MY SQL practice files

Author: basira-jafarova

sql.practice1.sql

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+--1
+Select  job_title,max_salary- min_salary as "Salary differences" from hr.jobs where max_salary BETWEEN 12000 and 18000;
+--2
+Select * from hr.employees where commission_pct is null and salary BETWEEN 7000 and 12000 and department_id not in(50,30,80);
+--3
+Select first_name||' '||last_name as "full_name", hire_date, commission_pct,email ||'-'|| phone_number,salary from hr.employees
+ where salary>11000 order by "full_name" desc;
+--4
+Select first_name, last_name, salary from hr.employees where first_name like '%m' and hire_date< '05-JUN-10';
+--5
+Select first_name||' '||last_name as "full_name",email ||'-'|| phone_number as "Contact_detail",salary as "Remunaration" from hr.employees 
+where salary not BETWEEN 9000 and 17000 and commission_pct is not null;
+--6
+Select * from hr.departments where department_id=20;
+--7
+Select * from hr.job_history order by employee_id desc, start_date asc;
+--8
+Select job_id, salary from hr.employees where phone_number like '515%' or phone_number like '590%' and hire_date>'31-DEC-03' order by hire_date,salary ;
+--9
+Select * from hr.employees where hire_date BETWEEN '01-Jan-01'and '31-Dec-01';
+--10
+Select first_name,last_name from hr.employees where hire_date not between '01-JAN-06' and '31-DEC-07';'31-dec-2005' and '01-jan-2008'
+
+--11
+Select email,job_id,first_name from hr.employees where hire_date like '%2007' or hire_date like '%JAN%';
+--12
+Select * from hr.employees where hire_date>'31-Dec-07' or salary<10000;

sql.practice2.sql

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+--1
+Select first_name||' '||last_name as "full name", salary, commission_pct*salary, hire_date from hr.employees where salary<10000;
+--2
+select DISTINCT city, location_id from hr.locations order by location_id asc;
+-- distinct ferqlendirir, location id e gore asc
+--3
+select first_name, hire_date, job_id from hr.employees where (job_id = 'IT_PROG' or job_id='SA_REP') and 
+(hire_date  between '01-JAN-2002' and '31-DEC-2005');
+--4
+Select job_title from jobs order by job_title desc;
+--5
+Select * from hr.employees where commission_pct is null and salary between 5000 and 10000 and department_id=30;
+--6
+Select first_name, last_name,hire_date from hr.employees where hire_date > '01-JAN-08';
+--7
+Select * from hr.employees where department_id in (150,160,170) ;
+--8
+Select * from hr.employees where first_name like 'S%' or last_name like 'S%';
+--9
+SELECT first_name, last_name FROM hr.employees WHERE INSTR(last_name,'b') > 3;
+--kecilmiyib 1 ci weekde

sql.practice3.sql

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+--1. Display employees who joined in the month of May.
+Select first_name,hire_date  from hr.employees where (extract(month from hire_date))=5 ;
+
+-----------2. Display employees who joined in the current year.
+Select first_name as Name, hire_date from hr.employees
+ where extract(year from hire_date)=extract(year from SYSDATE) ;
+
+-----------3. Display the number of days between system date and 1st January 2011.
+Select round( sysdate -to_date('01-JAN-11','DD-MM-YYYY')) "number of days" from hr.employees;
+
+--4. Display maximum salary of employees.
+Select max(salary) from hr.employees;
+
+--5. Display number of employees in each department.
+Select count(*), department_id from hr.employees group by department_id;
+
+---------6. Display number of employees who joined after 15th of month.
+Select count(*) from hr.employees
+ where extract( day from hire_date)>15;
+
+-----------7. Display average salary of employees in each department who have commission percentage.
+Select AVG(salary) "Average",department_id from hr.employees 
+WHERE commission_pct is not null group by department_id ;
+
+--8. Display job ID for jobs with average salary more than 10000.
+Select job_id from hr. employees group by job_id having avg(salary)>1000;
+
+/*9. Display job ID, number of employees, sum of salary, and difference between the highest 
+salary and the lowest salary of the employees for all jobs.*/
+Select job_id, count(*) NOE, sum(salary),max(salary)-min(salary)difference from hr.employees group by job_id;
+--10.Display manager ID and number of employees managed by the manager.
+Select manager_id, count(*) NOEm from hr.employees group by manager_id having manager_id is not null;
+
+--11.Search for the key differences between CHAR and VARCHAR�data�types.

