I am using SQL Server 2008 and I was wondering how to remove duplicate customers either from the table or exclude it in my query. An Account_ID can only have 1 product associated with it. And the account with the most recent purchase date is what should be showing. An example is below:
Account_ID, Account_Purchase, Purchase_Date
1 Product 1 1/1/2016
2 Product 1 1/2/2016
3 Product 2 1/5/2016
1 Product 3 3/12/2016
4 Product 3 1/5/2016
Ideally I would only see:
Account_ID, Account_Purchase, Purchase_Date
2 Product 1 1/2/2016
3 Product 2 1/5/2016
1 Product 3 3/12/2016
4 Product 3 1/5/2016
This should not show up because it is not the most recent purchase from account 1
Account_ID, Account_Purchase, Purchase_Date
1 Product 1 1/1/2016
Thank you all for help, folks!
Simply acquire the latest purchase_date using max and group by account_id. Then use inner join to get the other details from the acquired details.
SELECT TABLE_NAME.* FROM TABLE_NAME
INNER JOIN(
SELECT Account_ID, MAX(Purchase_Date) AS Purchase_Date
GROUP BY Account_ID
) LatestPurchases
ON TABLE_NAME.Account_ID = LatestPurchases.Account_ID
AND TABLE_NAME.Purchase_Date = LatestPurchases.Purchase_Date
Try below query, please replace TABLENAME with your table
WITH CTE
AS (
SELECT
*,
ROW_NUMBER() OVER (PARTITION BY Account_ID ORDER BY Purchase_Date DESC) AS RN
FROM TABLENAME
)
SELECT
*
FROM CTE
WHERE RN = 1
Here is another query
SELECT
t.Account_id,
t.Account_Purchase,
t.Purchase_Date
FROM
tablename t
WHERE
t.Purchase_Date = (SELECT MAX(Purchase_date) FROM Tablename WHERE Account_ID = t.Account_ID)
ORDER BY
t.Purchase_Date DESC
Related
I want to write a query to locate a group of clients whose purchased specific 2 product categories, at the same time, getting the information of first transaction date and first item they purchased. Since I used group by function, I could only get customer id but not first item purchase due to the nature of group by. Any thoughts to solve this problem?
What I have are transaction tables(t), customer_id tables(c) and product tables(p). Mine is SQL server 2008.
Update
SELECT t.customer_id
,t.product_category
,MIN(t.transaction_date) AS FIRST_TRANSACTION_DATE
,SUM(t.quantity) AS TOTAL_QTY
,SUM(t.sales) AS TOTAL_SALES
FROM transaction t
WHERE t.product_category IN ('VEGETABLES', 'FRUITS')
AND t.transaction_date BETWEEN '2020/01/01' AND '2022/09/30'
GROUP BY t.customer_id
HAVING COUNT(DISTINCT t.product_category) = 2
**Customer_id** **transaction_date** **product_category** **quantity** **sales**
1 2022-05-30 VEGETABLES 1 100
1 2022-08-30 VEGETABLES 1 100
2 2022-07-30 VEGETABLES 1 100
2 2022-07-30 FRUITS 1 50
2 2022-07-30 VEGETABLES 2 200
3 2022-07-30 VEGETABLES 3 300
3 2022-08-01 FRUITS 1 50
3 2022-08-05 FRUITS 1 50
4 2022-08-07 FRUITS 1 50
4 2022-09-05 FRUITS 2 100
In the above, what I want to show after executing the SQL query is
**Customer_id** **FIRST_TRANSACTION_DATE** **first_product_category** **TOTAL_QUANTITY** **TOTAL_SALES**
2 2022-07-30 VEGETABLES, FRUITS 4 350
3 2022-07-30 VEGETABLES 5 400
Customer_id 1 and 4 will not be shown as they only purchased either vegetables or fruits but not both
Check now, BTW need find logic with product_category
select CustomerId, transaction_date, product_category, quantity, sales
from(
select CustomerId, transaction_date, product_category , sum(quantity) over(partition by CustomerId ) as quantity , sum(sales) over(partition by CustomerId ) as sales, row_number() over(partition by CustomerId order by transaction_date ASC) rn
from(
select CustomerId, transaction_date, product_category, quantity, sales
from tablee t
where (product_category = 'FRUITS' and
EXISTS (select CustomerId
from tablee tt
where product_category = 'VEGETABLES'
and t.CustomerId = tt.CustomerId)) OR
(product_category = 'VEGETABLES' and
EXISTS (select CustomerId
from tablee tt
where product_category = 'FRUITS'
and t.CustomerId = tt.CustomerId)))x)over_all
where rn = 1;
HERE is FIDDLE
I'm working on a customers database and I want to get all data for their second purchase (for all of our customer weather they have 2 or more purchases).
