I have the table below.Using salary as condition I want to get multiple rows.
Below is current table call it employee.
empid name salary
-----------------------------------
1 A1 alex 20000
2 B2 ben 4500
3 C1 carl 14000
compare the salary to certain fixed values, and every time the salary is larger than the fixed value, show a record in output.My attempt condition case is close to this:
incometype= case When salary<6000 then 101 When salary Between 6000 And 18000 Then
102 Else 103 End
Desired ouput would be:
empid name salary incometype
------------------------------------------
1 A1 alex 20000 101
2 A1 alex 20000 102
3 A! alex 20000 103
4 B2 ben 4500 101
5 C1 carl 14000 101
6 C1 carl 14000 102
I have tried using union but union will give me 3 rows for each record even when value meets 1st condition.
Your question is unclear, because your logic implies that you should only have 3 output rows for 3 input rows. Your output however implies that you want to compare the salary to certain fixed values, and every time the salary is larger than the fixed value, show a record in output.
If the former is the case, Minh's query is all you need. In the latter case, you can do something like this:
select e.*, m.incometype
from employee e
left join
(
select 0 as threshold, 101 as incometype
union
select 5999 as threshold, 102 as incometype
union
select 17999 as threshold, 103 as incometype
) m
on e.salary > m.threshold
order by e.empid
If you want to add a calculate column i.e. one with values calculated using columns in this query, you can simply add it as a column in the select clause, like so:
select e.*,
m.incometype,
case
when <first condition> then <business logic here>
....
else <handle default case>
end as yourcomputedcolumn
from
...
This returns 3 rows and enough for your need:
SELECT empid, name, salary,
case When salary<6000 then 101
When salary Between 6000 And 18000 Then 102
Else 103 End as incometype
FROM employee;
Not very clear on the requirement, however the following worked for me:
Select
EmpId,Name,Sal,101 IncomeType
from Emp
Union all
Select
EmpId,Name,Sal,102
from Emp
Where Sal > 6000
union all
Select
EmpId,Name,Sal,103
from Emp
Where Sal > 18000;
Related
Consider following dataset:
id
name
mgr_id
salary
bonus
1
Paul
1
68000
10000
2
Lucas
2
29000
null
3
Max
1
50000
20000
4
Zack
2
30000
null
I now want to find the manager who pays his subordinates the highest average salary plus bonus. A manager is someone who is present in of the mgr_id cells. So in this example Paul and Lucas are managers because their id is present in the mgr_id column of themselves and Max for Paul and Zack for Lucas.
Basically I want MAX(AVG(salary + bonus)) and then grouped by the mgr_id. How can I do that?
I am using SQLite.
My expected output in this example would be simply the employee name 'Paul' because he has 2 subordinates (himself and Max) and pays them more than the other manager Lucas.
SELECT
mrg_id
, pay_avg
FROM
(
SELECT mrg_id
, AVG(salary + COALESCE(bonus,0)) pay_avg
FROM <table>
GROUP
BY mrg_id
) q
ORDER
BY pay_avg
desc
LIMIT 1
select top 1 t1.mgr_id,AVG((t1.salary)+(t1.bonus)) as tot_sal
from #tbl_emps as t1
group by t1.mgr_id
order by AVG((t1.salary)+(t1.bonus)) desc
This question already has answers here:
Select second most minimum value in Oracle
(3 answers)
Closed 4 years ago.
EMPID NAME DEPTID SALARY
---------- ------------------------------------------ --
101 surendra 201 1000
102 narendra 202 2000
103 rajesh 203 3000
104 ramesh 203 2000
105 hanumanth 202 10000
a) Write a Query to find least 5th (Least salary in 5th position from
least salary in the order) salaried employee?
b) Query to find the highest earning employee in each department
a) Write a Query to find least 5th (Least salary in 5th position from least salary in the order) salaried employee?
SELECT EMPID,NAME,DEPTID,SALARY
FROM
(
SELECT EMPID,NAME,DEPTID,SALARY, DENSE_RANK() OVER (ORDER BY SALARY) AS RN
FROM Table1
)
WHERE RN=5
b) Query to find the highest earning employee in each department
SELECT EMPID,NAME,DEPTID,SALARY
FROM
(
SELECT EMPID,NAME,DEPTID,SALARY,
DENSE_RANK() OVER (PARTITION BY DEPTID ORDER BY SALARY DESC) AS RN
FROM Table1
)
WHERE RN=1
Demo
http://sqlfiddle.com/#!4/63ce0/12
Let's assume we have a table named EmployeeSalary which has 3 columns:
Id
Name
Salary
Salary might be same for multiple employee according to the following table data
Id Name Salary
----------
6 Belalo 74
1 Karim 100
5 dIPU 100
4 Satter 102
9 Kiron 120
10 Rash 120
11 Harun 130
13 Baki 130
12 Munshi 132
2 Rahim 500
7 Kaif 987
8 Sony 987
3 Belal 4000
Now the query will be
SELECT A.Salary
FROM
(
SELECT DISTINCT Salary,
DENSE_RANK() OVER (ORDER BY Salary) AS Ranks
FROM [dbo].[EmployeeSalary]
) A
WHERE A.Ranks=5
I want to calculate highest nth salary in Oracle. I've already done with my solution but on google I found one query doing the same thing.