sql.practice4.sql

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+--1. Show minimum, average and maximum salary in last 15 years according to job id.
+Select MIN(salary),round(AVG(salary)),MAX(salary),job_id
+from hr.employees where EXTRACT(year from sysdate)-EXTRACT(year from hire_date)>15 group by job_id;
+
+--2. How many employees hired after 2005 for each department?
+Select count(*), department_id from hr.employees where hire_date> to_date('31-12-2005','dd.mm.yyyy') group by department_id;
+
+/*3. Write a query to show departments in which the difference between maximum and minimum 
+salary is greater than 5000.*/
+Select department_id from hr.employees group by department_id having max(salary)-min(salary)>5000;
+
+/*4. Display salaries of employees who has not commission pact according to departments 
+(without using where).*/
+Select sum(
+case 
+when commission_pct is null then salary
+else 0
+end)
+from hr.employees group by department_id;

sql.practice5.sql

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+--1. Display last name, job title of employees who have commission percentage and belongs to department 30.
+Select last_name,job_title from hr.employees emp left join hr.jobs jh on jh.job_id=emp.job_id
+where emp.commission_pct is not null and emp.department_id=30;
+--2. Display department name, manager name, and salary of the manager for all managers whose experience is more than 5 years.
+select dep.department_name, man.first_name || ' ' || man.last_name manager_name, man.salary
+from hr.departments dep
+left join hr.employees man on man.manager_id = dep.manager_id 
+where extract(year from sysdate) - extract(year from man.hire_date) >= 5;
+
+--3. Display employee name if the employee joined before his manager. 
+SELECT emp.first_name "Employee name",
+       emp.hire_date,
+       man.first_name "Mananger name",
+       man.hire_date
+FROM hr.employees emp
+LEFT JOIN hr.employees man ON emp.manager_id=man.employee_id
+where emp.hire_date < man.hire_date
+order by emp.employee_id;
+--4. Display employee name, job title for the jobs, employee did in the past where the job was done less than six months.
+select emp.first_name||' '|| emp.last_name "Employee name", jobs.job_title
+from hr.employees emp
+left join hr.job_history jh on emp.employee_id = jh.employee_id
+left join hr.jobs jobs on jh.job_id = jobs.job_id
+where months_between(sysdate, jh.start_date) < 6 ;
+--5. Display department name, average salary and number of employees with commission within the department.
+select dep.department_name, 
+       round(avg(emp.salary), 2) avg_salary, 
+       count(*)
+from hr.departments dep
+left join hr.employees emp on dep.department_id = emp.department_id
+where emp.commission_pct is not null
+group by dep.department_name;
+--6. Display employee name and country in which he is working.
+select emp.first_name||' '|| emp.last_name "Employee name",
+       cnt.country_name "Country name"
+from hr.employees emp
+left join hr.departments dep on emp.department_id = dep.department_id
+left join hr.locations loc on dep.location_id = loc.location_id
+left join hr.countries cnt on loc.country_id = cnt.country_id ;

