https://www.db-fiddle.com/f/rgLXTu3VysD3kRwBAQK3a4/3
My problem here is that I want function partition over to start counting the rows only from certain time range.
In this example, if I would add rn = 1 at the end, order_id = 5 would be excluded from the results (because partition is ordering by paid_date and there's order_id = 6 with earlier date) but it shouldn't be as I want that time range for partition starts from '2019-01-10'.
Adding condition rn = 1expected output should be order_id 3,5,11,15, now its only 3,11,15
it should include only orders with is_paid = 0 that are the first one within given time range (if there's preceeding order with is_paid = 1 it shouldn't be counted)
use correlated subquery with not exists
DEMO
SELECT order_id, customer_id, amount, is_paid, paid_date, rn FROM (
SELECT o.*,
ROW_NUMBER() OVER(PARTITION BY customer_id ORDER BY paid_date,order_id) rn
FROM orders o
WHERE paid_date between '2019-01-10'
and '2019-01-15'
) x where rn=1 and not exists (select 1 from orders o1 where x.order_id=o1.order_id
and is_paid=1)
OUTPUT:
order_id customer_id amount is_paid paid_date rn
3 101 30 0 10/01/2019 00:00:00 1
5 102 15 0 10/01/2019 00:00:00 1
11 104 31 0 10/01/2019 00:00:00 1
15 105 11 0 10/01/2019 00:00:00 1
If priority should be given to order_id then put that before paid date in the partition function order by clause, this will solve your issue.
SELECT order_id, customer_id, amount, is_paid, paid_date, rn FROM (
SELECT o.*,
ROW_NUMBER() OVER(PARTITION BY customer_id ORDER BY order_id,paid_date) rn
FROM orders o
) x WHERE is_paid = 0 and paid_date between
'2019-01-10' and '2019-01-15' and rn=1
Since you need the paid date to be ordered first you need to imply a where condition in the partitioning table in order to avoid unnecessary dates interrupting the partition function.
SELECT order_id, customer_id, amount, is_paid, paid_date, rn FROM (
SELECT o.*,
ROW_NUMBER() OVER(PARTITION BY customer_id ORDER BY paid_date, order_id) rn
FROM orders o
where paid_date between '2019-01-10' and '2019-01-15'
) x WHERE is_paid = 0 and rn=1
Related
I'm trying to query a dataset about user status changes. and I want to find out the time it takes for the status to change, and the steps in between(number of rows).
Example data:
user_id
Status
date
1
a
2001-01-01
1
a
2001-01-08
1
b
2001-01-15
1
b
2001-01-28
1
a
2001-01-31
1
b
2001-02-01
2
a
2001-01-08
2
a
2001-01-18
2
a
2001-01-28
3
b
2001-03-08
3
b
2001-03-18
3
b
2001-03-19
3
a
2001-03-20
Desired output:
user_id
From
to
days in between
Steps in between
1
a
b
14
2
1
b
a
16
2
1
a
b
1
1
3
b
a
12
3
You might consider below another approach.
WITH partitions AS (
SELECT *, COUNTIF(flag) OVER w AS part FROM (
SELECT *, ROW_NUMBER() OVER w AS rn, status <> LAG(status) OVER w AS flag,
FROM sample_data
WINDOW w AS (PARTITION BY user_id ORDER BY date)
) WINDOW w AS (PARTITION BY user_id ORDER BY date)
)
SELECT user_id,
LAG(ANY_VALUE(status)) OVER w AS `from`,
ANY_VALUE(status) AS `to`,
EXTRACT(DAY FROM MIN(date) - LAG(MIN(date)) OVER w) AS days_in_between,
MIN(rn) - LAG(MIN(rn)) OVER w AS steps_in_between
FROM partitions
GROUP BY user_id, part
QUALIFY `from` IS NOT NULL
WINDOW w AS (PARTITION BY user_id ORDER BY MIN(date));
Query results
with main as (
select
*,
dense_rank() over(partition by user_id order by date) as rank_,
row_number() over(partition by user_id, status order by date) as rank_2,
row_number() over(partition by user_id, status order by date) - dense_rank() over(partition by id order by date) as diff,
row_number() over(partition by user_id order by date) as row_num,
lag(status) over(partition by user_id order by date) as prev_status,
concat(lag(status) over(partition by user_id order by date) , ' to ' , status) as status_change
from table
),
new_rank as (
select
*,
rown_num - diff as row_num_diff,
min(date) over(partition by user_id, status, rown_num - diff) as min_date
from main
),
prev_date as (
select
*,
lag(min_date) over(partition by user_id order by date) as prev_min_date
from new_rank
)
select
status as from,
prev_status as to,
date_diff(prev_min_date, min_date, DAY) as days_in_between
from prev_date
where status !=prev_status and prev_status is not null
Does this seem to work? I tried to solve this but it's very hard to solve it without a fiddle plus:
you may remove the extra steps/ranks that I have added, I left them there so you can visually see what they are doing
I don't get your steps logic so it is missing from the code
I'm attempting to write a SQL query which returns every product where the most recent price on an order within the last 30 days is different than the most recent price in the previous 30 days, and that calculated variance. I'm currently using PostgreSQL 11.
