Hi I need to count the number of customers with subcategory=E grouped by seller (createdby). Once a customer has been counted by a seller, no other seller should be able to count that customer, eventhough a observation might exist.
Example
id customerID CreatedBy createdate subcategory
1 1111111111 EVAJEN 2014-03-14 E
2 1111111111 MICMAD 2014-04-15 E
3 9999999999 MICMAD 2014-02-10 E`
Here MICMAD shouldn't get a count for id=2 since EVAJEN already made a sale to that customer. Right now my code looks like this, but I'm not able to check if a customer already has been counted.
sel createdby, cast(createdate as date) as date1, count(distinct customerID)
from MyDatabase
where subcategory='E'
group by 1,2`
Thank you
You can use ROW_NUMBER to get one row per customer:
select createdby, cast(createdate as date) as date1, count(*)
from
(
select *
from tab
where subcategory = 'E'
qualify row_number() -- 1st row per customer
over (partition by customerId
order by createddate) = 1
) t
group by 1,2;
Use a subquery to get the first date and count that. In most databases (including Teradata), you can use window functions to get the first row for each customer:
select createdby, cast(createdate as date) as date1, count(*)
from (select t.*,
row_number() over (partition by customerId order by createddate asc) as seqnum
from MyDatabase t
where subcategory = 'E'
) t
where seqnum = 1
group by createdby, cast(createdate as date) ;
Related
I have a SQL database with a list of Customer IDs CustomerID and invoices, the specific product purchased in each invoice ProductID, the Date and the Income of each invoice . I need to write a query that will retrieve for each product, which was the second customer who made a purchase
How do I do that?
EDIT:
I have come up with the following query:
SELECT *,
LEAD(CustomerID) OVER (ORDER BY ProductID, Date) AS 'Second Customer Who Made A Purchase'
FROM a
ORDER BY ProductID, Date ASC
However, this query presents multiple results for products that have more than two purchases. Can you advise?
SELECT a2.ProductID,
(
SELECT a1.CustomerID
FROM a a1
WHERE a1.ProductID = a2.ProductID
ORDER BY Date asc
LIMIT 1,1
) as SecondCustomer
FROM a a2
GROUP BY a2.ProductID
I need to write a query that will retrieve for each product, which was the second customer who made a purchase
This sounds like a window function:
select a.*
from (select a.*,
row_number() over (partition by productid order by date asc) as seqnum
from a
) a
where seqnum = 2;
I've a table that has this information:
And need to get the following information:
If the country of the same person name (in this case Artur) is different, then I need to sum the two values of quantity from the max date (in this case 04/10) and return both person (Artur) and the qty (15k)
If the country of the same person name (in this case Joseph) is the same, then I need only the first row of the max date available.
I'm really struguling as I'm not sure how to implement the logic into my code:
Select
table.person,
table.quantity
From
(
Select
table.date,
table.person,
table.country,
table.quantity,
ROW_NUMBER () over (
PARTITION by table.code, table.person
ORDER by table.date DESC
) AS rn
FROM
table
WHERE table.date >= DATE '{2020-04-10}' -5
) a
WHERE a.RN IN (1,2)
Is it possible to create a rule to sum rows 1 and 2 when country is different (Artur case) and only return row number 1 when the country is the same for a name (Joseph case)?
Use dense_rank() or max() as a window function:
select person, sum(quantity)
from (select t.*,
max(date) over (partition by person) as max_date
from t
) t
where date = max_date
group by person;
EDIT:
Hmmm . . . I think you might want one row per country per person on the max date. If so:
select person, sum(quantity)
from (select t.*,
row_number() over (partition by person, country order by date desc) as seqnum_pc,
rank() over (partition by person order by date desc) as seqnum_p
from t
) t
where seqnum_p = 1 and seqnum_pc = 1
group by person;
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
I have three sets of information:
productID
date
seller
I need to get information of last product sold for each productID and sold by. I tried using max value of date but that forces me to use grouping for seller as well but I don't want to group by seller. I want to group by productID, get the date it was sold last and by who. How can I avoid grouping on seller?
Use Window function which will help you to find the Latest date in each group(productId)
SELECT ProductID,
[date],
seller
FROM (SELECT Row_number()
OVER(
partition BY ProductID
ORDER BY [date] desc) Rn,
*
FROM tablename) a
WHERE rn = 1
or use can also use Max aggregate with group by to get the result
SELECT ProductID,
[date],
seller
FROM tablename a
JOIN (SELECT Max([date]) [date],
productid
FROM tablename
group by productid) b
ON a.productid = b.productid
AND a.[date] = b.[date]
How do I query for distinct customers? Here's the table I have..
CustID DATE PRODUCT
=======================
1 Aug-31 Orange
1 Aug-31 Orange
3 Aug-31 Apple
1 Sept-24 Apple
4 Sept-25 Orange
This is what I want.
# of New Customers DATE
========================================
2 Aug-31
1 Sept-25
Thanks!
This is a bit tricky. You want to count the first date a customer appears and then do the aggregation:
select mindate, count(*) as NumNew
from (select CustId, min(Date) as mindate
from table t
group by CustId
) c
group by mindate
You could use a simple common table expression to find the first time a user id is used;
WITH cte AS (
SELECT date, ROW_NUMBER() OVER (PARTITION BY custid ORDER BY date) rn
FROM customers
)
SELECT COUNT(*)[# of New Customers], date FROM cte
WHERE rn=1
GROUP BY date
ORDER BY date
An SQLfiddle to test with.