I have a column for group name and a column for amount spent.
I need to sum the amounts group them based on the group name and then grab the highest five. After that, I need to combine the the rest into it's own group w/ a total of their amount spent. This is what i have right now
SELECT groupName, SUM(amount) AS theAmountSpent
FROM purchases
GROUP BY groupName
ORDER BY theAmountSpent DESC
This groups and orders them, but i dont know how to then grab the remaining groups to combine them. Any help would be appreciated.
Alternate CTE-approach using row_number() (SQL Server 2005+):
WITH cte AS (
SELECT ROW_NUMBER() OVER (ORDER BY (SUM(amount)) DESC) AS num,
groupName, SUM(amount) AS theAmountSpent
FROM purchases
GROUP BY groupName
)
SELECT groupName, theAmountSpent FROM cte WHERE num BETWEEN 1 AND 5 --top 5
UNION ALL
SELECT 'Sum rest', SUM(theAmountSpent) FROM cte WHERE num > 5 -- sum of rest
If I'm understanding you correctly, this should do it:
SELECT top 5 groupName, SUM(amount) AS theAmountSpent
into #tempSpent FROM purchases
GROUP BY groupName
ORDER BY theAmountSpent DESC
Select * from #tempSpent -- get the top 5
--get sum for the rest
SELECT SUM(amount) AS theAmountSpent
FROM purchases
where groupName not in (select groupName from #tempSpent)
Drop table #tempSpent
Another idea from Larsts code:
WITH cte
AS
(
SELECT case
when ROW_NUMBER() OVER (ORDER BY (SUM(amount)) DESC) <=5
then ROW_NUMBER() OVER (ORDER BY (SUM(amount)) DESC)
else 6 end AS num
, groupName
, SUM(amount) AS theAmountSpent
FROM purchases
GROUP BY groupName
)
SELECT num
, max(groupName)
, sum(theAmountSpent )
FROM cte
group by num
Related
I have a table Tabl1 : id, name, country, year, medal.
how can I find the top 10 countries by the number of medals for each year in 1 request?
thanks:)
You haven't told us anything about your table schema or the data, so this is a guess!
Going to assume your medal column contains the qty of medals for each Id/name, so you just need to rank by the sum of medals. Something along the lines of:
select [year], country, [Rank] from (
select [year], country, Rank() over(partition by [year] order by Sum(medal) desc ) [Rank]
from Tabl1
group by [year],country
)x
where [Rank]<=10
order by [year], [Rank]
here you can get the top 10 countries in each year:
select * from
(
select country,year,count(*),row_number() over (order by count(*) desc) as rn
from table
group by country, year
) tt
where tt.rn < 11
the sub query groups the data per country and year and gives you count() of each group, but at the same time It sorts them per count(*) desc and gives the a row number per each group ( it happanes using row_number() window funcion) , so the country with the most medal in eacg year is on top and it gets row number = 1 in each group , you need top 10 , so you filter them tt.rn < 11 in the main query.
If you want 10 countries per year:
with data as (
select country, "year" as yr,
rank() over (partition by "year" order by count(*) desc) as rnk
from T
group by country, "year"
)
select yr as "year", country from data
where rnk <= 10
order by yr, rnk;
Note that if ties are possible this could return more than ten rows for any given year.
Customer have ordered from different cities. Thus we have multiple cities against same customer_id. I want to display that city against customer id which has occurred maximum number of times , in case where customer has ordered same number of orders from multiple cities that city should be selected from where he has placed last order. I have tried something like
SELECT customer_id,delivery_city,COUNT(DISTINCT delivery_city)
FROM analytics.f_order
GROUP BY customer_id,delivery_city
HAVING COUNT(DISTINCT delivery_city) > 1
WITH cte as (
SELECT customer_id,
delivery_city,
COUNT(delivery_city) as city_count,
MAX(order_date) as last_order
FROM analytics.f_order
GROUP BY customer_id, delivery_city
), ranking as (
SELECT *, row_number() over (partition by customer_id
order by city_count DESC, last_order DESC) as rn
FROM cte
)
SELECT *
FROM ranking
WHERE rn = 1
select customer_id,
delivery_city,
amount
from
(
select t.*,
rank() over (partition by customer_id order by amount asc) as rank
from(
SELECT customer_id,
delivery_city,
COUNT(DISTINCT delivery_city) as amount
FROM analytics.f_order
GROUP BY customer_id,delivery_city
) t
)
where rank = 1
Please consider a table of vendors having two columns: VendorName and PayableAmount
I'm looking for a query which returns top ten vendors sorted by PayableAmount descending and sum of other payable amounts as "other" in 11th row.
Obviously, sum of PayableAmount from Vendors table should be equal to sum of PayableAmount from Query.
Technically, it's possible to do in one query:
declare #t table (
Name varchar(50) primary key,
Amount money not null
);
-- Dummy data
insert into #t (Name, Amount)
select top (20) sq.*
from (
select name, max(number) as [Amount]
from master.dbo.spt_values
where number between 100 and 100000
and name is not null
group by name
) sq
order by newid();
-- The table itself, for verification
select * from #t order by Amount desc;
-- Actual query
select top (11)
case when sq.RN > 10 then '<All others>' else sq.Name end as [VendorName],
case
when sq.RN > 10 then sum(sq.Amount) over(partition by case when sq.rn > 10 then 1 else 0 end)
else sq.Amount
end as [Value]
from (
select t.Name, t.Amount, row_number() over(order by t.Amount desc) as [RN]
from #t t
) sq
order by sq.RN;
It will even work on any SQL Server version starting with 2005. But, in real life, I would prefer to calculate these 2 parts separately and then UNION them.
