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    Given 3 tables: Employee (empid, empname), Employee_timesheet(empid, no_of_hours, date), Employee_hour_rate (empid, hour_rate). Calculate pay check of all employees on monthly basis.

    - Billa on January 30, 2011 Report Duplicate | Flag
    Amazon Analyst Database



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1
of 3 vote

The following a working SQL query. You can try it yourself.
-----------------------------------------------------------------------------------

SELECT     TEMP.EmpID, TEMP.EmpName, 
                     Employees_HoursRate.hour_rate *  TEMP.TotalHoursPerMonth AS SAL
FROM         Employees_HoursRate INNER JOIN
                    (SELECT     Employee.empid AS EmpID, Employee.empname AS EmpName, 
                     SUM(Employee_TimesSheet.no_of_hours) AS TotalHoursPerMonth
                     FROM        Employee INNER JOIN
                     Employee_TimesSheet ON Employee.empid = Employee_TimesSheet.empid
                     GROUP BY Employee.empid, Employee.empname, 
                                          DATEPART(month, Employee_TimesSheet.date)) AS TEMP ON 
                     Employees_HoursRate.empid = TEMP.EmpID

ORDER BY TEMP.EmpID

- Ayad Barsoum on January 02, 2012 | Flag Reply
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0
of 0 votes

You might want to include "year" in your group by , otherwise you are adding monthly salary of all years.

- Zhijing78 on April 05, 2013 | Flag
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1
of 1 vote

create table employee(eid smallint,ename varchar(10));

insert into employee values(1 ,'Abhishek');
insert into employee values(2, 'Nikhil');
insert into employee values(3,'Rama');
insert into employee values(4 , 'Vidyut');
insert into employee values(5 , 'Pavan');

select * from timesheet;

create table timesheet(eid smallint,no_hrs smallint,cdate date);

insert into timesheet values(1,100 , '2011-06-30');
insert into timesheet values(2 , 105 , '2011-06-30');
insert into timesheet values(3 , 90 , '2011-06-30');
insert into timesheet values(4 , 85 , '2011-06-30');
insert into timesheet values(5 , 110 , '2011-06-30');


insert into timesheet values(5 , 110 , '2011-04-30');
insert into timesheet values(5 , 110 , '2011-06-30');
insert into timesheet values(4 , 110 , '2010-06-30');
insert into timesheet values(3 , 110 , '2011-03-30');
insert into timesheet values(2 , 110 , '2011-06-30');
insert into timesheet values(4 , 110 , '2011-08-30');
insert into timesheet values(5 , 110 , '2011-06-30');
insert into timesheet values(2 , 110 , '2010-06-30');
insert into timesheet values(1 , 110 , '2011-06-30');
insert into timesheet values(1 , 110 , '2010-06-30');
insert into timesheet values(3 , 110 , '2011-06-30');
insert into timesheet values(4 , 110 , '2010-06-30');


select * from hr_rate;
+------+---------+
| eid | hr_rate |
+------+---------+
| 1 | 50 |
| 2 | 60 |
| 3 | 40 |
| 4 | 30 |
| 5 | 80 |

create table hr_rate(eid smallint,hr_rate smallint);

insert into hr_rate values(1,50);
insert into hr_rate values(2,60);
insert into hr_rate values(3,40);
insert into hr_rate values(4,30);
insert into hr_rate values(5,80);


select e.ename,month(t.cdate),year(t.cdate),sum(t.no_hrs*hr.hr_rate)
from employee e,timesheet t,hr_rate hr
where e.eid=t.eid and e.eid=hr.eid
group by e.ename,month(t.cdate),year(t.cdate)
order by ename,year(t.cdate),month(t.cdate)

- praveen on May 14, 2012 | Flag Reply
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SELECT 
    E1.empid,
    E1.empname,
    E2.Hours_rate*E3.total as SAL
FROM 
    PWM.EMP as E1, 
    PWM.emp_rate as E2, 
    ( 
    SELECT 
        EMPID, 
        SUM(HOURS) as total 
    FROM 
        PWM.EMP_TIMESHEET 
    WHERE 
        DATE     > '2011-01-01' 
        and DATE < '2011-10-01' 
    GROUP BY 
        EMPID 
    ) as E3 where E1.empid = E2.empid and E1.empid =E3.empid

- Amit on January 31, 2011 | Flag Reply
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of 0 votes

this looks to be wrong.. so is many other queries that are here...the hours and the hour rate of each employee can differ.. if so you can multiplication like this...dont write any SQL.. just try this in a spreadsheet.. you will get to know sum(hours)* hour_rate can lead you to wrong results if the hour rate is different....

