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Monday, April 19, 2010

Agile Record issue 2

NEWSLETTER
	The first printed issue of the Agile Record is out now! If you haven’t register yet, you can still do so and we will send you the next issue. I want to thank all the authors for their articles and columns. Please register here: www.agilerecord.com/subscribe.php Please download the issue from www.agilerecord.com/agilerecord_02.pdf I would really appreciate if you could forward this information to all your colleagues and interested contacts. Best regards José Díaz Editor
	Managing the Testing Process 3-day training course by Rex Black June 09 to 11, 2010 in Bad Homburg, Germany Test managers must take a potentially infinite job—testing a computer system—and accomplish it within tight time and resource restraints. It’s a tall order, but successful test managers have found proven ways to handle the challenges. This course will give attendees the tools they need to succeed as test managers. We’ll look at quality risk analysis, test estimation, and test planning. We’ll discuss developing high-quality test systems—test cases, test data, test tools, even automated test systems— that improve over time. We’ll talk about tracking bugs and test cases. We’ll discuss ways to derive and present metrics, charts, and graphs from the test results. We’ll also cover the human side of test management. We’ll look at ways to measure and manage the skills testers need. We’ll discuss hiring testers. We’ll talk about education and certification for testers. We’ll examine some ways to motivate and reward testers— and some ways not to! We’ll cover working effectively within the project organization, which is especially challenging when you’re the bearer of bad news. We’ll also look at the context of testing. We’ll discuss system development lifecycles and how they affect testing. We’ll cover testing as an investment. We’ll finish up by discussing test labs, test environments, and hardware issues. The materials presented in the course follow Rex Black’s book, Managing the Testing Process, which is the distillation of over two decades of software, hardware, and systems experience. www.agilerecord.com/knowledge_transfer.html
	Testing & Finance 2010 The Conference for Testing & Finance Professionals June 7th - 8th, 2010 in Bad Homburg (near Frankfurt am Main, Germany) Keynotes by Nitin Bhargava – Barclays Bank, UK BJ Rollison – Microsoft Corp Gernot Nolte – Fiducia Prof. Dr. Schulte-Mattler – Fachhochschule Dortmund Rex Black – RBCS Conference 4 tracks 46 speeches product fair full catering free internet access social event www.testingfinance.com
	Agile Testing Days 2010 October 04-07, 2010 in Berlin The Agile Testing Days is the European conference for the worldwide professionals involved in the agile world. We are very proud to count on the support and the work of Lisa Crispin, Michael Bolton, Janet Gregory, Linda Rising, Jennitta Andrea, Anko Tijman, Eric Jimmink, Stuart Reid, Isabel Evans, Elisabeth Hendrickson, Pekka Klärck, Juha Rantanen, Janne Härkönen and Brett Schuchert. We chose Berlin, one of the best-connected capitals in Europe to host this event due to the affordability/service ratio for hotels and flights. Berlin also offers a lot of fabulous sights to visit in your spare time. Please have a look at the program ( download the PDF file) and enjoy the conference! www.agiletestingdays.com
	OZ Agile Days 2010 - Call for Papers - the Australian conference for the professionals involved in the agile world (November 22 - 24, 2010 in Sydney, Australia) using our template at www.ozagile.com until May 30, 2010 to papers@ozagile.com

Friday, April 9, 2010

Agile Testing Days 2010

Dear Hemanth Bolem,

we are very proud to present the program for the Agile Testing Days 2010!

The Agile Testing Days will take place in Berlin, on the 4^th – 7^th of October 2010.

It was an honor to rely on Alessandro Collino as Program Chair, who set up the conference.

On October 4^th we have the tutorial of

1. Lisa Crispin, Making Test Automation Work on Agile Teams

2. Michael Bolton, A Rapid Introduction to Rapid Software Testing

3. Janet Gregory, A Tester’s Guide to Navigating an Iteration

4. Linda Rising, Patterns for Improved Customer Interaction & Influence Strategies for Practitioners

5. Jennitta Andrea, The Foundations of Agile Software Development

6. Anko Tijman & Eric Jimmink, Testing2.0 – agile testing in practice

7. Stuart Reid & Isabel Evans, Managing Testing in Agile Projects

8. Elisabeth Hendrickson, Agile Transitions

9. Pekka Klärck, Juha Rantanen & Janne Härkönen, Executable Requirements in Practice

40 Talks, 9 Keynotes and an Open Space are waiting for you.

They keynote speakers are:

· Lisa Crispin

· Linda Rising

· Elisabeth Hendrickson

· Michael Bolton

· Stuart Reid

· Janet Gregory

· Isabel Evans

· Jennitta Andrea

· Tom Gilb

The Open Space on October 7^th is hosted by Brett Schuchert.

