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1 CSE 480: Database Systems Lecture 12: SQL (Nested queries and Aggregate functions)
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2 NESTED QUERIES A nested query is specified within the WHERE-clause of the outer query
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3 NESTED QUERIES l Query: Retrieve the name and address of employees who work for the 'Research' department SELECT Fname, Lname, Address FROM Employee, Department WHERE Dname='Research' AND Dnumber=Dno; (Non-nested query approach)
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4 NESTED QUERIES Query: Retrieve the name and address of employees who work for the 'Research' department SELECTFNAME, LNAME, ADDRESS FROM EMPLOYEE WHEREDNO IN (SELECT DNUMBER FROM DEPARTMENT WHERE DNAME='Research' ); (Nested query approach)
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5 Example l Query: List the names of projects that involve an employee whose last name is 'Smith' as a worker or as a manager of the department that controls the project
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6 NESTED QUERIES l Query: List the names of projects that involve an employee whose last name is 'Smith' as a worker or as a manager of the department that controls the project (SELECT PNAME FROMPROJECT, DEPARTMENT, EMPLOYEE WHEREDNUM=DNUMBER AND MGRSSN=SSN AND LNAME='Smith') UNION (SELECT PNAME FROMPROJECT, WORKS_ON, EMPLOYEE WHEREPNUMBER=PNO AND ESSN=SSN AND NAME='Smith'); Non-nested query approach
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7 NESTED QUERIES l Query: List the names of projects that involve an employee whose last name is 'Smith' as a worker or as a manager of the department that controls the project SELECTPNAME FROMPROJECT WHEREPNUMBER IN (SELECT PNUMBER FROM PROJECT,DEPARTMENT,EMPLOYEE WHEREDNUM=DNUMBER AND MGRSSN=SSN AND LNAME='Smith') OR PNUMBER IN (SELECT PNO FROMWORKS_ON, EMPLOYEE WHEREESSN=SSN AND NAME='Smith') Nested query approach
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8 ANY, ALL, SOME v = ANY W (e.g., Id =ANY (1,2,3)) –evaluates to TRUE if v equals to one of the elements in W v = SOME W (e.g., Id =SOME (1,2,3)) –evaluates to TRUE if v equals to some elements in W v > ALL W (e.g., Age >ALL (23,19,34)) –evaluates to TRUE if v is greater than all the elements in W v < ALL W (e.g., Age <ALL (23,19,34)) –evaluates to TRUE if v is less than all the elements in W
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9 NESTED QUERIES l Query: List the names of employees whose salary is greater than the salary of all managers SELECTFNAME, LNAME FROMEMPLOYEE WHERESALARY > ALL (SELECT SALARY FROM DEPARTMENT,EMPLOYEE WHERE MGRSSN=SSN);
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10 THE EXISTS FUNCTION l EXISTS is used to check whether the result of a nested query is empty (contains no tuples) SELECT FROM WHERE EXISTS V –WHERE-clause evaluates to TRUE if V is not an empty table and FALSE otherwise
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1 THE EXISTS FUNCTION This query will return the names of students with GPA > 3.0
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12 THE EXISTS FUNCTION l Query: Retrieve the name of each employee who has a dependent with the same name as the employee’s first name. SELEC SELECT FNAME, LNAME FROM EMPLOYEE WHERE EXISTS (SELECT * FROM DEPENDENT WHERE SSN=ESSN AND FNAME=DEPENDENT_NAME); Correlated nested query
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13 THE EXISTS FUNCTION l Query: Retrieve the names of employees who have no dependents SELECT FNAME, LNAME FROMEMPLOYEE WHERENOT EXISTS (SELECT * FROM DEPENDENT WHERE SSN=ESSN); Correlated nested query because it involves attribute of outer query
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14 Division Query Student (Id, Name) Course (CrsCode, Dept) Transcript (StudId, CrsCode, Grade) Query: Find the names of students who have taken every course offered by the CS department Student 1234 2345 3456 4567 5678 Course CS CS101 CS234 Transcript CS101 CS234 1234 2345 4.0 3.5 2.5 Adam Bill Cathy Didi Eva AdamResult
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15 Division Query in SQL Student (Id, Name) Course (CrsCode, CrsName, Dept) Transcript (StudId, CrsCode, Semester, Grade) Query: Find the names of students who have taken every course offered by the CS department SELECT Name FROM Student WHERE ? Student has taken ALL the CS courses
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16 Division Query in SQL Student (Id, Name) Course (CrsCode, CrsName, Dept) Transcript (StudId, CrsCode, Semester, Grade) Query: Find the names of students who have taken every course offered by the CS department This will return names of students who have taken at least 1 CS course SELECT Name FROM Student WHERE EXISTS (SELECT * FROM Transcript, Course WHERE Transcript.StudId = Student.Id AND Transcript.CrsCode = Course.CrsCode AND Course.Dept = ‘CS’)
