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Chapter 3 Introduction to SQL

Chapter 3 Introduction to SQL. Yonsei University 1 st Semester, 2013 Sanghyun Park. Outline. History Basic Structure Aggregate Functions Nested Subqueries Database Modification. History. IBM Sequel language developed as part of System R project at the IBM San Jose Research Laboratory

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Chapter 3 Introduction to SQL

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  1. Chapter 3Introduction to SQL Yonsei University 1st Semester, 2013 Sanghyun Park

  2. Outline • History • Basic Structure • Aggregate Functions • Nested Subqueries • Database Modification

  3. History • IBM Sequel language developed as part of System R project at the IBM San Jose Research Laboratory • Renamed Structured Query Language (SQL) • ANSI and ISO standard SQL • SQL-86, SQL-89, SQL-92 • SQL:1999 (language name became Y2K compliant) • SQL:2003 • Commercial systems offer most SQL-92 features, plus varying feature sets from later standards and special proprietary features

  4. Schema Diagram forUniversity Database

  5. Basic Structure • SQL is based on set and relational operations with certain modifications and enhancements • A typical SQL query has the form:select A1, A2, ..., Anfrom r1, r2, ..., rmwhere P • Aisrepresent attributes • risrepresent relations • P is a predicate • This query is equivalent to the relational algebra expression: ∏A1, A2, ..., An (P (r1 x r2 x ... x rm)) • The result of an SQL query is a relation

  6. SELECT Clause • The select clause lists the attributes desired in the query result • An asterisk in the select clause denotes “all attributes” • The select clause can contain arithmetic expressions • SQL allows duplicates in relations as well as in query results • Find department names of all instructors w/o duplicates: select distinct dept_namefrom instructor • The keyword all specifies that duplicates not be removed: select alldept_namefrom instructor

  7. WHERE Clause • The where clause specifies conditions the result must satisfy • To find all instructors in the Comp. Sci. dept with salary > 80000:select namefrom instructorwhere dept_name =‘Comp. Sci’ and salary > 80000 • Comparison results can be combined using the logical connectives and, or, and not • Comparisons can be applied to results of arithmetic expressions • SQL includes a between comparison operator

  8. FROM Clause • The from clause lists the relations involved in the query • Find the Cartesian product instructor x teaches:select *from instructor, teaches • Find the course ID, semester, year and title of each course offered by the Comp. Sci. dept:select section.course_id, semester, year, titlefrom section, coursewhere section.course_id = course.course_id anddept_name = ‘Comp. Sci.’

  9. Aggregate Functions • These functions operate on the multiset of values of a column,and return a single value • avg, min, max, sum, count • Find the average salary of instructors in the Comp. Sci. dept:select avg(salary)from instructorwhere dept_name = ‘Comp. Sci’ • Find the names and average salaries of all departments whose average salary is greater than 42000:select dept_name, avg (salary)from instructorgroup by dept_namehaving avg(salary) > 42000

  10. Nested Subqueries • SQL provides a mechanism for the nesting of subqueries • A subquery is a select-from-where expression that is nested within another query • A common use of subqueries is to perform tests for set membership, set comparisons, and set cardinality • Find courses offered both in Fall 2009 and in Spring 2010:select distinct course_idfrom sectionwhere semester = ‘Fall’and year = 2009and course_id in (select course_idfrom sectionwhere semester =‘Spring’ and year =2010)

  11. Database Modification - Delete • Delete all instructors from the Finance dept:deletefrom instructorwhere dept_name = ‘Finance’ • Delete all instructors whose salary is less than the average salary of instructors:deletefrom instructorwhere salary < (select avg (salary)from instructor)

  12. Database Modification - Insert • Add a new tuple to course: insert into coursevalues(‘CS-437’, ‘Database Systems’, ‘Comp. Sci’, 4)or equivalentlyinsert into course(course_id, title, credits, dept_name)values(‘CS-437’, ‘Database Systems’, 4, ‘Comp. Sci’) • Add all instructors to the student relation with tot_creds set to 0:insert into studentselect ID, name, dept_name, 0from instructor

  13. Database Modification - Update • Increase salaries of instructors whose salary is over $100,000 by 3% and all others receive 5% raise: update instructorset salary = salary  1.03where salary > 100000; update instructorset salary = salary  1.05where salary  100000 • The order is important • Can be done better using the case statementupdateinstructorsetsalary =casewhensalary  100000 thensalary *1.05elsesalary * 1.03end

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