We want to thank all the tutorial speakers, keynote speakers, speakers, sponsors and people involved in this event for their hard work and support.

Please have a look at the program at www.agiletestingdays.com

The early bird ends on June 30^th, 2010. Don’t miss it!

I would really appreciate if you could forward this information on to all your colleagues and interested contacts.

Best regards

José Díaz

------------------------------------------------

José M. Díaz Delgado
Managing Director

Díaz & Hilterscheid Unternehmensberatung GmbH
Kurfürstendamm 179
10707 Berlin

Telefon: +49 (0)30 74 76 28 – 0
Telefax: +49 (0)30 74 76 28 - 99

Internet:
www.diazhilterscheid.de

Email:
jose.diaz@diazhilterscheid.de

José M. Díaz Delgado - Geschäftsführer, Julia Hilterscheid - Geschäftsführerin, Marc Ahlbach - Geschäftsführer

Ust-IdNr. DE195661616, Amtsgericht Berlin Charlottenburg HRB 67896

Internet: www.diazhilterscheid.de

----------------------------------------------------------------------------------------

If you don't want to receive information from Díaz & Hilterscheid please send an email with the subject "unsubscribe". We apologise for the trouble caused. This message was sent to hemanthbolem@gmail.com.

Thursday, January 28, 2010

SQL Queries

SQL Queries

Create the following Tables:

LOCATION
Location_ID	Regional_Group
122	NEW YORK
123	DALLAS
124	CHICAGO
167	BOSTON

DEPARTMENT
Department_ID	Name	Location_ID
10	ACCOUNTING	122
20	RESEARCH	124
30	SALES	123
40	OPERATIONS	167

JOB
Job_ID	Function
667	CLERK
668	STAFF
669	ANALYST
670	SALESPERSON
671	MANAGER
672	PRESIDENT

EMPLOYEE
EMPLOYEE_ID	LAST_NAME	FIRST_NAME	MIDDLE_NAME	JOB_ID	MANAGER_ID	HIREDATE	SALARY	COMM	DEPARTMENT_ID
7369	SMITH	JOHN	Q	667	7902	17-DEC-84	800	NULL	20
7499	ALLEN	KEVIN	J	670	7698	20-FEB-85	1600	300	30
7505	DOYLE	JEAN	K	671	7839	04-APR-85	2850	NULL	30
7506	DENNIS	LYNN	S	671	7839	15-MAY-85	2750	NULL	30
7507	BAKER	LESLIE	D	671	7839	10-JUN-85	2200	NULL	40
7521	WARK	CYNTHIA	D	670	7698	22-FEB-85	1250	500	30

Queries based on the above tables:

Simple Queries:

List all the employee details
List all the department details
List all job details
List all the locations
List out first name,last name,salary, commission for all employees
List out employee_id,last name,department id for all employees and rename employee id as "ID of the employee", last name as "Name of the employee", department id as "department ID"
List out the employees anuual salary with their names only.

Where Conditions:

List the details about "SMITH"
List out the employees who are working in department 20
List out the employees who are earning salary between 3000 and 4500
List out the employees who are working in department 10 or 20
Find out the employees who are not working in department 10 or 30
List out the employees whose name starts with "S"
List out the employees whose name start with "S" and end with "H"
List out the employees whose name length is 4 and start with "S"
List out the employees who are working in department 10 and draw the salaries more than 3500
list out the employees who are not receiving commission.

Order By Clause:

List out the employee id, last name in ascending order based on the employee id.
List out the employee id, name in descending order based on salary column
list out the employee details according to their last_name in ascending order and salaries in descending order
list out the employee details according to their last_name in ascending order and then on department_id in descending order.

Group By & Having Clause:

How many employees who are working in different departments wise in the organization
List out the department wise maximum salary, minimum salary, average salary of the employees
List out the job wise maximum salary, minimum salary, average salaries of the employees.
List out the no.of employees joined in every month in ascending order.
List out the no.of employees for each month and year, in the ascending order based on the year, month.
List out the department id having atleast four employees.
How many employees in January month.
How many employees who are joined in January or September month.
How many employees who are joined in 1985.
How many employees joined each month in 1985.
How many employees who are joined in March 1985.
Which is the department id, having greater than or equal to 3 employees joined in April 1985.