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17 Division Query in SQL Student (Id, Name) Course (CrsCode, CrsName, Dept) Transcript (StudId, CrsCode, Semester, Grade) Query: Find the names of students who have taken every course offered by the CS department SELECT Name FROM Student WHERE ? Student has taken ALL the CS courses
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18 Division Query in SQL Student (Id, Name) Course (CrsCode, CrsName, Dept) Transcript (StudId, CrsCode, Semester, Grade) Query: Find the names of students who have taken every course offered by the CS department SELECT Name FROM Student WHERE ( ? ) Equivalent to saying “there are no CS courses the student has not taken”
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19 Division Query in SQL Student (Id, Name) Course (CrsCode, CrsName, Dept) Transcript (StudId, CrsCode, Semester, Grade) Query: Find the names of students who have taken every course offered by the CS department SELECT Name FROM Student WHERE NOT EXISTS ( ? ) A CS course the student has not taken
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20 Division Query in SQL Student (Id, Name) Course (CrsCode, CrsName, Dept) Transcript (StudId, CrsCode, Semester, Grade) Query: Find the names of students who have taken every course offered by the CS department SELECT Name FROM Student WHERE NOT EXISTS ( SELECT * FROM Course WHERE Dept = ‘CS’ AND ? ) Student has not taken the CS course (Student, course) combination does not exist in Transcript
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21 Division Query in SQL Student (Id, Name) Course (CrsCode, CrsName, Dept) Transcript (StudId, CrsCode, Semester, Grade) Query: Find the names of students who have taken every course offered by the CS department SELECT Name FROM Student WHERE NOT EXISTS ( SELECT * FROM Course WHERE Dept = ‘CS’ AND NOT EXISTS ( SELECT * FROM Transcript WHERE Transcript.StudId = Student.Id AND Transcript.Crscode = Course.Crscode ) )
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2 Another Possible Solution Student (Id, Name) Course (CrsCode, CrsName, Dept) Transcript (StudId, CrsCode, Semester, Grade) Query: Find the names of students who have taken every course offered by the CS department SELECT Name FROM Student WHERE NOT EXISTS ( SELECT * FROM Course WHERE Dept = ‘CS’ AND CrsCode NOT IN ( SELECT CrsCode FROM Transcript WHERE Transcript.StudId = Student.Id) )
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23 Division Query l Retrieve the names of employees who work on every project controlled by the “Research” department
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24 Example: Division Query l Retrieve the names of employees who work on every project controlled by the “Research” department SELECT DISTINCT E.Lname, E.Fname FROMEMPLOYEE E WHERE NOT EXISTS ( A project controlled by the Research department in which the employee had NOT worked on )
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25 Example: Division Query l Retrieve the names of employees who work on every project controlled by the “Research” department SELECT DISTINCT E.Lname, E.Fname FROMEMPLOYEE E WHERE NOT EXISTS ( SELECT P.Pnumber FROM DEPARTMENT D, PROJECT P WHERE D.Dnumber = P.Dnum AND D.Dname=‘Research’ AND NOT EXISTS ( (employee,project) combination in the Works_On table )
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26 Example: Division Query l Retrieve the names of employees who work on every project controlled by the “Research” department SELECT DISTINCT E.Lname, E.Fname FROMEMPLOYEE E WHERE NOT EXISTS (SELECT P.Pnumber FROM DEPARTMENT D, PROJECT P WHERE D.Dnumber = P.Dnum AND D.Dname=‘Research’ AND NOT EXISTS ( SELECT * FROM WORKS_ON W WHERE W.ESSN=E.SSN AND P.Pnumber=W.Pno))
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27 Exercise l Find the names of projects worked on by every employee in the Research department –Exclude the projects worked on only by a few but not all employees in the Research department l Find the names of employees who work for the Research department but do not work on all the projects controlled by the Research department
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28 Aggregate Functions l Used to compute summary statistics for a group of tuples in a table Include COUNT, SUM, MAX, MIN, and AVG
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29 Aggregate Functions l Need to be careful when dealing with NULL values
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30 Aggregate Functions l Query: Find the sum of the salaries, maximum salary, minimum salary, and average salary of employees in the ‘Research’ department
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31 Aggregate Functions l Query: Count the number of employees who work for the 'Research' department
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32 Aggregate Functions Query: Count the number of employees in department number 3 who have worked on at least one project overestimate
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3 Aggregate Functions Query: Count the number of employees in Department number 3 who have worked on at least one project
