Consider this. An order management application has a screen to view all orders and another window to browse customers. It can also generate a number of reports, including one to analyze orders grouped by customer. But accountant Jerry is working on a report for his boss and needs to find the ten customers with the highest debt. He can request a new report from the IT department, but it will take days (or even weeks) because of bureaucratic routine, programmers' busyness, or something else. The knowledge of SQL can help Jerry to create his own ad-hoc query, get the data, and finish his report.
Overview
Previously we defined database as an organized collection of information. Not only does that mean that data have to be organized according to a company's business rules, but also the database organization should reflect the nature of the information. Databases can store dollar amounts, quantities, names, date and time values, binary files, and more. These can be further classified by type, which reflects the "nature" of the data: numbers, characters, dates, etc.
| Note | Data type is a characteristic of a database table column that determines what kind of data it can hold. |
One can ask: why do we need data types at all? Wouldn't it be easier simply have one uniform data type and store everything, let's say, in the form of character strings?
There are many reasons why we don't do that. Some of them are historical. For example, when relational databases were born in the late twentieth century, hard disk space and memory storage were at premium, so the idea was to store everything as efficiently as possible. Already existing programming languages had some built-in rules for how to store different types of data. For example, any English character (plus special characters and digits) could be represented using its ASCII equivalent and the necessary storage for it was one byte (more about ASCII later in this chapter). Numbers are traditionally stored in the form of binary strings (native to computer architecture). To represent a number from negative 32,768 to positive 32,767, two bytes (or sixteen bits) of storage are sufficient (2
16). But if we used ASCII characters to represent numbers, we would need six bytes to store any integer greater than 9,999 (five bytes for digits, and one for the plus or minus sign), five bytes for whole numbers greater than 999 (four bytes for digits, one byte for the sign), and so on. Now imagine — if we have a million of records, we could save four million bytes (about 4M). Sounds like almost nothing today, but back in the 1970s that was incredibly large storage space. The principle of effectiveness is still in place, of course, but now we are talking different scales.
| Note | One byte consists of eight bits. Each bit is a binary number that can either be 0 or 1. All information is stored in memory or on the hard disk in form of ones and zeroes, representing the only two states computers understand: zero means no signal, and one indicates the presence of the signal. |
Another reason is logical consistency. Every data type has its own rules, sort order, relations with other data types, implicit conversion rules, and so on. It is definitely easier to work with sets of like values, say dates, rather than with a mixture of dates, numbers, and character strings. Try comparing library shelves where all materials are sorted and classified (fiction is in one room, kids' literature in another, audio books in their special area, videotapes somewhere else) with a pile of chaotically mixed books, tapes, white papers, and CDs and think what would you prefer for finding information.
And the last thing to mention — some modern data types (particularly movie files) are too large and too complicated to store them in a traditional way. Now we are going to discuss existing SQL data types in more details
Queries based on the above tables:
Simple Queries:
1. List all the employee details
2. List all the department details
3. List all job details
4. List all the locations
5. List out first name,last name,salary, commission for all employees
6. 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”
7. List out the employees anuual salary with their names only.
Where Conditions:
8. List the details about “SMITH”
9. List out the employees who are working in department 20
10. List out the employees who are earning salary between 3000 and 4500
11. List out the employees who are working in department 10 or 20
12. Find out the employees who are not working in department 10 or 30
13. List out the employees whose name starts with “S”
14. List out the employees whose name start with “S” and end with “H”
15. List out the employees whose name length is 4 and start with “S”
16. List out the employees who are working in department 10 and draw the
salaries more than 3500
17. list out the employees who are not receiving commission.
Order By Clause:
18. List out the employee id, last name in ascending order based on the employee
id.
19. List out the employee id, name in descending order based on salary column
20. list out the employee details according to their last_name in ascending order
and salaries in descending order
21. 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:
22.How many employees who are working in different departments wise in the
organization
23. List out the department wise maximum salary, minimum salary, average
salary of the employees
24. List out the job wise maximum salary, minimum salary, average salaries of
the employees.
25. List out the no.of employees joined in every month in ascending order.
26. List out the no.of employees for each month and year, in the ascending order
based on the year, month.
27. List out the department id having atleast four employees.
28.How many employees in January month.
29.How many employees who are joined in January or September month.
30.How many employees who are joined in 1985.
31.How many employees joined each month in 1985.
32.How many employees who are joined in March 1985.
33.Which is the department id, having greater than or equal to 3 employees
joined in April 1985.