SELECT SUM(HOUR_RATE*number_of_hours) FROM EMPLOYEE where date between date_1 and date_2 would be the correct SQL

- Anon... on April 27, 2012 | Flag
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of 0 vote

select emp.empname, ab.yearmonth, total_hours*ehr.hour_rate
from employee emp
inner join
(
select b.yearmonth,b.empid, sum(b.no_of_hours) total_hours
(
select CAST(datepart(year,ts.date) AS VARCHAR(6)) + CAST(datepart(month,ts.date) AS VARCHAR(6)) as yearmonth,
ts.empid, ts.no_of_hours, date
from Employee_timesheet ts
) b
group by b.yearmonth, b.empid
) ab on ab.empid = emp.empid
inner join employee_hour_rate ehr on ehr.empid = emp.empid

- Swathi on January 31, 2011 | Flag Reply
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0
of 0 vote

I believe following query shud work. Natural join. Group by month

select EE.empname, ET.yearmonth, sum(ET.total_hours)*EHR.hour_rate payrate
from Employee EE, Employee_timesheet ET, Employee_hour_rate EHR
group by extract(month from date)

- AB on February 12, 2011 | Flag Reply
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0
of 0 vote

select e.empid,
e.empname,
to_date(date,'MM') as month,
sum(et.no_of_hours)*ehr.hour_rate
from Employee as e, Employee_timesheet as et, Employee_hour_rate as ehr
where e.empid = et.empid
and e.empid = ehr.empid
group by e.empid, e.empname, to_date(date,'MM'), ehr.hour_rate

- Anonymous on February 18, 2011 | Flag Reply
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0
of 0 vote

SELECT
e.empid,
e.empname,
CAST(YEAR(t.date) AS VARCHAR) + '/' + CAST(MONTH(t.date) AS VARCHAR) AS ReportingMonth,
SUM(no_of_hours) * r.hour_rate AS Total
FROM
Employee AS e INNER JOIN
Employee_timesheet AS t ON e.empid = t.empid INNER JOIN
Employee_hour_rate AS r ON e.empid = r.empid
GROUP BY
e.empid, e.empname, r.hour_rate, CAST(YEAR(t.date) AS VARCHAR) + '/' + CAST(MONTH(t.date) AS VARCHAR)

- thomas@zh-wang.me on April 04, 2011 | Flag Reply
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0
of 0 vote

select e.epname,v.total from Employee e JOIN
(select sum(empt.hours)*empr.rate TOTAL_SAL,empid
FROM employee_timesheet empt JOIN employee_hour_rate empr
ON empid
WHERE
empt.date between (select trunc(sysdate, 'MM')as start_date from dual) AND (select add_months(trunc(sysdate, 'MM'), 1) -1 as last_date from dual)
group by empid)
USING empid

- Meera on May 27, 2011 | Flag Reply
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of 0 votes

select e.epname,v.total from Employee e JOIN
(select sum(empt.hours)*empr.rate TOTAL_SAL,empid
FROM employee_timesheet empt JOIN employee_hour_rate empr
ON empid
WHERE
empt.date between (select trunc(sysdate, 'MM')as start_date from dual) AND (select add_months(trunc(sysdate, 'MM'), 1) -1 as last_date from dual)
group by empid)v
USING empid

- Anonymous on May 27, 2011 | Flag
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of 0 vote

SELECT E.ENAME,SUM(ET.NO_OF_HOURS*ER.HOUR_DATE) FROM EMPLOYEE E,EMPLOYEE_TIMESHEET ET,EMPLOYEE_HOR_RATE EH
WHERE E.EMPID=ET.EMPID AND EH.EMPID=E.EMPID
AND TO_CHAR(DATE,'MM')=TO_CHAR(SYSDATE,'MM')
GROUP BY E.ENAME