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34 Aggregate Functions Cannot use aggregate function directly in the WHERE clause Query: Find the id of employees who have two or more dependents
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35 Aggregate Functions Query: Find the name of employees who have two or more dependents This is ok
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36 Grouping Clause l Suppose we want to apply the aggregate functions to subgroups of tuples in a relation –Each subgroup consists of the set of tuples that have the same value for the grouping attribute(s) l SQL has a GROUP BY-clause for specifying the grouping attributes
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37 Grouping Clause Example l Query: For each department, retrieve the department number, the number of employees in the department, and their average salary. SELECT DNO, COUNT (*), AVG (SALARY) FROMEMPLOYEE GROUP BYDNO; –The EMPLOYEE tuples are divided into groups, each group having the same value for the grouping attribute DNO The COUNT and AVG functions are applied to each such group of tuples separately The SELECT-clause includes only the grouping attribute and the aggregate functions to be applied on each group of tuples
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38 Grouping Clause Example Effect of null value
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39 Grouping Clause Example l Query: For each project, retrieve the project number, project name, and the number of employees who work on that project. In this case, the grouping and aggregate functions are applied after joining the two relations
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40 Grouping Clause Example Query: For each project, retrieve the project number, project name, and number of employees from department 3 who work on the project
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41 HAVING Clause l The HAVING-clause is used to retrieve only groups whose values of the aggregate functions satisfy certain condition
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42 HAVING Clause Example l Query: For each project that has more than one employees work, retrieve the project number, project name, and the number of employees who work on that project.
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43 HAVING Clause Example l Query: For each project that has more than one employees work on it, retrieve the project number, project name, and the number of employees who work on that project.
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4 HAVING Clause Example Query: For each department that has more than 1 employee, retrieve the department name and the number of employees who are making more than $50,000 Answer should be: (Payroll, 1) and (Research, 1)
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45 HAVING Clause WRONG! This query looks for departments that have more than 1 employee making more than $50,000 Query: For each department that has more than 1 employee, retrieve the department name and the number of employees who are making more than $50,000
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46 HAVING Clause Query: For each department that has more than 1 employee, retrieve the department name and the number of employees who are making more than $50,000
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47 ORDER BY Clause l Sort tuples based on the values of some attribute(s) l Query: Retrieve a list of employees and the projects they work on, ordered by their department name, and within each department, ordered alphabetically by last name and first name. SELECT Dname, Lname, Fname, Pname FROM DEPARTMENT, EMPLOYEE, WORKS_ON, PROJECT WHEREDnumber=Dno AND SSN=ESSN AND Pno=Pnumber ORDER BY Dname, Lname, Fname; l The default order is in ascending order of values –We can also specify the keyword DESC or ASC
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48 ORDER BY Clause l Query: Retrieve a list of employees and the projects they work on, ordered by their department name, and within each department, ordered alphabetically by salary (in descending order) SELECT Dname, Lname, Fname, Pname FROM DEPARTMENT, EMPLOYEE, WORKS_ON, PROJECT WHEREDnumber=Dno AND SSN=ESSN AND Pno=Pnumber ORDER BY Dname ASC, Salary DESC;
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49 Summary of SQL Queries SELECT FROM [WHERE ] [GROUP BY ] [HAVING ] [ORDER BY ] l SELECT-clause lists the attributes/functions to be retrieved l FROM-clause specifies all relations (or aliases) needed in the query but not those needed in nested queries l WHERE-clause specifies the conditions for selection and join of tuples from the relations specified in the FROM-clause l GROUP BY specifies grouping attributes l HAVING specifies a condition for selection of groups l ORDER BY specifies an order for displaying the result of a query l A query is evaluated by first applying the WHERE-clause, then GROUP BY and HAVING, and finally the SELECT-clause
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50 Limit (in MySQL) l Find the name and salary of the highest paid employee
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