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
Related
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 ;
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;
Let's say I have a result from a query that looks like this:
ContactID LeadSalePrice
---------------------------
45 19.90
45 18.00
32 17.50
But, I want to eliminate duplicate ContactID's, always taking the higher price result. So what I want is:
ContactID LeadSalePrice
---------------------------
45 19.90
32 17.50
Here's (a simplified version of) the query:
SELECT
sc.ContactID
, c.LeadSalePrice
FROM
LeadSalesCampaignCriterias c
JOIN LeadSalesCampaigns sc ON c.LeadSalesCampaignID = sc.LeadSalesCampaignID
WHERE
...
ORDER BY
LeadSalePrice DESC
I've been playing around with DISTINCT and GROUP BY, but I'm not getting it.
Just use GROUP BY:
SELECT sc.ContactID, MAX(c.LeadSalePrice) as LeadSalePrice
FROM LeadSalesCampaignCriterias c JOIN
LeadSalesCampaigns sc
ON c.LeadSalesCampaignID = sc.LeadSalesCampaignID
WHERE ...
GROUP BY sc.ContactID;
Another option is the WITH TIES and Row_Number()
Select Top 1 with Ties *
From YourTable
Order By Row_Number() over (Partition By ContactID Order By LeadSalePrice Desc)
Returns
ContactID LeadSalePrice
32 17.50
45 19.90
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
I have the following table:
NAME | SCORE
ALICE | 100
BOB | 90
CHARLES| 90
DUKE | 80
EVE | 70
...
My question is the following:
How can I extract with one query the name of the three best players? In my example the query should return four rows (ALICE, BOB, CHARLES and DUKE) because there are two silver medalists (they both have 90 points).
Thank You in advance.
Oracle has the DENSE_RANK analytical function for that exact purpose:
select name, score from (
select name, score, dense_rank() over(order by score desc nulls last) rank
-- ^^^^^^^^^^
-- reject NULL score at the end
from t
) V
where rank < 4
order by rank, name
See http://sqlfiddle.com/#!4/88445/5
How about the following
select *
from table1
where score >=
(select score from (
select score, rownum r from (
select distinct score from table1 order by score desc
) where rownum <= 3
) where r = 3)
order by score desc
See also this SQLFiddle: http://sqlfiddle.com/#!4/23e68/1