How do you sum the results of a calculated column into one number in SQL?
SELECT
id, SUM(cost + r_cost) AS Revenue
FROM
revenue_table
WHERE
signup_date >= '2015-01-01'
GROUP BY
id
ORDER BY
Revenue DESC
LIMIT 20;
This query displays the revenue to date of the top 20 customers. How can I quickly do a total sum of the Revenue to get the total Revenue of the top 20 guys?
Assuming you're using MySQL:
-- Option 1: Simply put your query in the FROM clause and sum the result
select sum(Revenue)
from (select id, sum(cost + r_cost) as Revenue
from revenue_table
where signup_date >= '2015-01-01'
group by id
order by Revenue desc
limit 20) as a
-- Option 2: Use, as suggested by Siyual in his comment, ROLLUP.
-- You'll have to use a subquery too, because
-- LIMIT is applied after the ROLLUP
select id, sum(a.Revenue) as Revenue
from (select id, sum(cost + r_cost) as Revenue
from revenue_table
where signup_date >= '2015-01-01'
group by id
order by Revenue desc
limit 20) as a
GROUP BY id WITH ROLLUP
Related
I have the following query to return the number of users that booked a flight at least twice, but I need to identify those which have booked a flight more than once in the range of 12 months
SELECT COUNT(*)
FROM sales
WHERE customer in
(
SELECT customer
FROM sales
GROUP BY customer
HAVING COUNT(*) > 1
)
You would use window functions. The simplest method is lag():
select count(distinct customer)
from (select s.*,
lag(date) over (partition by customer order by date) as prev_date
from sales s
) s
where prev_date > s.date - interval '12 month';
At the cost of a self-join, #AdrianKlaver's answer can adapt to any 12-month period.
SELECT COUNT(DISTINCT customer) FROM
(SELECT customer
FROM sales s1
JOIN sales s2
ON s1.customer = s2.customer
AND s1.ticket_id <> s2.ticket_id
AND s2.date_field BETWEEN s1.date_field AND (s1.date_field + interval'1 year')
GROUP BY customer
HAVING COUNT(*) > 1) AS subquery;
A stab at it with a made up date field:
SELECT COUNT(*)
FROM sales
WHERE customer in
(
SELECT customer
FROM sales
WHERE date_field BETWEEN '01/01/2019' AND '12/31/2019'
GROUP BY customer
HAVING COUNT(*) > 1
)
I have a table with columns month, name and transaction_id. I would like to count the number of transactions per month and name. However, for each month I want to have the top N names with the highest transaction counts.
The following query groups by month and name. However the LIMIT is applied to the complete result and not per month:
SELECT
month,
name,
COUNT(*) AS transaction_count
FROM my_table
GROUP BY month, name
ORDER BY month, transaction_count DESC
LIMIT N
Does anyone have an idea how I can get the top N results per month?
Use row_number():
SELECT month, name, transaction_count
FROM (SELECT month, name, COUNT(*) AS transaction_count,
ROW_NUMBER() OVER (PARTITION BY month ORDER BY COUNT(*) DESC) as seqnum
FROM my_table
GROUP BY month, name
) mn
WHERE seqnum <= N
ORDER BY month, transaction_count DESC
select max(total), date
from (select sum(total) as total, date as date
from canteen
group by date) as max
i want to select the highest sale and the date of the highest sale from table.
With my query this error is shown.
Exception, Error code 8,120, SQLState S0001] Column 'max.date' is
invalid in the select list because it is not contained in either an
aggregate function or the GROUP BY clause.
You can use order by, set your data to order by sale descending and fetch first row.
If you wants result by date then you can use ROW_NUMBER()
select TOP(1) total, date
from
(
select sum(total) as total, date as date
from canteen
group by date
) as max
Order by todal desc
This will return all dates achieving the max sales total. Here is a demo.
; with x as (
select sum(total) as total, date from canteen group by date
)
, y as (
select dr = dense_rank() over(order by total desc), *
from x
)
select * from y where dr = 1
If you want to get all dates with max(total), here it is
;with temp as
(
select sum(total) as total, date as date
from canteen
group by date
)
SELECT TOP 1 WITH TIES *
FROM temp
ORDER BY temp.total desc
You have forgotten to add group by & order by in your outer query. I have modified it to display all the sales in descending order. So the highest sale will be at top.
select max(total) total, date
from (select sum(total) as total, date as date
from canteen
group by date) as max
group by date
order by total desc
I have a table table with columns date and total.
How can I aggregate the sum of the total column by week for the last 20 weeks to sysdate?
select trunc(date,'D'), sum(total)
from table
where date >= trunc(sysdate - 20*7, 'D')
group by trunc(date,'D')
order by 1
SELECT *, CONCAT(YEAR(`sysdate`), '/', WEEK(`sysdate`)) AS `year_week`
FROM `table`
GROUP BY `year_week`
ORDER BY `year_week` DESC
LIMIT 20;
I'm having an odd problem
I have a table with the columns product_id, sales and day
Not all products have sales every day. I'd like to get the average number of sales that each product had in the last 10 days where it had sales
Usually I'd get the average like this
SELECT product_id, AVG(sales)
FROM table
GROUP BY product_id
Is there a way to limit the amount of rows to be taken into consideration for each product?
