Can you let me know how to select only two employees from every department? The table has deptname, ssn, name . I am doing a sampling and I need only two ssns for every department name. Can someone help?
You can accomplish this with an "OLAP expression" row_number()
with e as
( select deptname, ssn, empname,
row_number() over (partition by dptname order by empname) as pick
from employees
)
select deptname, ssn, empname
from e
where pick < 3
order by deptname, ssn
This example will give you the two employees with the lowest order names, because that is what is specified in the row_number() (order by) expression.
Try this:
select *
from t t1
where (
select count(*)
from t t2
where
t2.deptname = t1.deptname
and
t2.ssn <= t1.ssn) <= 2
order by deptname, ssn,name;
The above will give "smallest" two ssn.
If you want top 2, change to t2.ssn >= t1.ssn
sqlfiddle
The data:
The result from query:
select * from
( select rank() over (partition by dptname order by empname) as count , *
from employees
)
where count<=2
order by deptname, ssn,name;
Related
I just started learning Postgres, and I'm trying to make an aggregation table that has the columns:
user_id
booking_sequence
booking_created_time
booking_paid_time
booking_price_amount
total_spent
All columns are provided, except for the booking_sequence column. I need to make a query that shows the first five flights of each user that has at least x purchases and has spent more than a certain amount of money, then sort it by the amount of money spent by the user, and then sort it by the booking sequence column.
I've tried :
select user_id,
row_number() over(partition by user_id order by user_id) as booking_sequence,
booking_created_time as booking_created_date,
booking_price_amount,
sum(booking_price_amount) as total_booking_price_amount
from fact_flight_sales
group by user_id, booking_created_time, booking_price_amount
having count(user_id) > 5
and total_booking_price_amount > 1000
order by total_booking_price_amount;
I got 0 when I added count(user_id) > 5, and total_booking_price_amount is not found when I add the second condition in the HAVING clause.
Edit:
I managed to make the code function correctly, for those who are curious:
select x.user_id, row_number() over(partition by x.user_id)
as booking_sequence, x.booking_created_time::date as booking_created_date, x.booking_price_amount,
sum(y.booking_price_amount) as total_booking_price_amount from
(
select user_id, booking_created_time, booking_price_amount from fact_flight_sales
group by user_id, booking_created_time, booking_price_amount
) as x
join
(
select user_id, booking_price_amount
from fact_flight_sales group by user_id, booking_price_amount
) as y
on x.user_id = y.user_id
group by x.user_id, x.booking_created_time, x.booking_price_amount
having count(x.user_id) >= 1 and sum(y.booking_price_amount) >250000
order by total_booking_price_amount desc, booking_sequence asc;
Big thanks to Laurenz for the help!
About count(user_id) > 5:
HAVING is calculated before window functions are evaluated, So result rows excluded by the HAVING clause will not be used to calculate the window function.
About total_booking_price_amount in HAVING:
You cannot use aliases from the SELECT list in the HAVING clause. You will have to repeat the expression (or use a subquery).
Hi guys I have two tables dbo.Sales (customer_id, order_date, product_id) and dbo.Menu (Product_id, product_name, price). The question is
What was the first item from the menu purchased by each customer?
My solution is
select A.customer_id,m.product_id, m.product_name
from dbo.menu m
cross apply
(select top 1 * from dbo.sales s
where s.product_id=m.product_id
group by s.customer_id,s.order_date, s.product_id
order by s.order_date) A
customer_id product_id product_name
A 1 sushi
A 2 curry
C 3 ramen
Missing customer is B. Instead of B it gives me the second first order by A.
I need for each customer
Murat
You could use a ROW_NUMBER() window function to get the earliest product_id per customer and then join to the Menu table to get your product details.
Edit: Updated ORDER to ASC.
;with cte
as (
select customer_id, product_id, row_number() over (partition by customer_id order by order_date acs) RN
from dbo.Sales)
select c.customer_id, c.product_id, m.product_name
from cte c
join dbo.menu m on c.product_id=m.product_id
where RN = 1
SELECT distinct s.customer_id,
FIRST_VALUE(m.product_name) OVER (partition by s.customer_id order by order_date )
as FirstItem_Customer
FROM [dbo].[sales] S
join [dbo].[menu] M on M.product_id=s.product_id
Name Value AnotherColumn
-----------
Pump 1 8000.0 Something1
Pump 1 10000.0 Something2
Pump 1 10000.0 Something3
Pump 2 3043 Something4
Pump 2 4594 Something5
Pump 2 6165 Something6
My table looks something like this. I would like to know how to select max value for each pump.
select a.name, value from out_pumptable as a,
(select name, max(value) as value from out_pumptable where group by posnumber)g where and g.value = value
this code does the job, but i get two entries of Pump 1 since it has two entries with same value.
