryridge ' order by customer_name desc desc for descending order, asc for ascending order Ascending order is the default

18. 3.18 Set Operations union, intersect, and except correspond to the relational algebra operations    To retain duplicates use union all, intersect all and except all Find the names of all customers who have a loan, an account, or both: (select customer_name from depositor) union (select customer_name from borrower) Find all customers who have both a loan and an account: (replace union with intersect) Find all customers who have an account but no loan: (replace union with except)

19. 3.19 Aggregate Functions These functions operate on the values of a column of a relation avg: average value min: minimum value max: maximum value sum: sum of values count: number of values Find the average account balance at the Aurora branch select avg (balance) from account where branch_name = “Aurora” Find the number of tuples in the customer relation select count (*) from customer Find the number of depositors in the bank. select count (distinct customer_name) from depositor

20. 3.20 Aggregate Functions – Group and Having Sometimes we want to calculate over a group select branch_name, avg (balance) from account group by branch_name finds the average for each group To restrict within those group calculations: branches where the average account balance is more than $1,200 select branch_name, avg (balance) from account group by branch_name having avg (balance) > 1200

21. 3.21 Null Values Tuples may be null for some of their attributes null signifies unknown or that a value does not exist Use is null to check for null values: find all loan numbers in loan with null values for amount select loan_number from loan where amount is null Aggregate operations (except count(*)) ignore tuples with null values on the aggregated attributes E.g. select avg (amount) from loan ignores null amounts

22. 3.22 Nested Subqueries Find all customers who have both an account and a loan select distinct customer_name from borrower where customer_name in (select customer_name from depositor) Can we do it without nesting? (select distinct customer_name from borrower) intersect (select distinct customer_name from depositor)

23. 3.23 Nested Subqueries Find the names of all customers who have a loan at the bank but do not have an account at the bank select distinct customer_name from borrower where customer_name not in (select customer_name from depositor) Can we do it without nesting? (select distinct customer_name from borrower) except (select distinct customer_name from depositor)

24. 3.24 Set Comparison Find all branches that have greater assets than some branch located in Brooklyn Same query using > some 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 '

25. 3.25 Example Find the names of all branches that have greater assets than all branches located in Brooklyn select branch_name from branch where assets > all (select assets from branch where branch_city = 'Brooklyn')

26. 3.26 Test for Absence of Duplicate Tuples unique tests whether a subquery has duplicate tuples in its result Find all customers with at most one account at Perryridge branch select T.customer_name from depositor as T where 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') Relational algebra?

27. 3.27 Derived Relations SQL allows a subquery in the from clause Find the average account balance of those branches where the average account balance is greater than $1200 select branch_name, avg_balance from (select branch_name, avg (balance) from account group by branch_name ) as branch_avg ( branch_name, avg_balance ) where avg_balance > 1200

28. 3.28 Views A relation that is not of the conceptual model but is made visible to a user as a “virtual relation” A way to hide certain data from some users Defined using create view create view v as where is any legal SQL expression The view name is represented by v. When a view is created  the query expression is stored in the database  it is substituted into queries using the view

29. 3.29 Example A view consisting of branches and their customers Find all customers of the Aurora branch create view all_customer as (select branch_name, customer_name from depositor, account where depositor.account_number = account.account_number ) union (select branch_name, customer_name from borrower, loan where borrower.loan_number = loan.loan_number ) select customer_name from all_customer where branch_name = 'Aurora'

30. 3.30 Modification of the Database – Deletion Delete all account records at the Aurora branch delete from account where branch_name = “Aurora” Delete all accounts at every branch located in Needham. delete from account where branch_name in (select branch_name from branch where branch_city = “Needham”) What does this mean? delete from depositor where account_number in (select account_number from branch, account where branch_city = “Needham” and branch.branch_name = account.branch_name)

31. 3.31 Example Delete the record of all accounts with balances below the average at the bank. delete from account where balance < (select avg (balance) from account) Problem: deleting tuples from deposit changes the average balance SQL's solution: 1.first, compute avg (balance) 2.next, delete all tuples found above it (without recomputing avg)

32. 3.32 Modification of the Database – Insertion Add a new tuple to account insert into account values ('A-9732', 'Perryridge', 1200) Equivalently insert into account (branch_name, balance, account_number) values ('Perryridge', 1200, 'A-9732') Add a new tuple to account with balance set to null insert into account values ('A-777','Perryridge', null )

33. 3.33 Modification of the Database – Insertion Provide a $200 savings account to all loan customers of the Perryridge branch. Let the loan number serve as the account number for the new savings account What tables need to be modified? insert into account select loan_number, branch_name, 200 from loan where branch_name = 'Perryridge' insert into depositor select customer_name, loan_number from loan, borrower where branch_name = 'Perryridge' and loan.account_number = borrower.account_number

34. 3.34 Modification of the Database – Updates Increase all accounts with balances over $10,000 by 6%, all other accounts receive 5%  Write two update statements: update account set balance = balance  1.06 where balance > 10000 update account set balance = balance  1.05 where balance  10000 Why is the order is important? Can be done better using case: update account set balance = case when balance <= 10000 then balance *1.05 else balance * 1.06 end

35. 3.35 Update through a View Create a view of all loan data in the loan relation, hiding the amount attribute create view loan_branch as select loan_number, branch_name from loan Add a new tuple to branch_loan insert into loan_branch values ('L-37‘, 'Perryridge‘) Result: tuple ('L-37', 'Perryridge', null ) is inserted into the loan relation