I'm afraid it's not possible but I wanted to check if someone has an idea
Update to clarify:
Product may be sold on days 1,3,5,10,15,17,20.
Since I don't want to get an the average of all days but only the average of the days where the product did actually get sold doing something like
SELECT product_id, AVG(sales)
FROM table
WHERE day > '01/01/2009'
GROUP BY product_id
won't work
If you want the last 10 calendar day since products had a sale:
SELECT product_id, AVG(sales)
FROM table t
JOIN (
SELECT product_id, MAX(sales_date) as max_sales_date
FROM table
GROUP BY product_id
) t_max ON t.product_id = t_max.product_id
AND DATEDIFF(day, t.sales_date, t_max.max_sales_date) < 10
GROUP BY product_id;
The date difference is SQL server specific, you'd have to replace it with your server syntax for date difference functions.
To get the last 10 days when the product had any sale:
SELECT product_id, AVG(sales)
FROM (
SELECT product_id, sales, DENSE_RANK() OVER
(PARTITION BY product_id ORDER BY sales_date DESC) AS rn
FROM Table
) As t_rn
WHERE rn <= 10
GROUP BY product_id;
This asumes sales_date is a date, not a datetime. You'd have to extract the date part if the field is datetime.
And finaly a windowing function free version:
SELECT product_id, AVG(sales)
FROM Table t
WHERE sales_date IN (
SELECT TOP(10) sales_date
FROM Table s
WHERE t.product_id = s.product_id
ORDER BY sales_date DESC)
GROUP BY product_id;
Again, sales_date is asumed to be date, not datetime. Use other limiting syntax if TOP is not suported by your server.
Give this a whirl. The sub-query selects the last ten days of a product where there was a sale, the outer query does the aggregation.
SELECT t1.product_id, SUM(t1.sales) / COUNT(t1.*)
FROM table t1
INNER JOIN (
SELECT TOP 10 day, Product_ID
FROM table t2
WHERE (t2.product_ID=t1.Product_ID)
ORDER BY DAY DESC
)
ON (t2.day=t1.day)
GROUP BY t1.product_id
BTW: This approach uses a correlated subquery, which may not be very performant, but it should work in theory.
I'm not sure if I get it right but If you'd like to get the average of sales for last 10 days for you products you can do as follows :
SELECT Product_Id,Sum(Sales)/Count(*) FROM (SELECT ProductId,Sales FROM Table WHERE SaleDAte>=#Date) table GROUP BY Product_id HAVING Count(*)>0
OR You can use AVG Aggregate function which is easier :
SELECT Product_Id,AVG(Sales) FROM (SELECT ProductId,Sales FROM Table WHERE SaleDAte>=#Date) table GROUP BY Product_id
Updated
Now I got what you meant ,As far as I know it is not possible to do this in one query.It could be possible if we could do something like this(Northwind database):
select a.CustomerId,count(a.OrderId)
from Orders a INNER JOIN(SELECT CustomerId,OrderDate FROM Orders Order By OrderDate) AS b ON a.CustomerId=b.CustomerId GROUP BY a.CustomerId Having count(a.OrderId)<10
but you can't use order by in subqueries unless you use TOP which is not suitable for this case.But maybe you can do it as follows:
SELECT PorductId,Sales INTO #temp FROM table Order By Day
select a.ProductId,Sum(a.Sales) /Count(a.Sales)
from table a INNER JOIN #temp AS b ON a.ProductId=b.ProductId GROUP BY a.ProductId Having count(a.Sales)<=10
If this is a table of sales transactions, then there should not be any rows in there for days on which there were no Sales. I.e., If ProductId 21 had no sales on 1 June, then this table should not have any rows with productId = 21 and day = '1 June'... Therefore you should not have to filter anything out - there should not be anything to filter out
Select ProductId, Avg(Sales) AvgSales
From Table
Group By ProductId
should work fine. So if it's not, then you have not explained the problem completely or accurately.
Also, in yr question, you show Avg(Sales) in the example SQL query but then in the text you mention "average number of sales that each product ... " Do you want the average sales amount, or the average count of sales transactions? And do you want this average by Product alone (i.e., one output value reported for each product) or do you want the average per product per day ?
If you want the average per product alone, for just thpse sales in the ten days prior to now? or the ten days prior to the date of the last sale for each product?
If the latter then
Select ProductId, Avg(Sales) AvgSales
From Table T
Where day > (Select Max(Day) - 10
From Table
Where ProductId = T.ProductID)
Group By ProductId
If you want the average per product alone, for just those sales in the ten days with sales prior to the date of the last sale for each product, then
Select ProductId, Avg(Sales) AvgSales
From Table T
Where (Select Count(Distinct day) From Table
Where ProductId = T.ProductID
And Day > T.Day) <= 10
Group By ProductId