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 ;
Related
I have a list of orders, I need to find which ones occur with code 47 more than once with different users. For example:
ORDER_ID CODE USER
111 47 1
111 47 2
222 47 1
333 47 1
333 47 2
444 47 1
The expected result is 111 and 333.
How can I accomplish this?
Regards
I think you want aggregation and having:
select order_id
from orders o
where code = 47
group by order_id
having min(user) <> max(user);
You can also express the having as:
having count(distinct user) >= 2
You can try below -
select order_id from tablename
group by order_id
having count(distinct user)>=1
You can do it via row_number() as well
Select distinct order_id from
(select order_id, code, row_number()
over
( Partition by order_id, code
Order by order_id, code) rn
from
tablename
where user in (1,2)
) where rn>=1
But I guess you already have a user column hence i dont think you require extra manipulation
Select orderid, code from table
Group by orderid, code having
max(distinct user) >=1
What I need to select is total number of trips made by every 'id_customer' from table user and their id, dispatch_seconds, and distance for first order. id_customer, customer_id, and order_id are strings.
It should looks like this
+------+--------+------------+--------------------------+------------------+
| id | count | #1order id | #1order dispatch seconds | #1order distance |
+------+--------+------------+--------------------------+------------------+
| 1ar5 | 3 | 4r56 | 1 | 500 |
| 2et7 | 2 | dc1f | 5 | 100 |
+------+--------+------------+--------------------------+------------------+
Cheers!
Original post was edited as during discussion S-man helped me to find exact problem solution. Solution by S-man https://dbfiddle.uk/?rdbms=postgres_10&fiddle=e16aa6008990107e55a26d05b10b02b5
db<>fiddle
SELECT
customer_id,
order_id,
order_timestamp,
dispatch_seconds,
distance
FROM (
SELECT
*,
count(*) over (partition by customer_id), -- A
first_value(order_id) over (partition by customer_id order by order_timestamp) -- B
FROM orders
)s
WHERE order_id = first_value -- C
https://www.postgresql.org/docs/current/static/tutorial-window.html
A window function which gets the total record count per user
B window function which orders all records per user by timestamp and gives the first order_id of the corresponding user. Using first_value instead of min has one benefit: Maybe it could be possible that your order IDs are not really increasing by timestamp (maybe two orders come in simultaneously or your order IDs are not sequential increasing but some sort of hash)
--> both are new columns
C now get all columns where the "first_value" (aka the first order_id by timestamp) equals the order_id of the current row. This gives all rows with the first order by user.
Result:
customer_id count order_id order_timestamp dispatch_seconds distance
----------- ----- -------- ------------------- ---------------- --------
1ar5 3 4r56 2018-08-16 17:24:00 1 500
2et7 2 dc1f 2018-08-15 01:24:00 5 100
Note that in these test data the order "dc1f" of user "2et7" has a smaller timestamp but comes later in the rows. It is not the first occurrence of the user in the table but nevertheless the one with the earliest order. This should demonstrate the case first_value vs. min as described above.
You are on the right track. Just use conditional aggregation:
SELECT o.customer_id, COUNT(*)
MAX(CASE WHEN seqnum = 1 THEN o.order_id END) as first_order_id
FROM (SELECT o.*,
ROW_NUMBER() OVER (PARTITION BY customer_id ORDER BY order_timestamp ASC) as seqnum
FROM orders o
) o
GROUP BY o.customer_id;
Your JOIN is not necessary for this query.
You can use window function :
select distinct customer_id,
count(*) over (partition by customer_id) as no_of_order
min(order_id) over (partition by customer_id order by order_timestamp) as first_order_id
from orders o;
I think there are many mistakes in your original query, your rank isn't partitioned, the order by clause seems incorrect, you filter out all but one "random" order, then apply the count, the list goes on.
Something like this seems closer to what you seem to want?
SELECT
customer_id,
order_count,
order_id
FROM (
SELECT
a.customer_id,
a.order_count,
a.order_id,
RANK() OVER (PARTITION BY a.order_id, a.customer_id ORDER BY a.order_count DESC) AS rank_id
FROM (
SELECT
customer_id,
order_id,
COUNT(*) AS order_count
FROM
orders
GROUP BY
customer_id,
order_id) a) b
WHERE
b.rank_id = 1;
I am trying to develop a query to pull out the top 2 months of sales by customer id. Here is a sample table:
Customer_ID Sales Amount Period
144567 40 2
234567 50 5
234567 40 7
144567 80 10
144567 48 2
234567 23 7
desired output would be
Customer_ID Sales Sum Period
144567 80 10
144567 48 2
234567 50 5
234567 40 7
I've tried
select sum(net_sales_usd_spot), valid_period, customer_id
from sales_trans_price_output
where valid_period in (select valid_period, sum(net_sales_usd_spot)
from sales_trans_price_output
where rank<=2)
group by valid_period, customer_id
error is
too many values ORA-00913.
I see why, but not sure how to rework it.
Try:
SELECT *
FROM (
SELECT t.*,
row_number() over (partition by customer_id order by sales_amount desc ) rn
FROM sales_trans_price t
)
WHERE rn <= 2
ORDER BY 1,2 desc
Demo: http://sqlfiddle.com/#!4/882888/3
what if you change your where clause to:
where valid_period in
(
select p.valid_period from sales_trans_price_output p
join (select valid_period, sum(net_sales_usd_spot)
from sales_trans_price_output
where rank<=2) s on s.valid_period = p.valid_period
)
It might be ugly and need refactoring, but I think this is the logic you're after.
The error is because of this.
where valid_period in (select valid_period, sum(net_sales_usd_spot)
from sales_trans_price_output
where rank<=2)
The subquery can only contain one field.
You are on the right track using rank, but you might not be using it correctly. Google oracle rank to find the correct syntax.
Back to what you are looking to achieve, a derived table is the approach I would use. That's simply a subquery with an alias. Or, if you use the keyword with, it might be called a CTE - Computed Table Expression.
Try it
SELECT * FROM (
SELECT T.*,
RANK () OVER (PARTITION BY CUSTOMER_ID
ORDER BY VALID_PERIOD DESC) FN_RANK
FROM SALES_TRANS_PRICE_OUTPUT T
) A
WHERE A.FN_RANK <= 2
ORDER BY CUSTOMER_ID ASC, VALID_PERIOD DESC, FN_RANK DESC
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
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