This would perform the query you're looking for. Firstly extracting those in the top 10, then UNION ing that result with the higher ranked vendors, but calling those 'Other'
WITH rank AS (SELECT
VendorName,
PayableAmount,
ROW_NUMBER() OVER (ORDER BY PayableAmount DESC) AS rn
FROM vendors)
SELECT VendorName,
rn,
PayableAmount
FROM
rank WHERE rn <= 10
UNION
SELECT VendorName, 11 AS rn, PayableAmount
FROM
(
SELECT 'Other' AS VendorName,
SUM(PayableAmount) AS PayableAmount
FROM
rank WHERE rn > 10
) X11
ORDER BY rn
This has been tested in SQLFiddle.
this is for the 11th row
i didnt check it
declare #i int
set #i=
(select sum(x.PayableAmount)
from
(select * from table
except
select top 10 *from table
order by PayableAmount desc) as x)
select 'another',#i
In SQL Server, suppose we have a SALES_HISTORY table as below.
CustomerNo PurchaseDate ProductId
1 20120411 12
1 20120330 13
2 20120312 14
3 20120222 16
3 20120109 16
... and many records for each purchase of each customer...
How can I write the appropriate query for finding:
For each customer,
find the product he bought at MOST,
find the percentage of this product over all products he bought.
The result table must have columns like:
CustomerNo,
MostPurchasedProductId,
MostPurchasedProductPercentage
Assuming SQL Server 2005+, you can do the following:
;WITH CTE AS
(
SELECT *,
COUNT(*) OVER(PARTITION BY CustomerNo, ProductId) TotalProduct,
COUNT(*) OVER(PARTITION BY CustomerNo) Total
FROM YourTable
), CTE2 AS
(
SELECT *,
RN = ROW_NUMBER() OVER(PARTITION BY CustomerNo
ORDER BY TotalProduct DESC)
FROM CTE
)
SELECT CustomerNo,
ProductId MostPurchasedProductId,
CAST(TotalProduct AS NUMERIC(16,2))/Total*100 MostPurchasedProductPercent
FROM CTE2
WHERE RN = 1
You still need to deal when you have more than one product as the most purchased one. Here is a sqlfiddle with a demo for you to try.
Could do a lot prettier, but it works:
with cte as(
select CustomerNo, ProductId, count(1) as c
from SALES_HISTORY
group by CustomerNo, ProductId)
select CustomerNo, ProductId as MostPurchasedProductId, (t.c * 1.0)/(select sum(c) from cte t2 where t.CustomerNo = t2.CustomerNo) as MostPurchasedProductPercentage
from cte t
where c = (select max(c) from cte t2 where t.CustomerNo = t2.CustomerNo)
SQL Fiddle
I need to return the following statement but I only want to return the TOP 5 of each Sale value only.....not all the records.
Select ID, Code, sum(Sale) as Sale from TableName
Where Code = 11
Group By ID, code
I do not want this!
Select TOP 5 ID, Code, sum(Sale) as Sale from TableName
Where Code = 11
Group By ID, code
With Cte as
( Select ID, Code, sale as Sales ,
row_number() over (partition by ID,code order by sale desc) as row_num
from TableName where code=11
)
Select Id,code,sum(sales) from cte
GROUP BY ID, code
WHERE row_num < 6
WITH TopSales AS (
SELECT *, RANK() OVER (PARTITION BY ID, Code ORDER BY Sale DESC) saleRank
FROM TableName
)
SELECT ID, Code, SUM(Sale) AS Sale
FROM TopSales
WHERE (Code = 11) AND (saleRank <= 5)
GROUP BY ID, code
select id, code, SUM (sale)
from
(
select id, code, sale,
ROW_NUMBER() over(partition by id, code order by sale desc) rn
from tablename
) v
where rn<=5
group by id, code
Probably you need something like:
;WITH sales (
SELECT
id,
code,
sale,
ROW_NUMBER() OVER (PARTITION BY id, code ORDER BY sales DESC) n
FROM
TableName
WHERE
Code = 11
)
SELECT
id, code, sum(sale) sale
FROM
sales
WHERE
n <= 5
GROUP BY
id,
code
ROW_NUMBER() and PARTITION BY help to find last 5 sales. Then you SUM only top (highest) 5.
This query returns sum of top 5 sales for each (id, code) group.
If you want to return just the top 5 results for each group you could do this:
with cte as
(ID, Code, Sale,ROW_NUMBER() over(partition by ID,
Code order by (select 0)) rownum
from TableName)
Select ID, Code, sum(Sale) as Sale from cte
Where Code = 11
and rownum<=5
Group By ID, code
If you want to return top 5 results with highest salary for each group you could do this:
with cte as
(ID, Code, Sale,ROW_NUMBER() over(partition by ID,
Code order by Sale desc) rownum
from TableName)
Select ID, Code, sum(Sale) as Sale from cte
Where Code = 11
and rownum<=5
Group By ID, code
select id, code, sum(sale) as sale from tablename
where code = 11
group by id, code
order by sum(sale) desc
limit 5