Sub-Queries
34. Display the employee who got the maximum salary.
35. Display the employees who are working in Sales department
36. Display the employees who are working as “Clerk”.
37. Display the employees who are working in “New York”
38. Find out no.of employees working in “Sales” department.
39.Update the employees salaries, who are working as Clerk on the basis of
10%.
40.Delete the employees who are working in accounting department.
41. Display the second highest salary drawing employee details.
42. Display the Nth highest salary drawing employee details
Sub-Query operators: (ALL,ANY,SOME,EXISTS)
43. List out the employees who earn more than every employee in department
30.
44. List out the employees who earn more than the lowest salary in department
30.
45. Find out whose department has not employees.
46. 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
1. SQL > Select * from employee;
2. SQL > Select * from department;
3. SQL > Select * from job;
4. SQL > Select * from loc;
5. SQL > Select first_name, last_name, salary, commission from employee;
6. SQL > Select employee_id “id of the employee”, last_name “name",
department id as “department id” from employee;
7. SQL > Select last_name, salary*12 “annual salary” from employee
8. SQL > Select * from employee where last_name=’SMITH’;
9. SQL > Select * from employee where department_id=20
10.SQL > Select * from employee where salary between 3000 and 4500
11.SQL > Select * from employee where department_id in (20,30)
12.SQL > Select last_name, salary, commission, department_id from employee
where department_id not in (10,30)
13.SQL > Select * from employee where last_name like ‘S%’
14.SQL > Select * from employee where last_name like ‘S%H’
15.SQL > Select * from employee where last_name like ‘S___’
16.SQL > Select * from employee where department_id=10 and salary>3500
17.SQL > Select * from employee where commission is Null
18.SQL > Select employee_id, last_name from employee order by employee_id
19.SQL > Select employee_id, last_name, salary from employee order by salary
desc
20.SQL > Select employee_id, last_name, salary from employee order by
last_name, salary desc
21.SQL > Select employee_id, last_name, salary from employee order by
last_name, department_id desc
22.SQL > Select department_id, count(*), from employee group by
department_id
23.SQL > Select department_id, count(*), max(salary), min(salary), avg(salary)
from employee group by department_id
24.SQL > Select job_id, count(*), max(salary), min(salary), avg(salary) from
employee group by job_id
25.SQL > Select to_char(hire_date,’month’)month, count(*) from employee
group by to_char(hire_date,’month’) order by month
26.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’)
27.SQL > Select department_id, count(*) from employee group by
department_id having count(*)>=4
28.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’
29.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’)
30.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
31.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’)
32.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’)
33.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
34.SQL > Select * from employee where salary=(select max(salary) from
employee)
35.SQL > Select * from employee where department_id IN (select
department_id from department where name=’SALES’)
36.SQL > Select * from employee where job_id in (select job_id from job where
function=’CLERK’
37.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’))
38.SQL > Select * from employee where department_id=(select department_id
from department where name=’SALES’ group by department_id)
39.SQL > Update employee set salary=salary*10/100 wehre job_id=(select
job_id from job where function=’CLERK’)
40.SQL > delete from employee where department_id=(select department_id
from department where name=’ACCOUNTING’)
41.SQL > Select * from employee where salary=(select max(salary) from
employee where salary <(select max(salary) from employee))
42.SQL > Select distinct e.salary from employee where & no-1=(select
count(distinct salary) from employee where sal>e.salary)
43.SQL > Select * from employee where salary > all (Select salary from
employee where department_id=30)
44.SQL > Select * from employee where salary > any (Select salary from
employee where department_id=30)
45.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)
46.SQL > Select name from department d where not exists (select last_name
from employee e where d.department_id=e.department_id)
47.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)
48.SQL > Select employee_id, last_name, name from employee e, department d
where e.department_id=d.department_id
49.SQL > Select employee_id, last_name, function from employee e, job j where
e.job_id=j.job_id
50.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
51.SQL > Select name, count(*) from employee e, department d where
d.department_id=e.department_id group by name
52.SQL > Select name, count(*) from employee e, department d where
d.department_id=e.department_id group by name having name=’SALES’
53.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
54.SQL > Select function, count(*) from employee e, job j where
j.job_id=e.job_id group by function
55.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
56.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
57.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
58.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
59.SQL > Select e.last_name emp_name, m.last_name, mgr_name from
employee e, employee m where e.manager_id=m.employee_id
60.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
61.SQL > Select m.manager_id, count(*) from employee e, employee m where
e.employee_id=m.manager_id group by m.manager_id
62.SQL > Select last_name, d.department_id, d.name from employee e,
department d where e.department_id(+)=d.department_id
63.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’))
64.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’))
65.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’))
66.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
