SQL for highest sold Products - sql

Question

Order Sample data:
ORDER_DAY ORDER_ID PRODUCT_ID QUANTITY PRICE
---------- --------- ----------- ---------- ---------
01-JUL-11 O1 P1 5 5
01-JUL-11 O2 P2 2 10
01-JUL-11 O3 P3 10 25
01-JUL-11 O4 P1 20 5
02-JUL-11 O5 P3 5 25
02-JUL-11 O6 P4 6 20
02-JUL-11 O7 P1 2 5
02-JUL-11 O8 P5 1 50
02-JUL-11 O9 P6 2 50
02-JUL-11 O10 P2 4 10
Q: Get me highest sold Products (Qty* Price) on both days
Desired output :
DATE PRODUCT_ID SOLD_AMOUNT
01-JUL-11 P3 250
02-JUL-11 P3 125

Answer 1

Try following query:
select order_day, product_id, totalsale
from (select order_day,
product_id,
nvl(QUANTITY, 0) * PRICE as totalsale,
dense_rank() over(partition by ORDER_DAY order by(nvl(QUANTITY, 0) * PRICE) desc) as maxsum
from orders )
where maxsum = 1;

Answer 2

You would start with getting the sold amount per day and product. With this data you first select the maximum profit per day and then select those entries that match this:
WITH PRODUCT_PER_DAY AS
(
SELECT ORDER_DAY, PRODUCT_ID, SUM(QUANTITY * PRICE) AS SOLD_AMOUNT
FROM MYTABLE
GROUP BY ORDER_DAY, PRODUCT_ID
)
SELECT ORDER_DAY, PRODUCT_ID, SOLD_AMOUNT
FROM PRODUCT_PER_DAY
WHERE (ORDER_DAY, SOLD_AMOUNT) IN
(
SELECT ORDER_DAY, MAX(SOLD_AMOUNT)
FROM PRODUCT_PER_DAY
GROUP BY ORDER_DAY
)
ORDER BY ORDER_DAY, PRODUCT_ID;

Answer 3

select b.product_id,a.order_day,a.total_price from(select order_day, max(quantity*price)as total_price from order group by order_day)a
join (select product_id , quantity * price as total_amount from order)b on
a.total_price= b.total_price

Answer 4

Try this out (modify the table and column names as per your needs):
Select a.order_day, b.product_id, a.sales
from
(select order_day, max(quantity*price) as sales
from ordr
group by order_day) a
inner join
(select order_day, product_id, quantity*price as sales
from ordr) b
on a.order_day = b.order_day and a.sales = b.sales;

Answer 5

with temp as
(select order_day,product_id,sum(quantity*price) qp
from orders_amazn
group by order_day,product_id
)
select * from (select order_day,product_id,qp,
dense_rank() over (partition by order_day order by qp desc) rnk
from temp)
where rnk=1;

Answer 6

select date,product_id,max(product_id*price) as sold_amount form order
group by order_day
order by order_day

Answer 7

select t.ORDER_DAY as date1 ,t.PRODUCT_ID,max(t.quantity * t.price)
as sold_amount from table t group by t.ORDER_DAY,t.PRODUCT_ID

Answer 8

Kindly use this query you will be the desired output.I have tried in MySQL.
SELECT
order_date,
Product_id,
sold_amount
FROM
(
SELECT
date_format(order_day,'%d-%b-%y') AS order_date,
Product_id,
SUM(price*quantity) AS sold_amount
FROM
ecommerce
GROUP BY
date_format(order_day,'%d-%b-%y'),
product_id) a
WHERE
(
order_date,sold_amount) IN
(
SELECT
DATE,
MAX(sold_amt) AS sold_amount
FROM
(
SELECT
date_format(order_day,'%d-%b-%y') AS DATE,
product_id,
SUM(price*quantity) AS sold_amt
FROM
ecommerce
GROUP BY
date_format(order_day,'%d-%b-%y'),
product_id)a
GROUP BY
DATE)

Answer 9

select order_day as orderdate,product_id,max(quantity * price) as sold_amount
from order
group by order_day
order by product_id