1. SQL Operations Aggregate Functions Having Clause Database Access Layer A2 Teacher Up skilling LECTURE 5

2. What’s to come today? SQL Operations Aggregate Functions Having Clause Database Access Layer

3. SQL Operations Over the next few slides, we will take a look at a few useful operations that we can perform using SQL. Firstly, we will look at the rename operation which allows for the renaming of relations and attributes in queries. Then we will proceed to look a string operations within SQL, that allow us to craft SQL queries that search for specific character patterns.

4. Rename Operation SQL allows renaming relations and attributes using the as clause: old-name as new-name Example: select ID, name, salary/12 as monthly_salary from instructor; Sometimes we may rename a relation just to get another copy. Example: Find the names of all instructors who have a higher salary than some instructor in ‘Comp. Sci’. select distinct T. name from instructor as T, instructor as S where T.salary > S.salary and S.dept_name = ‘Comp. Sci.’; Keyword as is optional and may be omitted.

5. String Operations SQL includes a string-matching operator for comparisons on character strings. The operator “like” uses patterns that are described using two special characters. percent (%) - The % character matches any substring. underscore (_) - The _ character matches any character. Example: Find the names of all instructors whose name includes the substring “dar”. select name from instructor where name like '%dar%‘;

6. String Operations (cont.) Pattern matching examples ‘Perry%’ matches any string beginning with “Perry”. ‘%idge%’ matches any string containing “idge” as a substring, for example, ‘Perryeidge’, ‘Rock Ridge’, ‘Mianus Bridge’, and ‘Ridgeway’. ‘---’ matches any string of exactly three characters. ‘---%’ matches any string of at least three characters. Match the string “100 %” like ‘100 \%' escape '\’

7. Ordering display of rows List in alphabetic order the names of all instructors select distinct name from instructor order by name; The rows in the result table will be ordered according to the values of the name attribute. We may specify desc for descending order or asc for ascending order, for each attribute; ascending order is the default. Example: order by name desc Can sort on multiple attributes Example: order by dept_name, name

8. Set Operations A set is an unordered collection of elements, that contains no duplicates. A table is basically a set of rows. So a set operator can be applied to two tables (two sets of rows) just as two sets are in set theory. We will take a look at the following set operations: Union Intersection Difference

9. Set Union Given two sets A and b, an Union operation between these sets returns distinct elements that belong to either A or B or both. Example Find courses that ran in Fall 2009 or in Spring 2010 (select course_id from section where sem = ‘Fall’ and year = 2009) union (select course_id from section where sem = ‘Spring’ and year = 2010);

10. Set Intersection Given two sets A and B, a intersection operation between these sets returns the distinct elements that belong to both A and B. Example Find courses that ran in Fall 2009 and in Spring 2010 (select course_id from section where sem = ‘Fall’ and year = 2009) intersect (select course_id from section where sem = ‘Spring’ and year = 2010);

11. Set Difference Given two sets A and B, a difference operation (A\B) between these sets returns the elements that belong to A but don’t belong to B. Example Find courses that ran in Fall 2009 but not in Spring 2010 (select course_id from section where sem = ‘Fall’ and year = 2009) except (select course_id from section where sem = ‘Spring’ and year = 2010);

12. Aggregate Functions These functions operate on the multi-set of values of a column of a relation or tuples in a relation, and return a value. A multi-set is a set which allows duplicates. Below we have listed a number of commonly used aggregate functions. avg: average value min: minimum value max: maximum value sum: sum of values count: number of values

13. Avg() Aggregate Function Find the average salary of instructors in the Computer Science department select avg (salary) from instructor where dept_name= ’Comp. Sci.’; Note that a set of salaries is first generated and then the aggregate function avg is applied to the set.

14. Avg() Aggregate Function Example IDnamedept_namesalary 10101SrinivasanComp. Sci.65000 45565KazeComp. Sci.75000 83821BrandtComp. Sci.92000 select avg (salary) from instructor where dept_name= ’Comp. Sci.’ avg (salary) 77333

15. Count() Aggregate Function Find the total number of instructors who teach a course in the Spring 2010 semester select count (distinct ID) from teaches where semester = ’Spring’ and year = 2010; The Count() function can be used to calculate the number of rows in a table. select count (*) from course;

16. Group By Aggregate Function The Group By function group the result set by one or more columns. Find the average salary of instructors in each department select dept_name, avg (salary) from instructor group by dept_name;

17. Group By Aggregate Function (cont.) Attributes in select clause outside of aggregate functions must appear in group by list Example: erroneous query as the group by statement should include both the dept_name and ID attributes. select dept_name, ID, avg (salary) from instructor group by dept_name;

18. Having Clause The having clause is similar to where clause. Note: The conditions in the having clause are applied after the formation of groups whereas conditions in the where clause are applied before forming groups. The having clause applies to the result of an aggregate function. Find the names and average salaries of all departments whose average salary is greater than 42000 select dept_name, avg (salary) from instructor group by dept_name having avg (salary) > 42000;

19. Having Clause Example select dept_name, avg (salary) from instructor group by dept_name having avg (salary) > 42000; dept_namesalary Biology72000 Comp. Sci.77333 Elec. Eng.80000 Finance85000 History61000 Physics91000

20. Data Access Layer You may have noticed in some of the practical classes, that we created classes called ProjectDAL or CustomerDAL. The DAL postfix in this class name, is just a reminder to the programmer that this class is the layer that performs SQL operations on the data in the database. A Data Access Layer is a good idea, as it helps to separate database logic from the business and presentation logic of the application. By separating layers, this allows our application to be more flexible to change.

21. What’s to come next time Week 6 System Methodologies Roundup