Sub-Queries

Display the employee who got the maximum salary.
Display the employees who are working in Sales department
Display the employees who are working as "Clerk".
Display the employees who are working in "New York"
Find out no.of employees working in "Sales" department.
Update the employees salaries, who are working as Clerk on the basis of 10%.
Delete the employees who are working in accounting department.
Display the second highest salary drawing employee details.
Display the Nth highest salary drawing employee details

Sub-Query operators: (ALL,ANY,SOME,EXISTS)

List out the employees who earn more than every employee in department 30.
List out the employees who earn more than the lowest salary in department 30.
Find out whose department has not employees.
Find out which department does not have any employees.

Co-Related Sub Queries:

47.Find out the employees who earn greater than the average salary for their department.

Joins

Simple join

48.List our employees with their department names

49.Display employees with their designations (jobs)

50.Display the employees with their department name and regional groups.

51.How many employees who are working in different departments and display with department name.

52.How many employees who are working in sales department.

53.Which is the department having greater than or equal to 5 employees and display the department names in ascending order.

54.How many jobs in the organization with designations.

55.How many employees working in "New York".

Non – Equi Join:

56.Display employee details with salary grades.

57.List out the no. of employees on grade wise.

58.Display the employ salary grades and no. of employees between 2000 to 5000 range of salary.

Self Join:

59.Display the employee details with their manager names.

60.Display the employee details who earn more than their managers salaries.

61.Show the no. of employees working under every manager.

Outer Join:

61.Display employee details with all departments.

62.Display all employees in sales or operation departments.

Set Operators:

63.List out the distinct jobs in Sales and Accounting Departments.

64.List out the ALL jobs in Sales and Accounting Departments.

65.List out the common jobs in Research and Accounting Departments in ascending order.

