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COP5725 Database Management System Final Review

COP5725 Database Management System Final Review. Reviewed by Ramakrishna . Parts of this are taken from Fernando Farfan’s presentation. AGENDA. Exercises to do…. Ch6 JDBC Ch7 3-Tier Architecture Ch8. Storage and Indexing Ch10. Tree-Structured Indexing Ch7. XML Documents

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COP5725 Database Management System Final Review

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  1. COP5725 Database Management SystemFinal Review Reviewed by Ramakrishna. Parts of this are taken from Fernando Farfan’s presentation

  2. AGENDA • Exercises to do…. • Ch6 JDBC • Ch7 3-Tier Architecture • Ch8. Storage and Indexing • Ch10. Tree-Structured Indexing • Ch7. XML Documents • Ch27. XQUERY: Querying XML Data

  3. Exercises to do…. • Remember to practice even numbered exercises of the book • Solutions are available online at http://www.cs.wisc.edu/~dbbook/openAccess/thirdEdition/supporting_material.htm • Important Recommended exercises: 8.3, 8.4, 8.5, 8.7, 8.10, 8.11

  4. JDBC • What is JDBC ? Explain its purpose. • JDBC is Java DataBase Connectivity used to enable integration of SQL with a general purpose programming language. • Explain JDBC Architecture • 4 components : • Application, • Driver Manager, • Data Source Specific Drivers, • Data Sources (stored in MySql,Oracle,DB2,MSSQL,Access and so on).

  5. JDBC • Explain the individual steps required to submit a query to a data source and to retrieve results in JDBC ? • JDBC Driver Management (class.forName(..)), • Establishing Connection (Connection Object), • Executing SQL Statements: • Statement, • PreparedStatement, • CallableStatement. • Retrieving results ( Examining ResultSets).

  6. Stored Procedures • Explain the term stored procedure, and give examples why stored procedures are useful. • Stored procedures are programs that run on the database server and can be called with a single SQL statement. • Runs inside the process space of the database server. • Stored procedures can also be used to reduce network communication. • They are locally-executed and results are packaged in to one big result. • Different users can re-use the stored procedure.

  7. 3-Tier Application Architecture • What is a 2-tier architecture ? Explain different types of 2-tier architectures. • It’s a client-server architecture. Client Application Logic Network DBMS Client Architecture 1: Thin Clients ( web browsers)

  8. 3-Tier Application Architecture • Architecture 2: Thick Clients Client Application Logic Network DBMS Client Application Logic • Disadvantages ? • No central place to update • Need to trust client

  9. 3-tier architecture • What are the advantages of 3-tier architecture ? • Heterogeneous Systems • Thin Clients • Integrated Data Access • Scalable Client (user interface) Network Network DBMS (database) Client (web browser) Application Logic (in C++ or Java..)

  10. CH8. OVERVIEW OF STORAGE AND INDEXING • DBMS stores vast quantities of data • Data must persist across program executions • Data is stored on external storage devices. • The unit of information read from or written to disk is a page. • The cost of page I/O dominates the cost of typical database operations. • Input from disk to main memory • Output from main memory to disk

  11. CH8. OVERVIEW OF STORAGE AND INDEXING • Simplest file structure is an unordered file, or heap file. • An index is a data structure to optimize certain kinds of retrieval operations. • CLUSTERED - When a file is organized so that the ordering of data records is the same as the ordering of data entries in some index • UNCLUSTERED - It is un-clustered otherwise • There is UTMOST one clustered index and several un-clustered ones for a table.

  12. Example • Employee (Eid: integer, name: String, age: integer…..) • Assume data records sorted by name • Assume an index on name • Index on name  CLUSTERED • Assume an index on age • Index on age  UNCLUSTERED

  13. Example: Using Indexes • Employee Data Records : Sorted by Name • Page 1 : • (1,Alex,30) • (2,Amy,21) • Page 2: • (3, Bob,31) • (4, Brenda,21) • Select * from Employee where name like “A%”; • Use index on name  Retrieve Page 1. • Select * from Employee where age = 21; • Use index on age  Retrieve Page 1 & Page 2.

  14. Clustered and UnClustered Indexes • Retrieval using a Clustered index • Data retrieval using minimum number of Data page I/Os. • Retrieval using an Unclustered index • In worst case, 1 page I/O for every qualifying tuple.

  15. Another Example • Consider “Select E.dno from Employee E where E.age > 40”; • Assume B+ Tree index on age. • Is it better to use this index ? Or just do a segment scan ? • Answer : Depends on several factors: • Depends on Selectivity of the condition. • Whether the index is clustered or unclustered.