- finance_oracle_support on March 04, 2012 | Flag Reply
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of 0 vote

create table employee (
empid int auto_increment primary key,
name varchar(20))TYPE=InnoDB;

create table employee_time(
timeid int(10) auto_increment primary key,
emp_id int,
no_of_hours int(10),
edate datetime,
index(emp_id),
foreign key (emp_id) references employee(empid))TYPE=InnoDB;

create table employee_rate(
rateid int auto_increment primary key,
emp_id int,
rate int,
foreign key (emp_id) references employee(empid))TYPE=InnoDB;

insert into employee(empid, name) values(1, 'adam');
insert into employee(empid, name) values(2, 'eve');

insert into employee_time(emp_id,no_of_hours,edate) values(1,8,'2013-02-26');
insert into employee_time(emp_id,no_of_hours,edate) values(1,8,'2013-02,25');
insert into employee_time(emp_id,no_of_hours,edate) values(1,8,'2013-02-24');
insert into employee_time(emp_id,no_of_hours,edate) values(1,8,'2013-03-01');

insert into employee_time(emp_id,no_of_hours,edate) values(2,8,'2013-02-26');
insert into employee_time(emp_id,no_of_hours,edate) values(2,8,'2013-02,25');

insert into employee_rate(emp_id,rate) values(1,10);
insert into employee_rate(emp_id,rate) values(2,10);

select * from employee;
select * from employee_time;
select * from employee_rate;

select name, sum(no_of_hours*rate) as salary from employee e,employee_time t, employee_rate r
where e.empid = t.emp_id and t.emp_id = r.emp_id and t.edate between '2013-02-01' and '2013-02-28' group by name

- bhat.86 on March 04, 2013 | Flag Reply
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of 0 vote

select e.empid, e.empname, T.yearR, T.monthR, T.salary
from guol.emp e left join
(
select h.empid as eid, extract(year from t.dateE) as yearR, extract(month from t.dateE) as monthR, sum(t.no_of_hours*h.rate) as salary
from guol.emp_h h, guol.emp_t t
where h.empid=t.empid
group by h.empid, extract(year from t.dateE), extract(month from t.dateE) ) T on e.empid=T.eid
;

- aileen on April 20, 2013 | Flag Reply
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-1
of 1 vote

mysql> select * from employee;
+-----+----------+
| eid | ename |
+-----+----------+
| 1 | Abhishek |
| 2 | Nikhil |
| 3 | Rama |
| 4 | Vidyut |
| 5 | Pavan |
+-----+----------+
5 rows in set (0.00 sec)

mysql> select * from timesheet;
+------+--------+------------+
| eid | no_hrs | cdate |
+------+--------+------------+
| 1 | 100 | 2011-06-30 |
| 2 | 105 | 2011-06-30 |
| 3 | 90 | 2011-06-30 |
| 4 | 85 | 2011-06-30 |
| 5 | 110 | 2011-06-30 |
+------+--------+------------+
5 rows in set (0.00 sec)

mysql> select * from hr_rate;
+------+---------+
| eid | hr_rate |
+------+---------+
| 1 | 50 |
| 2 | 60 |
| 3 | 40 |
| 4 | 30 |
| 5 | 80 |
+------+---------+
5 rows in set (0.00 sec)

mysql> select Employee.ename,timesheet.no_hrs,hr_rate.hr_rate,timesheet.no_hrs*hr_rate.hr_rate as TOTAL
-> from Employee inner join timesheet using (eid)
-> inner join hr_rate using (eid)
-> where DAY(timesheet.cdate) = DAY(LAST_DAY(CURDATE()));
+----------+--------+---------+-------+
| ename | no_hrs | hr_rate | TOTAL |
+----------+--------+---------+-------+
| Abhishek | 100 | 50 | 5000 |
| Nikhil | 105 | 60 | 6300 |
| Rama | 90 | 40 | 3600 |
| Vidyut | 85 | 30 | 2550 |
| Pavan | 110 | 80 | 8800 |
+----------+--------+---------+-------+
5 rows in set (0.00 sec)

- Abhishek on June 05, 2011 | Flag Reply
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0
of 0 votes

understanding the question should have saved u a lot of time.

- anon on August 21, 2011 | Flag


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