Answers

SQL > Select * from employee;
SQL > Select * from department;
SQL > Select * from job;
SQL > Select * from loc;
SQL > Select first_name, last_name, salary, commission from employee;
SQL > Select employee_id "id of the employee", last_name "name", department id as "department id" from employee;
SQL > Select last_name, salary*12 "annual salary" from employee
SQL > Select * from employee where last_name='SMITH';
SQL > Select * from employee where department_id=20
SQL > Select * from employee where salary between 3000 and 4500
SQL > Select * from employee where department_id in (20,30)
SQL > Select last_name, salary, commission, department_id from employee where department_id not in (10,30)
SQL > Select * from employee where last_name like 'S%'
SQL > Select * from employee where last_name like 'S%H'
SQL > Select * from employee where last_name like 'S___'
SQL > Select * from employee where department_id=10 and salary>3500
SQL > Select * from employee where commission is Null
SQL > Select employee_id, last_name from employee order by employee_id
SQL > Select employee_id, last_name, salary from employee order by salary desc
SQL > Select employee_id, last_name, salary from employee order by last_name, salary desc
SQL > Select employee_id, last_name, salary from employee order by last_name, department_id desc
SQL > Select department_id, count(*), from employee group by department_id
SQL > Select department_id, count(*), max(salary), min(salary), avg(salary) from employee group by department_id
SQL > Select job_id, count(*), max(salary), min(salary), avg(salary) from employee group by job_id
SQL > Select to_char(hire_date,'month')month, count(*) from employee group by to_char(hire_date,'month') order by month
SQL > Select to_char(hire_date,'yyyy') Year, to_char(hire_date,'mon') Month, count(*) "No. of employees" from employee group by to_char(hire_date,'yyyy'), to_char(hire_date,'mon')
SQL > Select department_id, count(*) from employee group by department_id having count(*)>=4
SQL > Select to_char(hire_date,'mon') month, count(*) from employee group by to_char(hire_date,'mon') having to_char(hire_date,'mon')='jan'
SQL > Select to_char(hire_date,'mon') month, count(*) from employee group by to_char(hire_date,'mon') having to_char(hire_date,'mon') in ('jan','sep')
SQL > Select to_char(hire_date,'yyyy') Year, count(*) from employee group by to_char(hire_date,'yyyy') having to_char(hire_date,'yyyy')=1985
SQL > Select to_char(hire_date,'yyyy')Year, to_char(hire_date,'mon') Month, count(*) "No. of employees" from employee where to_char(hire_date,'yyyy')=1985 group by to_char(hire_date,'yyyy'),to_char(hire_date,'mon')
SQL > Select to_char(hire_date,'yyyy')Year, to_char(hire_date,'mon') Month, count(*) "No. of employees" from employee where to_char(hire_date,'yyyy')=1985 and to_char(hire_date,'mon')='mar' group by to_char(hire_date,'yyyy'),to_char(hire_date,'mon')
SQL > Select department_id, count(*) "No. of employees" from employee where to_char(hire_date,'yyyy')=1985 and to_char(hire_date,'mon')='apr' group by to_char(hire_date,'yyyy'), to_char(hire_date,'mon'), department_id having count(*)>=3
SQL > Select * from employee where salary=(select max(salary) from employee)
SQL > Select * from employee where department_id IN (select department_id from department where name='SALES')
SQL > Select * from employee where job_id in (select job_id from job where function='CLERK'
SQL > Select * from employee where department_id=(select department_id from department where location_id=(select location_id from location where regional_group='New York'))
SQL > Select * from employee where department_id=(select department_id from department where name='SALES' group by department_id)
SQL > Update employee set salary=salary*10/100 wehre job_id=(select job_id from job where function='CLERK')
SQL > delete from employee where department_id=(select department_id from department where name='ACCOUNTING')
SQL > Select * from employee where salary=(select max(salary) from employee where salary <(select max(salary) from employee))
SQL > Select distinct e.salary from employee where & no-1=(select count(distinct salary) from employee where sal>e.salary)
SQL > Select * from employee where salary > all (Select salary from employee where department_id=30)
SQL > Select * from employee where salary > any (Select salary from employee where department_id=30)
SQL > Select employee_id, last_name, department_id from employee e where not exists (select department_id from department d where d.department_id=e.department_id)
SQL > Select name from department d where not exists (select last_name from employee e where d.department_id=e.department_id)
SQL > Select employee_id, last_name, salary, department_id from employee e where salary > (select avg(salary) from employee where department_id=e.department_id)
SQL > Select employee_id, last_name, name from employee e, department d where e.department_id=d.department_id
SQL > Select employee_id, last_name, function from employee e, job j where e.job_id=j.job_id
SQL > Select employee_id, last_name, name, regional_group from employee e, department d, location l where e.department_id=d.department_id and d.location_id=l.location_id
SQL > Select name, count(*) from employee e, department d where d.department_id=e.department_id group by name
SQL > Select name, count(*) from employee e, department d where d.department_id=e.department_id group by name having name='SALES'
SQL > Select name, count(*) from employee e, department d where d.department_id=e.department_id group by name having count (*)>=5 order by name
SQL > Select function, count(*) from employee e, job j where j.job_id=e.job_id group by function
SQL > Select regional_group, count(*) from employee e, department d, location l where e.department_id=d.department_id and d.location_id=l.location_id and regional_group='NEW YORK' group by regional_group
SQL > Select employee_id, last_name, grade_id from employee e, salary_grade s where salary between lower_bound and upper_bound order by last_name
SQL > Select grade_id, count(*) from employee e, salary_grade s where salary between lower_bound and upper_bound group by grade_id order by grade_id desc
SQL > Select grade_id, count(*) from employee e, salary_grade s where salary between lower_bound and upper_bound and lower_bound>=2000 and lower_bound<=5000 group by grade_id order by grade_id desc
SQL > Select e.last_name emp_name, m.last_name, mgr_name from employee e, employee m where e.manager_id=m.employee_id
SQL > Select e.last_name emp_name, e.salary emp_salary, m.last_name, mgr_name, m.salary mgr_salary from employee e, employee m where e.manager_id=m.employee_id and m.salary<e.salary
SQL > Select m.manager_id, count(*) from employee e, employee m where e.employee_id=m.manager_id group by m.manager_id
SQL > Select last_name, d.department_id, d.name from employee e, department d where e.department_id(+)=d.department_id
SQL > Select last_name, d.department_id, d.name from employee e, department d where e.department_id(+)=d.department_id and d.department_idin (select department_id from department where name IN ('SALES','OPERATIONS'))
SQL > Select function from job where job_id in (Select job_id from employee where department_id=(select department_id from department where name='SALES')) union Select function from job where job_id in (Select job_id from employee where department_id=(select department_id from department where name='ACCOUNTING'))
SQL > Select function from job where job_id in (Select job_id from employee where department_id=(select department_id from department where name='SALES')) union all Select function from job where job_id in (Select job_id from employee where department_id=(select department_id from department where name='ACCOUNTING'))
SQL > Select function from job where job_id in (Select job_id from employee where department_id=(select department_id from department where name='RESEARCH')) intersect Select function from job where job_id in (Select job_id from employee where department_id=(select department_id from department where name='ACCOUNTING')) order by function

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Friday, May 14, 2010

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Monday, April 19, 2010

Agile Record issue 2

NEWSLETTER

Managing the Testing Process

3-day training course by Rex Black

Testing & Finance 2010

The Conference for Testing & Finance Professionals

Agile Testing Days 2010

OZ Agile Days 2010 - Call for Papers

Friday, April 9, 2010

Agile Testing Days 2010

Thursday, January 28, 2010

SQL Queries

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