For example:
Customer_ID Order_ID Order_Date
1 259 09/05/2020
1 644 03/11/2020
1 617 18/04/2022
4 834 22/09/2021
4 995 07/02/2022
I want to display the second order which is:
Customer_ID Order_ID Order_Date
1 644 03/11/2020
4 995 07/02/2022
I'm facing some difficulties in finding the right logic, any idea how I can achieve my end goal? :)
*Note: I'm using snowflake
You can use a ROW_NUMBER and filter using QUALIFY clause:
select * from table qualify row_number() over(partition by customer_id order by order_date) = 2;
You can use common table expression
with CTE_RS
AS (
SELECT Customer_ID,ORDER_ID,Order_Date,ROW_NUMBER() OVER(PARTITION BY Customer_ID ORDER BY Order_Date ) ORDRNUM FROM *TABLE NAME*
)
SELECT Customer_ID,ORDER_ID,Order_Date
FROM CTE_RS
WHERE ORDRNUM = 2 ;
I have a table with customers and their purchase behaviour that looks as follows:
customer shop time
----------------------------
1 5 13.30
1 5 14.33
1 10 22.17
2 3 12.15
2 1 13.30
2 1 15.55
2 3 17.29
Since I want the shift in shop I need the following output
customer shop time
----------------------------
1 5 13.30
1 10 22.17
2 3 12.15
2 1 13.30
2 3 17.29
I have tried using
ROW_NUMBER() OVER (PARTITION BY customer, shop ORDER BY time ASC) AS a counter
and then only keeping all counter=1. However, this troubles me when the customer visits the same shop again later on, as with customer=2 and shop=3 in my example.
I came up with this:
WITH a AS
(
SELECT
customer, shop, time,
ROW_NUMBER() OVER (PARTITION BY customer ORDER BY time ASC) AS counter
FROM
db
)
SELECT a1.*
FROM a a1
JOIN a AS a2 ON (a1.device = a2.device AND a2.counter1 + 1 = a1.counter1 AND a2.id <> a1.id)
UNION
SELECT a.*
FROM a
WHERE counter1 = 1
However, this is very inefficient and running it in AWS where my data is located results in a error telling me that
Query exhausted resources at this scale factor
Is there any way to make this query more efficient?
This is a gaps-and-islands problem. But the simplest solution uses lag():
select customer, shop, time
from (select t.*, lag(shop) over (partition by customer order by time) as prev_shop
from t
) t
where prev_shop is null or prev_shop <> shop;
In a firebird database with a table "Sales", I need to select the first sale of all customers. See below a sample that show the table and desired result of query.
---------------------------------------
SALES
---------------------------------------
ID CUSTOMERID DTHRSALE
1 25 01/04/16 09:32
2 30 02/04/16 11:22
3 25 05/04/16 08:10
4 31 07/03/16 10:22
5 22 01/02/16 12:30
6 22 10/01/16 08:45
Result: only first sale, based on sale date.
ID CUSTOMERID DTHRSALE
1 25 01/04/16 09:32
2 30 02/04/16 11:22
4 31 07/03/16 10:22
6 22 10/01/16 08:45
I've already tested following code "Select first row in each GROUP BY group?", but it did not work.
In Firebird 2.5 you can do this with the following query; this is a minor modification of the second part of the accepted answer of the question you linked to tailored to your schema and requirements:
select x.id,
x.customerid,
x.dthrsale
from sales x
join (select customerid,
min(dthrsale) as first_sale
from sales
group by customerid) p on p.customerid = x.customerid
and p.first_sale = x.dthrsale
order by x.id
The order by is not necessary, I just added it to make it give the order as shown in your question.
With Firebird 3 you can use the window function ROW_NUMBER which is also described in the linked answer. The linked answer incorrectly said the first solution would work on Firebird 2.1 and higher. I have now edited it.