Here is the query
SELECT *
FROM Employee Emp1
WHERE (N - 1) = (SELECT COUNT(DISTINCT(Emp2.orig_salary))
FROM Employee Emp2
WHERE emp2.orig_salary > emp1.orig_salary)
Data
ID Name Birth Orig_Salary
2 John 15-JUL-97 2341
3 Joe 25-JAN-86 4321
4 Tom 13-SEP-06 2413
5 Jane 17-APR-05 7654
6 James 18-JUL-04 5679
7 Jodd 20-JUL-03 5438
8 Joke 01-JAN-02 8765
9 Jack 29-AUG-01 7896
I'm not not able to understand this query.
After running inner query it will always gives me count of 8 after doing this it will go to where clause where it will pick salary which is higher than the outer query salary. How equal operator is working here in between inner and outer query and how comparison is happening.
Could any help me to understand how this query is working technically in back-end..?
SELECT *
FROM Employee Emp1
WHERE (N - 1) = (SELECT COUNT(DISTINCT(Emp2.orig_salary))
FROM Employee Emp2 <--- cartesian product with same table
WHERE emp2.orig_salary > emp1.orig_salary) <---- but do the cartesian product only if line of salary of emp 2 is greater than the current line of Emp1 'salary
e.g assume there are only 3 lines in the table:
ID Name Birth Orig_Salary
2 John 15-JUL-97 2341
3 Joe 25-JAN-86 4321
4 Tom 13-SEP-06 5413
the main query will look at the first line --> 2 John 15-JUL-97 2341 <---, and subquery will return 2 because the salaries 4321 (emp2.orig_salary) and 5413 (emp2.orig_salary) are greater than 2341 (emp1.orig_salary)
the main query will then look at the second line --> 3 Joe 25-JAN-86 4321 <---, and subquery will return 1 because the salaries 5413 (emp2.orig_salary) is greater than 2341 (emp1.orig_salary)
when i say subquery, it is the
=(SELECT COUNT(DISTINCT(Emp2.orig_salary))
FROM Employee Emp2 <--- cartesian product with same table
WHERE emp2.orig_salary > emp1.orig_salary)
and the main query is
SELECT *
FROM Employee Emp1
WHERE
the returned value from the subquery is then compare to the where condition n-1, if the condition is satisfied, then it retrieves the line.
There is no need to understand that query. The correct formulation is:
SELECT Emp1.*
FROM (SELECT Emp1.*, DENSE_RANK() OVER (ORDER BY Emp2.orig_salary) as seqnum
FROM Employee Emp1
) Emp1
WHERE seqnum = <n>;
This gives the details for the employees. If you just want the salary:
SELECT orig_salary
FROM (SELECT Emp1.*, DENSE_RANK() OVER (ORDER BY Emp2.orig_salary) as seqnum
FROM Employee Emp1
) Emp1
WHERE seqnum = <n> AND rownum = 1;
I should note that a simpler version of this is:
select distinct orig_salary
from employees
order by orig_salary desc
offset <n - 1>
fetch first 1 row only;
The use of a correlated subquery for this is a pleasant anachronism from the days when relational databases were not as powerful as they are now. It is of historical interest.
I have the following table with data:
ID SAL
_______________
1 1500
1 1500
1 2000
1 1500
2 2500
2 2500
3 1500
I want to query the same data BUT with new column that gives me a sequence of each duplicated row (based on id and sal columns).
My desired query results should be like this:
ID SAL ROW_SEQ
_______________________
1 1500 1
1 1500 2
1 2000 1
1 1500 3
2 2500 1
2 2500 2
3 1500 1
Any one has any idea PLEASE!..
You want to use a ROW_NUMBER() function here.
SELECT ID, SAL, ROW_NUMBER() OVER (PARTITION BY ID, SAL, ORDER BY SAL) AS ROW_SEQ
FROM MyTable
ORDER BY ID, SAL
By partitioning over your two desired fields, it will give you a ranking within each group in a sequential manner.
The documentation on it can be found here:
https://docs.oracle.com/cd/B19306_01/server.102/b14200/functions137.htm
I have the table below.Using salary as condition I want to get multiple rows. Below is current table call it employee.
empid | name | salary
-----------------------------------
1 A1 alex 20000
2 B2 ben 4500
3 C1 carl 14000
compare the salary to certain fixed values, and every time the salary is larger than the fixed value, show a record in output.Also output a calculated tax column.Below is a step closer to my final result
select e.*,tax= (case when salary<6000 then tax=0.06 *salary,when salary between 6000 and 18000 then tax= 0.06 *(salary -6000),else tax =0 ),m.incometype,
from employee e
left join
(
select 0 as threshold, 101 as incometype
union
select 5999 as threshold, 102 as incometype
union
select 17999 as threshold, 103 as incometype
) m
on e.salary > m.threshold
order by e.empid
Desired ouput would be:
empid | name | salary | incometype | tax
----------------------------------------------
1 A1 alex 20000 101 360
2 A1 alex 20000 102 720
3 A! alex 20000 103 0
4 B2 ben 4500 101 270
5 C1 carl 14000 101 360
6 C1 carl 14000 102 480
this is a question further to Select statement with multiple rows from condition on values in single column
this may help you
select e.*,
case
when salary<6000 then (0.06*salary)
when salary between 6000 and 1800 then (0.06*(salary -6000))
when m.incometype=103 then 0
end as tax
,m.incometype,
from employee e
left join
(
select 0 as threshold, 101 as incometype
union
select 5999 as threshold, 102 as incometype
union
select 17999 as threshold, 103 as incometype
) m
on e.salary > m.threshold
order by e.empid