  16. CH8. OVERVIEW OF STORAGE AND INDEXING • 8.3 Consider a relation stored as a randomly ordered file for which the only index is an unclustered index on a field called sal. If you want to retrieve all records with sal > 20, is using the index always the best alternative? Explain. • No. In this case, the index is unclustered, each qualifying data entry could contain an rid that points to a distinct data page, leading to as many data page I/Os as the number of data entries that match the range query. In this situation, using index is actually worse than file scan.

  17. CH8. OVERVIEW OF STORAGE AND INDEXING • 8.5 Explain the difference between Hash indexes and B+-tree indexes. In particular, discuss how equality and range searches work. • Hash Index: Hashing function. Quickly maps a search key value to a bucket. Inserts/Deletes are simple. Linked list for collisions. Good at equality searches. Terrible for range queries. • B+-tree Index: Hierarchical search data structure. Maintenance is costly. Costly for individual record lookup. Efficient for range queries.

  18. CH8. OVERVIEW OF STORAGE AND INDEXING • Consider the following relation Answer: Update the salary of an employee using the employee id.

  19. CH8. OVERVIEW OF STORAGE AND INDEXING Update the age or employee id for some department id Update the salaries of all employees in some department

  20. Constraints

  21. Constraints

  22. Constraints

  23. Constraints Assume tables are already created. ALTER TABLE ENROLLED add constraint cons1 check((select count(*) from ENROLLED group by cname)>=5) ALTER TABLE ENROLLED add constraint cons2 check((select count(*) from ENROLLED group by cname)<=30)

  24. Constraints

  25. Views

  26. Views

  27. Tree-Structured Indexing • Explain ISAM and B+ Trees ? What are the differences, advantages and disadvantages ? • ISAM – Indexes Sequential access Method • Effective for static files • Unsuitable for dynamically changing files • B+ Tree • Dynamic index structure • Adjusts well to changes • Does well for both range and equality selections

  28. Tree-Structured Indexing • Insertions and deletions in ISAM • Happen only in leaf pages • Insertions – adding records to the overflow chain • Search – inefficient as the chain grows • In B+ tree even after several insertions and deletions, • The tree is kept balanced • Height of the tree  length of the path from root to leaf

  29. Query Evaluation • Consider a relation R(a,b,c,d,e) containing 5,000,000 records, where each data page of the relation holds 10 records. R is organized as a sorted file with secondary indexes. Assume that R.a is a candidate key for R, with values lying in the range 0 to 4,999,999, and that R is stored in R.a order. For each of the following relational algebra queries, state which of the following three approaches is most likely to be the cheapest: • Access the sorted file for R directly. • Use a (clustered) B+ tree index on attribute R.a. • Use a linear hashed index on attribute R.a.

  30. Query Evaluation

  31. CH7. XML DOCUMENTS • 7.1 When is an XML document well-formed? When is an XML document valid? • An XML document is valid if it has an associated DTD and the document follows the rules of the DTD. An XML document is well-formed if it follows three guidelines: • It starts with an XML declaration. • It contains a root element that contains all other elements. • All elements are properly nested.

  32. CH27. XQUERY • <bookstore> <book category="COOKING"> <title lang="en">Everyday Italian</title> <author>Giada De Laurentiis</author> <year>2005</year> <price>30.00</price> </book> <book category="CHILDREN"> <title lang="en">Harry Potter</title> <author>J K. Rowling</author> <year>2005</year> <price>29.99</price> </book> <book category="WEB"> <title lang="en">XQuery Kick Start</title> <author>James McGovern</author> <author>Per Bothner</author> <author>Kurt Cagle</author> <author>James Linn</author> <author>Vaidyanathan Nagarajan</author> <year>2003</year> <price>49.99</price> </book> <book category="WEB"> <title lang="en">Learning XML</title> <author>Erik T. Ray</author> <year>2003</year> <price>39.95</price> </book></bookstore>

  33. CH27. XQUERY • Query:doc("books.xml")/bookstore/book/title • Result:<title lang="en">Everyday Italian</title><title lang="en">Harry Potter</title><title lang="en">XQuery Kick Start</title><title lang="en">Learning XML</title>

  34. CH27. XQUERY • Query:doc("books.xml")/bookstore/book[price<30] • Result:<book category="CHILDREN"> <title lang="en">Harry Potter</title> <author>J K. Rowling</author> <year>2005</year> <price>29.99</price></book>

  35. CH27. XQUERY • Query:for $x in doc("books.xml")/bookstore/book where $x/price>30 order by $x/title return $x/title • Result:<title lang="en">Learning XML</title><title lang="en">XQuery Kick Start</title>

  36. CH27. XQUERY • Query:<ul> { for $x in doc("books.xml")/bookstore/book/title order by $x return <li>{$x}</li> } </ul> • Result:<ul> <li><title lang="en">Everyday Italian</title></li> <li><title lang="en">Harry Potter</title></li> <li><title lang="en">Learning XML</title></li> <li><title lang="en">XQuery Kick Start</title></li></ul>

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