sql.practice6.sql

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+--1. Display the first promotion year for each employee.
+Select emp.first_name, emp.employee_id,to_char (min(end_date),'YYYY') FPY
+from hr.employees emp left join hr.job_history jb on emp.employee_id=jb.employee_id group by emp.first_name, emp.employee_id;
+--2. Display location, city and department name of employees who have been promoted more than 
+--once.
+Select dep.location_id, loc.city, dep.department_name from hr.employees emp inner join hr.departments dep on emp.department_id=dep.department_id
+inner join hr.locations loc on dep.location_id=loc.location_id where emp.employee_id in(
+Select employee_id from hr.job_history jh group by employee_id having count(*)>1);
+--3. Display minimum and maximum “hire_date” of employees work in IT and HR departments.
+Select  min(hire_date),max(hire_date) from hr.employees emp left join hr.departments dep on dep.department_id=emp.department_id
+where department_name in ('Human Resources', 'IT') 
+group by department_name;
+--4. Find difference between current date and hire dates of employees after sorting them by hire 
+--date, then show difference in days, months and years.
+select first_name||' '||last_name name, hire_date, sysdate, 
+extract(year from sysdate) - extract(year from hire_date) year_diff,
+round(MONTHS_BETWEEN(SYSDATE, hire_date), 2) month_diff,
+round(sysdate-hire_date, 2) day_diff
+from hr.employees;
+
+--5. Find which departments used to hire earliest/latest.
+Select distinct department_name from hr.departments dep left join hr.employees emp on dep.department_id=emp.department_id where hire_date = (select min(hire_date) from hr.employees) or
+      hire_date = (select max(hire_date) from hr.employees);
+
+--6. Find the number of departments with no employee for each city. 
+select loc.city, count(employee_id) no_employee
+from hr.locations loc
+left join hr.departments dep on loc.location_id = dep.location_id
+left join hr.employees emp on dep.department_id = emp.department_id
+group by loc.city
+having count(employee_id)=0;
+--7. Create a category called “seasons” and find in which season most employees were hired.
+select case when extract(month from hire_date) in(12, 1, 2) then 'Winter'
+            when extract(month from hire_date) in(3, 4, 5) then 'Spring'
+            when extract(month from hire_date) in(6, 7, 8) then 'Summer'
+            when extract(month from hire_date) in(9, 10, 11) then 'Autumn'
+            else 'N/A' end as seasons,
+            count(distinct employee_id) 
+from hr.employees
+group by case when extract(month from hire_date) in(12, 1, 2) then 'Winter'
+            when extract(month from hire_date) in(3, 4, 5) then 'Spring'
+            when extract(month from hire_date) in(6, 7, 8) then 'Summer'
+            when extract(month from hire_date) in(9, 10, 11) then 'Autumn'
+            else 'N/A' end ;
+
+--8. Find the cities of employees with average salary more than 5000.
+Select loc.city,emp.first_name from hr.employees emp inner join hr.departments dep on emp.department_id=dep.department_id
+inner join hr.locations loc on dep.location_id=loc.location_id where avg(salary)>5000;
+