select name, max(value)
from out_pumptable
group by name
select name, value
from( select name, value, ROW_NUMBER() OVER(PARTITION BY name ORDER BY value desc) as rn
from out_pumptable ) as a
where rn = 1
SELECT
b.name,
MAX(b.value) as MaxValue,
MAX(b.Anothercolumn) as AnotherColumn
FROM out_pumptabl
INNER JOIN (SELECT
name,
MAX(value) as MaxValue
FROM out_pumptabl
GROUP BY Name) a ON
a.name = b.name AND a.maxValue = b.value
GROUP BY b.Name
Note this would be far easier if you had a primary key. Here is an Example
SELECT * FROM out_pumptabl c
WHERE PK in
(SELECT
MAX(PK) as MaxPK
FROM out_pumptabl b
INNER JOIN (SELECT
name,
MAX(value) as MaxValue
FROM out_pumptabl
GROUP BY Name) a ON
a.name = b.name AND a.maxValue = b.value)
select Name, Value, AnotherColumn
from out_pumptable
where Value =
(
select Max(Value)
from out_pumptable as f where f.Name=out_pumptable.Name
)
group by Name, Value, AnotherColumn
Try like this, It works.
select * from (select * from table order by value desc limit 999999999) v group by v.name
Using analytic function is the easy way to find max value of every group.
Documentation : https://learn.microsoft.com/en-us/sql/t-sql/functions/row-number-transact-sql?view=sql-server-ver15
Select name,
value,
AnotherColumn
From(
SELECT Row_Number() over(partition by name order by value desc)as
row_number, *
FROM students
)
Where row_number = 1
SELECT t1.name, t1.Value, t1.AnotherColumn
FROM mytable t1
JOIN (SELECT name AS nameMax, MAX(Value) as valueMax
FROM mytable
GROUP BY name) AS t2
ON t2.nameMax = t1.name AND t2.valueMax = t1.Value
WHERE 1 OR <anything you would like>
GROUP BY t1.name;
SELECT DISTINCT (t1.ProdId), t1.Quantity FROM Dummy t1 INNER JOIN
(SELECT ProdId, MAX(Quantity) as MaxQuantity FROM Dummy GROUP BY ProdId) t2
ON t1.ProdId = t2.ProdId
AND t1.Quantity = t2.MaxQuantity
ORDER BY t1.ProdId
this will give you the idea.
Have a table with 3 columns: ID, Signature, and Datetime, and it's grouped by Signature Having Count(*) > 9.
select * from (
select s.Signature
from #Sigs s
group by s.Signature
having count(*) > 9
) b
join #Sigs o
on o.Signature = b.Signature
order by o.Signature desc, o.DateTime
I now want to select the 1st and 10th records only, per Signature. What determines rank is the Datetime descending. Thus, I would expect every Signature to have 2 rows.
Thanks,
I would go with a couple of common table expressions.
The first will select all records from the table as well as a count of records per signature, and the second one will select from the first where the record count > 9 and add row_number partitioned by signature - and then just select from that where the row_number is either 1 or 10:
With cte1 AS
(
SELECT ID, Signature, Datetime, COUNT(*) OVER(PARTITION BY Signature) As NumberOfRows
FROM #Sigs
), cte2 AS
(
SELECT ID, Signature, Datetime, ROW_NUMBER() OVER(PARTITION BY Signature ORDER BY DateTime DESC) As Rn
FROM cte1
WHERE NumberOfRows > 9
)
SELECT ID, Signature, Datetime
FROM cte2
WHERE Rn IN (1, 10)
ORDER BY Signature desc
Because I don't know what your data looks like, this might need some adjustment.
The simplest way here, since you already know your sort order (DateTime DESC) and partitioning (Signature), is probably to assign row numbers and then select the rows you want.
SELECT *
FROM
(
select o.Signature
,o.DateTime
,ROW_NUMBER() OVER (PARTITION BY o.Signature ORDER BY o.DateTime DESC) [Row]
from (
select s.Signature
from #Sigs s
group by s.Signature
having count(*) > 9
) b
join #Sigs o
on o.Signature = b.Signature
order by o.Signature desc, o.DateTime
)
WHERE [Row] IN (1,10)
I want to select rows for all employess without repeating the data in one column.
For example I have two rows where salary (before raise) is displayed, how can I display only the largest figure without duplication.
You can use Row_Number function
Here is a sample code
select * from (
select *,
row_number() over (partition by empid, name, department order by salary desc) as rn
from employee
) employee where rn = 1
You can find Row_Number() with Partition By clause sample at http://www.kodyaz.com
If I'm understanding the question correctly, then a simple MAX function and GROUP BY would work.
SELECT EmployeeId, OtherColumns, MAX(Salary)
FROM tblEmployees
GROUP BY EmployeeId, OtherColumns