Search for the sales with no earlier sales:
SELECT S1.*
FROM SALES S1
LEFT JOIN SALES S2 ON S2.CUSTOMERID = S1.CUSTOMERID AND S2.DTHRSALE < S1.DTHRSALE
WHERE S2.ID IS NULL
Define an index over (customerid, dthrsale) to make it fast.
in Firebird 3 , get first row foreach customer by min sales_date :
SELECT id, customer_id, total, sales_date
FROM (
SELECT id, customer_id, total, sales_date
, row_number() OVER(PARTITION BY customer_id ORDER BY sales_date ASC ) AS rn
FROM SALES
) sub
WHERE rn = 1;
İf you want to get other related columns, This is where your self-answer fails.
select customer_id , min(sales_date)
, id, total --what about other colums
from SALES
group by customer_id
So simple as:
select CUSTOMERID min(DTHRSALE) from SALES group by CUSTOMERID
There is a table tbl_products that contains data as shown below:
Id Name
----------
1 P1
2 P2
3 P3
4 P4
5 P5
6 P6
And another table tbl_inputs that contains data as shown below:
Id Product_Id Price Register_Date
----------------------------------------
1 1 10 2010-01-01
2 1 20 2010-10-11
3 1 30 2011-01-01
4 2 100 2010-01-01
5 2 200 2009-01-01
6 3 500 2011-01-01
7 3 270 2010-10-15
8 4 80 2010-01-01
9 4 50 2010-02-02
10 4 92 2011-01-01
I want to select all products(id, name, price, register_date) with maximum date in each group.
For Example:
Id Name Price Register_Date
----------------------------------------
3 P1 30 2011-01-01
4 P2 100 2010-01-01
6 P3 500 2011-01-01
10 P4 92 2011-01-01
select
id
,name
,code
,price
from tbl_products tp
cross apply (
select top 1 price
from tbl_inputs ti
where ti.product_id = tp.id
order by register_date desc
) tii
Although is not the optimum way you can do it like:
;with gb as (
select
distinct
product_id
,max(register_date) As max_register_date
from tbl_inputs
group by product_id
)
select
id
,product_id
,price
,register_date
from tbl_inputs ti
join gb
on ti.product_id=gb.product_id
and ti.register_date = gb.max_register_date
But as I said earlier .. this is not the way to go in this case.
;with cte as
(
select t1.id, t1.name, t1.code, t2.price, t2.register_date,
row_number() over (partition by product_id order by register_date desc) rn
from tbl_products t1
join tbl_inputs t2
on t1.id = t2.product_id
)
select id, name, code, price, register_date
from cte
where rn = 1
Something like this..
select id, product_id, price, max(register_date)
from tbl_inputs
group by id, product_id, price
you can use the max function and the group by clause. if you only need results from the table tbl_inputs you even don't need a join
select product_id, max(register_date), price
from tbl_inputs
group by product_id, price
if you need field from the tbl_prducts you have to use a join.
select p.name, p. code, i.id, i.price, max(i.register_date)
from tbl_products p join tbl_inputs i on p.id=i.product_id
grooup by p.name, p. code, i.id, i.price
Try this:
SELECT id, product_id, price, register_date
FROM tbl_inputs T1 INNER JOIN
(
SELECT product_id, MAX(register_date) As Max_register_date
FROM tbl_inputs
GROUP BY product_id
) T2 ON(T1.product_id= T2.product_id AND T1.register_date= T2.Max_register_date)
This is, of course, assuming your dates are unique. if they are not, you need to add the DISTINCT Keyword to the outer SELECT statement.
edit
Sorry, I didn't explain it very well. Your dates can be duplicated, it's not a problem as long as they are unique per product id. if you can have duplicated dates per product id, then you will have more then one row per product in the outcome of the select statement I suggested, and you will have to find a way to reduce it to one row per product.
i.e:
If you have records like that (when the last date for a product appears more then once in your table with different prices)
id | product_Id | price | register_date
--------------------------------------------
1 | 1 | 10.00 | 01/01/2000
2 | 1 | 20.00 | 01/01/2000
it will result in having both of these records as outcome.
However, if the register_date is unique per product id, then you will get only one result for each product id.