I am hoping I can get some clarification on how to best handle getting the data set correct and efficiently.
Here are three queries from three different tables I am working with. The Donor_ID is key between the tables, but as you can see - there are multiple records associated with each Donor_ID - with the runid_gmt column having differing dates.
Ideally, I would like use the max(runid_gmt) for each record - and join the EMAIL and ADDRESSES tables on the Donor_ID but only select the max(runid_gmt) record in each of those tables as well.
I believe that is what I need to do - but not sure. Any suggestions on how to tackle this problem?
SELECT donor_id, last_name, birthdate, runid_gmt
FROM [dbo].TBL_DONORS where donor_id = '51999441' order by runid_gmt desc;
SELECT donor_id, city, state, zip, runid_gmt
FROM [dbo].TBL_ADDRESSES where donor_id = '51999441' order by runid_gmt desc;
SELECT donor_id, donor_email, runid_gmt
FROM [dbo].TBL_EMAIL where donor_id = '51999441' order by runid_gmt desc;
Try with row_number window function:
select * from
(select *, row_number() over(partition by donorid order by gmt desc) rn
from donors) t1 join
(select *, row_number() over(partition by donorid order by gmt desc) rn
from addresses) t2 on t1.donorid = t2.donorid join
(select *, row_number() over(partition by donorid order by gmt desc) rn
from emails) t3 on t1.donorid = t3.donorid
where t1.rn = 1 and t2.rn = 1 and t3.rn = 1
Related
I am writing a query in Redshift to answer the question "Give the average lifetime spend of users who spent more on their first order than their second order." This is based off of an order_items table which has one row for every item ordered (so an order with 3 items would be represented in 3 rows). Here's a snapshot of the first 10 rows:
First 10 rows of order_items:
Here is my solution:
with
cte1_lifetime as (
select oi.user_id, sum(oi.sale_price) as lifetime_spend
from order_items as oi
group by oi.user_id
),
cte2_order as (
select oi.user_id, oi.order_id, sum(oi.sale_price) as order_total, rank() over(partition by oi.user_id order by oi.created_at) as order_rank
from order_items as oi
group by oi.user_id, oi.order_id, oi.created_at
order by oi.user_id, oi.order_id
),
cte3_first_order as (
select user_id, order_id, order_total
from cte2_order
where order_rank=1
order by user_id, order_id
),
cte4_second_order as (
select user_id, order_id, order_total
from cte2_order
where order_rank=2
order by user_id, order_id
)
select avg(cte1.lifetime_spend) as average_lifetime_spend
from cte1_lifetime as cte1
where exists (
select *
from cte3_first_order as cte3, cte4_second_order as cte4
where cte3.user_id=cte4.user_id
and cte1.user_id=cte3.user_id
and cte3.order_total > cte4.order_total)
And here is the answer key:
WITH
table1 AS
(SELECT user_id, order_id,
SUM(sale_price) OVER (PARTITION BY order_id ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) as order_total,
RANK() OVER (PARTITION BY user_id ORDER BY created_at) AS "sequence"
FROM order_items)
,
table2 AS
(SELECT user_id, SUM(sale_price) AS lifetime_spend
FROM order_items
WHERE EXISTS
(SELECT t1.user_id
FROM table1 t1, table1 t2
WHERE t1.user_id = t2.user_id AND t1.sequence = 1 AND t2.sequence = 2 AND t1.order_total>t2.order_total
AND t1.user_id = order_items.user_id)
GROUP BY 1
ORDER BY 1)
SELECT AVG(lifetime_spend)
FROM table2
These answers yield slightly different results on the same data- an average lifetime spend of $215 vs $220. I'd really like to understand why they are different but so far I can't figure it out. Any ideas?
Hi guys I have two tables dbo.Sales (customer_id, order_date, product_id) and dbo.Menu (Product_id, product_name, price). The question is
What was the first item from the menu purchased by each customer?
My solution is
select A.customer_id,m.product_id, m.product_name
from dbo.menu m
cross apply
(select top 1 * from dbo.sales s
where s.product_id=m.product_id
group by s.customer_id,s.order_date, s.product_id
order by s.order_date) A
customer_id product_id product_name
A 1 sushi
A 2 curry
C 3 ramen
Missing customer is B. Instead of B it gives me the second first order by A.
I need for each customer
Murat
You could use a ROW_NUMBER() window function to get the earliest product_id per customer and then join to the Menu table to get your product details.
