I want to select items and sum their sales and price like:
Produkt Sales Price
Car1 1 100
Car1 2 120
Car1 3 110
Car2 1 200
Car2 2 210
Result should be
Produkt Sales Price
Car1 6 330
Car2 3 410
I know the sum function, but I cant get out how I select the simular Cars to group them to sum the numbers.
You just need to GROUP your results on column produkt and use sum() aggregate function to calculate sales and prices for each group.
This is ANSI SQL, so it works not only in Firebird.
SELECT produkt, SUM(sales) as sales, SUM(price) as price
FROM yourtable
GROUP BY produkt
ORDER BY produkt
SELECT Produkt, SUM(Sales) AS Total_sales, SUM(Price) as Total_price
FROM table_name --Your table name here
GROUP BY Produkt
ORDER BY Produkt
Try this SQL
Related
product
quantity
price
milk
3
10
bread
7
3
bread
5
2
And my output table should be
product
total_price
milk
30
bread
31
I can't seem to get my code to work. Here is my code
SELECT product, (SELECT (quantity*unit_price)
FROM shopping_history AS sh ) AS total_price
FROM shopping_history
GROUP BY product
You are looking for the aggregate function SUM (which doesn't require a sub-query) e.g.
SELECT product, SUM(quantity*unit_price) AS Total_Price
FROM shopping_history
GROUP BY product
I have the following transactions table:
customer_id purchase_date product category department quantity store_id
1 2020-10-01 Kit Kat Candy Food 2 store_A
1 2020-10-01 Snickers Candy Food 1 store_A
1 2020-10-01 Snickers Candy Food 1 store_A
2 2020-10-01 Snickers Candy Food 2 store_A
2 2020-10-01 Baguette Bread Food 5 store_A
2 2020-10-01 iPhone Cell phones Electronics 2 store_A
3 2020-10-01 Sony PS5 Games Electronics 1 store_A
I would like to calculate the average number of products purchased (for each product in the table). I'm also looking to calculate averages across each category and each department by accounting for all products within the same category or department respectively. Care should be taken to divide over unique customers AND the product quantity being greater than 0 (a 0 quantity indicates a refund, and should not be accounted for).
So basically, the output table would like below:
...where store_id and average_level_type are partition columns.
Is there a way to achieve this in a single pass over the transactions table? or do I need to break down my approach into multiple steps?
Thanks!
How about using “union all” as below -
Select store_id, 'product' as average_level_type,product as id, sum(quantity) as total_quantity,
Count(distinct customer_id) as unique_customer_count, sum(quantity)/count(distinct customer_id) as average
from transactions
where quantity > 0
group by store_id,product
Union all
Select store_id, 'category' as average_level_type, category as id, sum(quantity) as total_quantity,
Count(distinct customer_id) as unique_customer_count, sum(quantity)/count(distinct customer_id) as average
from transactions
where quantity > 0
group by store_id,category
Union all
Select store_id, 'department' as average_level_type,department as id, sum(quantity) as total_quantity,
Count(distinct customer_id) as unique_customer_count, sum(quantity)/count(distinct customer_id) as average
from transactions
where quantity > 0
group by store_id,department;
If you want to avoid using union all in that case you can use something like rollup() or group by grouping sets() to achieve the same but the query would be a little more complicated to get the output in the exact format which you have shown in the question.
EDIT : Below is how you can use grouping sets to get the same output -
Select store_id,
case when G_ID = 3 then 'product'
when G_ID = 5 then 'category'
when G_ID = 6 then 'department' end As average_level_type,
case when G_ID = 3 then product
when G_ID = 5 then category
when G_ID = 6 then department end As id,
total_quantity,
unique_customer_count,
average
from
(select store_id, product, category, department, sum(quantity) as total_quantity, Count(distinct customer_id) as unique_customer_count, sum(quantity)/count(distinct customer_id) as average, GROUPING__ID As G_ID
from transactions
group by store_id,product,category,department
grouping sets((store_id,product),(store_id,category),(store_id,department))
) Tab
order by 2
;
have the following table and data
partno price qty
A0001 10 2
A0001 8 6
A0001 15 10
How can I issue a query to get the following result
partno. price. qty.
A0001 15 250
unique partno, highest price in the list and sum(qty )* highest price.
A simple aggregation shows the result you want:
select
partno,
max(price) as max_price,
max(price) * sum(qty) as total
from t
group by partno
/*Just replace "Table_Name" with your first table's name in the database */
select
distinct
partno,
max(price) price,
max(price) * sum(qty) qty
from
Table_Name
group by
partno
suppose i have a product is and sales column
product id sales
1 1000
2 10000
3 50000
4 12000
5 8000
write an sql query to get all product ids that contribute to top 80 % of sales?
For this, you want a cumulative sum. Presumably, you want the top selling such products, so:
select p.*
from (select p.*,
sum(sales) over (order by sales desc) as running_sales,
sum(sales) over () as total_sales,
from products
) p
where running_sales - sales < 0.8 * total_sales;
This returns the product that reaches or first exceeds 80% of the total sales.
I have one table with following data..
saleId amount date
-------------------------
1 2000 10/10/2012
2 3000 12/10/2012
3 2000 11/12/2012
2 3000 12/10/2012
1 4000 11/10/2012
4 6000 10/10/2012
From my table I want result with max of sum amount between dates 10/10/2012 and 12/10/2012 which for the data above will be:
saleId amount
---------------
1 6000
2 6000
4 6000
Here 6000 is the max of the sums (by saleId) so I want ids 1, 2 and 4.
You have to use Sub-queries like this:
SELECT saleId , SUM(amount) AS Amount
FROM Table1
GROUP BY saleId
HAVING SUM(amount) =
(
SELECT MAX(AMOUNT) FROM
(
SELECT SUM(amount) AS AMOUNT FROM Table1
WHERE date BETWEEN '10/10/2012' AND '12/10/2012'
GROUP BY saleId
) AS A
)
See this SQLFiddle
This query goes through the table only once and is fairly optimised.
select top(1) with ties saleid, amount
from (
select saleid, sum(amount) amount
from tbl
where date between '20121010' and '20121210'
group by saleid
) x
order by amount desc;
You can produce the SUM with the WHERE clause as a derived table, then SELECT TOP(1) in the query using WITH TIES to show all the ones with the same (MAX) amount.
When presenting dates to SQL Server, try to always use the format YYYYMMDD for robustness.