I have an employee table with columns
employee_ID, employee_name, employee_DOB, emp_Email, Emp_Phone, Emp_Position
and a transaction table with columns
transaction_ID, employee_ID, distribution_ID, Invoice_Number, transaction_Date
I want to show these columns using a SQL query:
Emp_Position, Transaction_done
Transaction_done is a number consisting of how many transactions employees in each position has done.
I've tried to use select count(transaction_ID) but it didn't show a correct count.
Try this:
SELECT Emp_Position
,count(transaction_ID)
FROM employee e
INNER JOIN [transaction] t
ON e.[employee_ID] = t.[employee_ID]
GROUP BY Emp_Position;
Related
Given a table with two columns, department and employee, where every employee belongs to 1 department.
Given a list of employee ids, how do I select departments where all employees are in the list?
Department
Employee
finance
1
finance
2
marketing
3
marketing
4
IT
5
IT
6
given (2,3,4,5,6), returns ('marketing', 'IT')
(Note: DB flavor does not matter to me, you may use standard or DB-specific SQL)
Put your input id's into a table, join to the existing table, and check counts between input and the existing employees. If they match, output.
Change this to whatever RDBMS you're using
CREATE TEMP TABLE input_vals (id int);
insert into input_vals
values
(2),
(3),
(4),
(5),
(6);
with cte_joins AS (
select ot.department,
count(ot.employee) AS employee_count1,
sum(case when iv.id is not null then 1 else 0 end) AS employee_count2
from orig_table ot
left join input_vals iv
on ot.employee = iv.id
group by ot.department
)
select department
from cte_joins
where employee_count1 = employee_count2
You can use a couple of derived tables - one to get the count of employees in each department and one to get the count of employees in each department limited by your list of employees. Return the ones that match.
select
distinct full_list.department
from (
select
department,
count (*) as cnt
from
<your table>
group by department) full_list
inner join (
select
department,
count (*) as cnt
from
<your table>
group by department
where
employee in (2,3,4,5,6)
) limited_list
on full_list.department = limited_list.department
and full_list.cnt = limited_list.cnt
-- departments
select department from t where employee in (2,3,4,5,6);
-- departments containing extra persons
select department from t
where department in (select department from t where employee in (2,3,4,5,6))
and employee not in (2,3,4,5,6);
--- departments, employees from the list, with no extra
select department from t
where employee in (2,3,4,5,6) and department not in (
select department from t
where department in (
select department from t where employee in (2,3,4,5,6)
) and employee not in (2,3,4,5,6)
)
Assume you employee list is captured as a column in a table. Since you didn't provide any detail on the tool set you are using.
Something like this is a common pattern, depending on your dbms there's other more simple syntax.
select distinct
de.department
from department_employee de
where not exists
( select 1
from department_employee de2
where not exists
( select 1
from employee_list el
where de2.employee = el.employee
)
and de.department = de2.department
);
I want to write a query against this table, such that it will return the list of employees in an order of date of joining as per the sequence of Manager_ID.
It should be able to tell the reportee that joined most recently for each manager.
You can try this:
SELECT *
FROM
<TABLE>
INNER JOIN
(
SELECT MANAGER_ID
, MAX(DATE_OF_JOINING) AS MAX_DATE
FROM <TABLE>
GROUP BY MANAGER_ID
) MAX_MANAGER
ON <TABLE>.MANAGER_ID = MAX_MANAGER.MANAGER_ID
AND <TABLE>.DATE_OF_JOINING = MAX_MANAGER.MAX_DATE
If two employees were hired the same day for the same manager you have two rows.
I have employee table with emp id (emp_id) and department (dep_id) fields. An employee could be working in more than one Department. I want to write a sql query to display unique emp_ids who work in more than one department.
Pl help me to write sql query.
Thx
Answered here: SQL query for finding records where count > 1
You need to use count, group by and having like this.
select emp_id, count(dep_id)
from employee_department
group by emp_id
having count(dep_id)>1
Query
SELECT COUNT(*)
FROM
(
SELECT id_employee, COUNT(*) AS CNT
FROM Department_Employee
GROUP BY id_employee
) AS T
WHERE CNT > 1
I have the following example tables
Employee (EmpID, DepID)
Order (OrderID, EMpID, description)
What I'm trying to achieve is to select employees with most orders by department. I'm on it for like 4 hours already and can't find resolution to this perhaps easy problem.
All I get is either number of order by employee or max number of orders by one employee in one department but I'm struggling to get result as:
DepID, EmpID, Number of orders
Here's my solution for you :
WITH Temp AS (
SELECT
emp.EMpID
,emp.DepID
,COUNT(OrderId) nb_order
,ROW_NUMBER() OVER(PARTITION BY emp.DepID ORDER BY COUNT(OrderId) DESC) Ordre
FROM
Order ord
INNER JOIN
Employee emp
ON emp.EmpID = ord.EmpID
GROUP BY
emp.EMpID
,emp.DepID)
SELECT *
FROM Temp
WHERE Ordre = 1
I hope this will help you :)
I want to find the largest sale for each of my employees (and display the name of the employee). In MySQL, it's pretty straightforward:
select *
from employee, sale
where employee.id = sale.employee_id
group by employee_id
order by sale.total desc
This does pretty much what one would expect, it would return a list of employees and end up returning the largest sale record with the employee row.
But, Oracle does not allow you to return columns which are not group by expressions when a group by clause is used. Do this make what I do in MySQL "impossible" in Oracle? Or is there some workaround? I suppose I could perform some sort of subquery, but not sure if there is another way to do this that wouldn't quite be so complicated to construct.
Get rid of your select * and replace it with just the columns you need and then group by all the "non-processed" columns.
You'll end up with something like:
select employee.id, employee.name, max(sale.total)
from employee, sale
where employee.id = sale.employee_id
group by employee.id, employee.name
order by max(sale.total) desc
It's a pain - I've had to do this many times before - but just add all the related columns to your group by
To get the largest sale you can use group by with the max function.
select e.name, max (s.total)
from employee e, sale s
where e.id = s.employee_id
group by e.name
order by s.total desc
I have made an assumption that the employee name is in the name column of the employee table. I have also aliased the employee table and sales tables.
If you would prefer to see the total sales for an employee, you can swap out max() and use sum() instead.
Congratulations, you've learned just enough to be dangerous!
What you really want is each employee's largest sale. Now it happens that sorting them by sales amount desc and then grouping them works in MySQL, even though that isn't legal according to ANSI SQL. (Basically, MySQL is arbitrarily grabbing the first row for each employee, and that "works" because of the sort.)
The right way to do this is not to rely on the side of effect of the sort doing what you want; instead you should explicitly ask for what you want: the largest sale for each employee. In SQL that's:
select employee.id, max( sale.total)
from employee, sale
where employee.id = sale.employee_id
group by employee.id
order by 2
If you want to select an employee with a highest sale, you don't need GROUP BY here at all.
All you need is to select the highest sale and join it back to the employees:
SELECT *
FROM (
SELECT sale.*, ROW_NUMBER() OVER (ORDER BY total DESC) AS rn
FROM sale
) s
JOIN employee e
ON e.id = s.employee_id
AND s.rn = 1
This will select a single row with a total highest sale.
If you want to select per-employee highest sale, just add a PARTITION BY clause to your query:
SELECT *
FROM (
SELECT sale.*, ROW_NUMBER() OVER (PARTITION BY employee_id ORDER BY total DESC) AS rn
FROM sale
) s
JOIN employee e
ON e.id = s.employee_id
AND s.rn = 1