Edit: Updated ORDER to ASC.
;with cte
as (
select customer_id, product_id, row_number() over (partition by customer_id order by order_date acs) RN
from dbo.Sales)
select c.customer_id, c.product_id, m.product_name
from cte c
join dbo.menu m on c.product_id=m.product_id
where RN = 1
SELECT distinct s.customer_id,
FIRST_VALUE(m.product_name) OVER (partition by s.customer_id order by order_date )
as FirstItem_Customer
FROM [dbo].[sales] S
join [dbo].[menu] M on M.product_id=s.product_id
I have a SELECT in Postgres:
SELECT DISTINCT ON (price) price, quantity, is_ask, final_update_id
FROM (SELECT *
FROM ((SELECT price, quantity, is_ask, book_depth.final_update_id
FROM order_depth
LEFT JOIN book_depth ON book_depth_id = book_depth.id
WHERE book_depth_id IN (SELECT id
FROM book_depth
WHERE final_update_id > (SELECT last_update_id
FROM order_book
WHERE symbol_name = 'XRPRUB'
ORDER BY last_update_id DESC
LIMIT 1)
AND symbol_name = 'XRPRUB'))
UNION
(SELECT price, quantity, is_ask, order_book_id
FROM "order"
WHERE order_book_id = (SELECT id
FROM order_book
WHERE symbol_name = 'XRPRUB'
ORDER BY last_update_id DESC
LIMIT 1))
ORDER BY final_update_id DESC) AS t) AS t1
ORDER BY price, final_update_id DESC;
It works for about 20 seconds.
But when I create function with this select this function works for about 1 min 40 seconds. Can someone explain me is it normal or I make mistake somewhere?
Have a table with 3 columns: ID, Signature, and Datetime, and it's grouped by Signature Having Count(*) > 9.
select * from (
select s.Signature
from #Sigs s
group by s.Signature
having count(*) > 9
) b
join #Sigs o
on o.Signature = b.Signature
order by o.Signature desc, o.DateTime
I now want to select the 1st and 10th records only, per Signature. What determines rank is the Datetime descending. Thus, I would expect every Signature to have 2 rows.
Thanks,
I would go with a couple of common table expressions.
The first will select all records from the table as well as a count of records per signature, and the second one will select from the first where the record count > 9 and add row_number partitioned by signature - and then just select from that where the row_number is either 1 or 10:
With cte1 AS
(
SELECT ID, Signature, Datetime, COUNT(*) OVER(PARTITION BY Signature) As NumberOfRows
FROM #Sigs
), cte2 AS
(
SELECT ID, Signature, Datetime, ROW_NUMBER() OVER(PARTITION BY Signature ORDER BY DateTime DESC) As Rn
FROM cte1
WHERE NumberOfRows > 9
)
SELECT ID, Signature, Datetime
FROM cte2
WHERE Rn IN (1, 10)
ORDER BY Signature desc
Because I don't know what your data looks like, this might need some adjustment.
The simplest way here, since you already know your sort order (DateTime DESC) and partitioning (Signature), is probably to assign row numbers and then select the rows you want.
SELECT *
FROM
(
select o.Signature
,o.DateTime
,ROW_NUMBER() OVER (PARTITION BY o.Signature ORDER BY o.DateTime DESC) [Row]
from (
select s.Signature
from #Sigs s
group by s.Signature
having count(*) > 9
) b
join #Sigs o
on o.Signature = b.Signature
order by o.Signature desc, o.DateTime
)
WHERE [Row] IN (1,10)
Can you let me know how to select only two employees from every department? The table has deptname, ssn, name . I am doing a sampling and I need only two ssns for every department name. Can someone help?
You can accomplish this with an "OLAP expression" row_number()
with e as
( select deptname, ssn, empname,
row_number() over (partition by dptname order by empname) as pick
from employees
)
select deptname, ssn, empname
from e
where pick < 3
order by deptname, ssn
This example will give you the two employees with the lowest order names, because that is what is specified in the row_number() (order by) expression.
Try this:
select *
from t t1
where (
select count(*)
from t t2
where
t2.deptname = t1.deptname
and
t2.ssn <= t1.ssn) <= 2
order by deptname, ssn,name;
The above will give "smallest" two ssn.
If you want top 2, change to t2.ssn >= t1.ssn
sqlfiddle
The data:
The result from query:
select * from
( select rank() over (partition by dptname order by empname) as count , *
from employees
)
where count<=2
order by deptname, ssn,name;