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C H A P T E R 5 Advanced SQL Practice Exercises 5.1 Describe the circumstances in which you would choose to use embed- ded SQL rather than SQL alone or only a general-purpose programming language. Answer: Writing queries in SQL is typically much easier than coding the same queries in a general-purpose programming language. However not all kinds of queries can be written in SQL. Also nondeclarative actions such as printing a report, interacting with a user, or sending the results of a query to a graphical user interface cannot be done from within SQL. Under circumstances in which we want the best of both worlds, we can choose embedded SQL or dynamic SQL, rather than using SQL alone or using only a general-purpose programming language. Embedded SQL has the advantage of programs being less complicated since it avoids the clutter of the ODBC or JDBC function calls, but requires a specialized preprocessor. 5.2 Write a Java function using JDBC metadata features that takes a Result- Set as an input parameter, and prints out the result in tabular form, with appropriate names as column headings. Answer: public class ResultSetTable implements TabelModel { ResultSet result; ResultSetMetaData metadata; int num cols; ResultSetTable(ResultSet result) throws SQLException { this.result = result; metadata = result.getMetaData(); num cols = metadata.getColumnCount(); for(int i = 1; i <= num cols; i++) { System.out.print(metadata.getColumnName(i) + ‘‘ ‘‘); 1 2 Chapter 5 Advanced SQL } System.out.println(); while(result.next()) { for(int i = 1; i <= num cols; i++) { System.out.print(result.getString( metadata.getColumnName(i) + ‘‘ ‘‘)); } System.out.println(); } } } 5.3 Write a Java function using JDBC metadata features that prints a list of all relations in the database, displaying for each relation the names and types of its attributes. Answer: DatabaseMetaData dbmd = conn.getMetaData(); ResultSet rs = dbmd.getTables(); while (rs.next()) { System.out.println(rs.getString(‘‘TABLE NAME’’); ResultSet rs1 = dbmd.getColumns(null, ‘‘schema-name’’, rs.getString(‘‘TABLE NAME’’), ‘‘%’’); while (rs1.next()) { System.out.println(rs1.getString(‘‘COLUMN NAME’’), rs.getString(‘‘TYPE NAME’’); } } 5.4 Show how to enforce the constraint “an instructor cannot teach in two different classrooms in a semester in the same time slot.” using a trigger (remember that the constraint can be violated by changes to the teaches relation as well as to the section relation). Answer: FILL 5.5 Write triggers to enforce the referential integrity constraint from section to time slot, on updates to section, and time slot. Note that the ones we wrote in Figure 5.8 do not cover the update operation. Answer: FILL 5.6 To maintain the tot cred attribute of the student relation, carry out the following: a. Modify the trigger on updates of takes, to handle all updates that can affect the value of tot cred. b. Write a trigger to handle inserts to the takes relation. Exercises 3 c. Under what assumptions is it reasonable not to create triggers on the course relation? Answer: FILL 5.7 Consider the bank database of Figure 5.25. Let us define a view branch cust as follows: create view branch cust as select branch name, customer name from depositor, account where depositor.account number = account.account number Suppose that the view is materialized; that is, the view is computed and stored. Write triggers to maintain the view, that is, to keep it up-to-date on insertions to and deletions from depositor or account. Do not bother about updates. Answer: For inserting into the materialized view branch cust we must set a database trigger on an insert into depositor and account. We assume that the database system uses immediate binding for rule execution. Fur- ther, assume that the current version of a relation is denoted by the relation name itself, while the set of newly inserted tuples is denoted by qualifying the relation name with the prefix – inserted. The active rules for this insertion are given below – define trigger insert into branch cust via depositor after insert on depositor referencing new table as inserted for each statement insert into branch cust select branch name, customer name from inserted, account where inserted.account number = account.account number define trigger insert into branch cust via account after insert on account referencing new table as inserted for each statement insert into branch cust select branch name, customer name from depositor, inserted where depositor.account number = inserted.account number Note that if the execution binding was deferred (instead of immediate), then the result of the join of the set of new tuples of account with the set of new tuples of depositor would have been inserted by both active rules, leading to duplication of the corresponding tuples in branch cust. The deletion of a tuple from branch cust is similar to insertion, exce pt that a deletion from either depositor or account will cause the natural join of these relations to have a lesser number of tuples. We denote the newly 4 Chapter 5 Advanced SQL deleted set of tuples by qualifying the relation name with the keyword deleted. define trigger delete from branch cust via depositor after delete on depositor referencing old table as deleted for each statement delete from branch cust select branch name, customer name from deleted, account where deleted.account number = account.account number define trigger delete from branch cust via account after delete on account referencing old table as deleted for each statement delete from branch cust select branch name, customer name from depositor, deleted where depositor.account number = deleted.account number 5.8 Consider the bank database of Figure 5.25. Write an SQL trigger to carry out the following action: On delete of an account, for each owner of the account, check if the owner has any remaining accounts, and if she does not, delete her from the depositor relation. Answer: create trigger check-delete-trigger after delete on account referencing old row as orow for each row delete from depositor where depositor.customer name not in ( select customer name from depositor where account number <> orow.account number ) end 5.9 Show how to express group by cube(a , b, c, d) using rollup; your answer should have only one group by clause. Answer: groupby rollup(a), rollup(b), rollup(c ), rollup(d) 5.10 Given a relation S(student, sub ject, marks), write a query to find the top n students by total marks, by using ranking. Answer: We assume that multiple students do not have the same marks since otherwise the question is not deterministic; the query below deter- ministically returns all students with the same marks as the n student, so it may return more than n students. Exercises 5 select student, sum(marks) as total, rank() over (order by (total) desc ) as trank from S groupby student having trank ≤ n 5.11 Consider the sales relation from Section 5.6. Write an SQL query to com- pute the cube operation on the relation, giving the relation in Figure 5.21. Do not use the cube construct. Answer: (select color, size, sum(number) from sales groupby color, size ) union (select color, ’all’, sum(number) from sales groupby color ) union (select ’all’, size, sum(number) from sales groupby size ) union (select ’all’, size, sum(number) from sales groupby size ) union (select ’all’, ’all’, sum(number ) from sales )