sql.practice7.sql

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+-- 1. According to the given diagram create STUDENTS, ACTIVITIES and SCHEDULE tables. 
+
+
+-- 2. Insert data into students table from employees table.
+
+insert into hr.students (s_id, first_name, last_name, phone_number, email)
+select employee_id, first_name, last_name, phone_number, email
+from hr.employees;
+
+
+-- 3. Change phone number to ‘***’ for students with s_id > 200.
+
+update students
+set phone_number = '***'
+where s_id > 200;
+
+
+-- 4. Update first name and last names of students in Upper cases.
+
+update students
+set first_name = upper(first_name),
+    last_name = upper(last_name);
+
+
+-- 5. Based on the students table populated with the following data, update the email to 'DSA' for all records whose s_id 
+--    is greater than 150.
+
+update students
+set email = 'DSA'
+where s_id > 150;
+
+
+-- 6. Create PROGRAMMERS table using records from EMPLOYEES where job_id contains ‘PROG’ substring
+
+create table PROGRAMMERS as
+select *
+from hr.employees
+where job_id like '%PROG%';
+
+
+-- 7. Delete records from students table where s_id is between 150 and 160.
+
+delete from students
+where s_id between 150 and 160;
+
+
+-- 8. 
+--    a) Insert some date into SCHEDULE, then truncate and see results.
+
+insert into schedule (s_id, a_id, s_date)
+values (123, 321, to_date('2023-05-25', 'YYYY-MM-DD'));
+insert into schedule (s_id, a_id, s_date)
+values (234, 432, to_date('2023-05-26', 'YYYY-MM-DD'));
+
+select * from schedule;
+
+truncate table SCHEDULE;
+
+select * from schedule;
+
+
+--    b) Drop schedule table
+
+drop table schedule;
+
+select * from schedule;
+
+
+-- 9. For any date given, write a script to find:
+
+--    a) The first and the last days of the next year;
+
+select '01-01-'|| to_char(extract(year from sysdate)+1) as "First Day of Next Year",  
+       '31-12-'|| to_char(extract(year from sysdate)+1) as "Last Day of Next Year"
+       from dual;
+
+--    b) The first and the last days of the next month;
+
+select last_day(trunc(sysdate, 'mm')) + 1 as "First_Day_Next_Month",
+       last_day(add_months(trunc(sysdate, 'mm'), 1)) as "Last_Day_Next_Month"     
+       from dual;
+
+--    c) The first and the last days of the previous month.
+
+select last_day(add_months(trunc(sysdate, 'mm'), -2)) + 1 as "First_Day_Previous_Month", 
+       last_day(add_months(trunc(sysdate, 'mm'), -1)) as "Last_Day_Previous_Month" 
+       from dual;
+
+
+-- 10. Create a table named “Participants” which consists of first_name, last_name and salary (have to more than 10000).
+
+create table PARTICIPANTS as
+select first_name, last_name, salary
+from hr.employees
+where salary > 10000;
+
+select * from participants;
+
+

sql.practice8.sql

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+-- 1. Return the name of the employee with the lowest salary in department 90.
+
+select min(salary)
+from hr.employees
+where department_id=90;
+
+select* from hr.employees emp 
+where salary = (select min(salary) from hr.employees where department_id=90)
+and department_id = 90 ;
+
+
+--2. Select the department name, employee name, and salary of all employees who work in the human resources or purchasing 
+--   departments. Compute a rank for each unique salary in both departments.
+
+SELECT 
+department_id, last_name, salary, 
+DENSE_RANK() OVER (PARTITION BY department_id ORDER BY salary) DENSE_RANK, 
+RANK() OVER (PARTITION BY department_id ORDER BY salary) RANK 
+FROM employees 
+WHERE department_id in (40, 60) 
+ORDER BY DENSE_RANK, last_name;
+
+
+--3. Select the 3 employees with minimum salary for department id 50.
+
+SELECT * from (select
+department_id, last_name, salary, 
+DENSE_RANK() OVER (PARTITION BY department_id ORDER BY salary) DRANK
+FROM employees 
+WHERE department_id = 50) 
+WHERE drank in (1, 2, 3);
+
+
+--4. Show first name, last name, salary and previously listed employee’s salary who works in “IT_PROG” over hire date.
+
+select 
+first_name, last_name, hire_date, salary, job_id, 
+lag(salary, 1, 0) over (partition by job_id order by hire_date) previous
+from employees 
+where job_id = 'IT_PROG';
+
+
+--5. Display details of current job for employees who worked as IT Programmers in the past.
+
+select * from hr.jobs
+where job_id in
+(select job_id from employees where employee_id in
+(select employee_id from job_history where job_id='IT_PROG'));
+
+-- Diger usul
+select * from JOBS J
+inner join hr.employees emp on j.job_id=emp.job_id
+inner join hr.job_history jh on jh.employee_id=emp.employee_id
+and jh.job_id='IT_PROG';
+
+
+--6. Make a copy of the employees table and update the salaries of the employees in the new table with the maximum salary 
+--   in their departments.
+
+update employees_copy ec set salary=(select max(salary) from hr.employees emp
+where emp.department_id=ec.department_id);
+
+
+--7. Make a copy of the employees table and update the salaries of the employees in the new table with a 30 percent increase.
+
+update employees_copy ec set salary=1.3*salary;
+