Data Model
Right now, the data is structured into three tables: orders, products, and a pivot table, order_product. Here is the simplified version of the table structure:
Orders
id
order_date
1
2022-01-15
2
2022-02-15
3
2022-03-08
Products
id
name
1
Some product
2
Another product
3
Yet another product
Order_Product
order_id
product_id
unit_price
1
1
10
1
2
20
1
3
10
2
1
12
2
2
20
2
3
5
3
1
15
Desired Output
The desired output would be something like the following:
id
name
order_date
latest_unit_price
previous_unit_price
variance
1
Some product
2022-03-08
15
10
5
3
Yet another product
2022-02-15
5
10
-5
What I've done so far
I've been able to write a join that combines the Orders and Products via the order_product table, within the 60-day window, which is seemingly the easy part:
SELECT
"products"."id",
"products"."name",
"order_product"."unit_price",
"orders"."order_date"
FROM
products
JOIN order_product ON products.id = order_product.product_id
JOIN orders ON order_product.order_id = orders.id
WHERE
order_date BETWEEN now() - INTERVAL '60 days'
AND now()
I've been trying to work with RANK() and LAG(); however, where I'm getting stuck is being able to find the rank the rows within the 30-day time windows, and then calculate the variance between the two windows.
Any help would be much appreciated!
Update: Added solution
Building off of the answer by D-Shih, I had to tweak this to work based on the time window starting from the current date:
WITH CTE AS (
SELECT
"products"."id",
"products"."name",
"order_product"."unit_price",
"orders"."order_date"
FROM
products
JOIN order_product ON products.id = order_product.product_id
JOIN orders ON order_product.order_id = orders.id
WHERE
order_date BETWEEN now() - INTERVAL '60 days' AND now()
),
CTE2 AS (
SELECT
*,
EXTRACT(DAYS FROM now() - order_date :: timestamp) gap_days
FROM
CTE
),
CTE3 AS (
SELECT
*,
(CASE WHEN gap_days < 30 THEN 1 ELSE 0 END) grp
FROM
CTE2
)
SELECT
id,
name,
MAX(CASE WHEN grp = 1 THEN order_date END) order_date,
MAX(CASE WHEN grp = 1 THEN unit_price END) latest_unit_price,
MAX(CASE WHEN grp = 0 THEN unit_price END) previous_unit_price,
SUM(CASE WHEN grp = 1 THEN unit_price ELSE - unit_price END) variance
FROM
(
SELECT
*,
ROW_NUMBER() OVER (PARTITION BY ID, grp ORDER BY order_date DESC) rn
FROM
CTE3
) t1
WHERE
rn = 1
GROUP BY
id,
name
HAVING
MAX(CASE WHEN grp = 1 THEN unit_price END) <> MAX(CASE WHEN grp = 0 THEN unit_price END)
sqlfiddle
You can try to use EXTRACT with LAG window function to get days difference from order_date and previous order_date each productId.
Then use SUM aggregate condition window function to calculate the group
grp = 0 within the last 30 days
grp = 1 most recent price in the previous 30 days,
the query would be look like as below.
WITH CTE AS (
SELECT "products"."id",
"products"."name",
"order_product"."unit_price",
"orders"."order_date"
FROM
products
JOIN order_product ON products.id = order_product.product_id
JOIN orders ON order_product.order_id = orders.id
WHERE
order_date BETWEEN now() - INTERVAL '60 days'
AND now()
), CTE2 AS (
SELECT *,EXTRACT(DAYS FROM order_date - LAG(order_date,1,order_date) OVER(PARTITION BY id ORDER BY order_date)) gap_seconds
FROM CTE
), CTE3 AS (
SELECT *,(CASE WHEN SUM(gap_seconds) OVER(PARTITION BY id ORDER BY order_date) > 30 THEN 1 ELSE 0 END) grp
FROM CTE2
)
SELECT id,
name,
MAX(CASE WHEN grp = 1 THEN order_date END) order_date,
MAX(CASE WHEN grp = 1 THEN unit_price END) latest_unit_price,
MAX(CASE WHEN grp = 0 THEN unit_price END) previous_unit_price,
SUM(CASE WHEN grp = 1 THEN unit_price ELSE - unit_price END) variance
FROM (
SELECT *,ROW_NUMBER() OVER(PARTITION BY ID,grp ORDER BY order_date DESC) rn
FROM CTE3
) t1
WHERE rn = 1
GROUP BY id,
name
HAVING MAX(CASE WHEN grp = 1 THEN unit_price END) <> MAX(CASE WHEN grp = 0 THEN unit_price END)
sqlfiddle
suppose I have the following data frame in Reradata SQL.
How can I get the variation between the highest and lowest date, at user level? Regards
Initial table
user date price
1 1-1 10
1 2-1 20
1 3-1 30
2 1-1 12
2 2-1 22
2 3-1 32
3 1-1 13
3 2-1 23
3 3-1 33
Final table
user var_price
1 30/10-1
2 32/12-1
3 33/13-1
Try this-
SELECT B.[user],
CAST(SUM(B.max_price) AS VARCHAR)+'/'+CAST(SUM(B.min_price) AS VARCHAR)+ '-1' var_price,
SUM(B.max_price)/SUM(B.min_price) -1 calculated_var_price
FROM
(
SELECT * FROM
(
SELECT [user],0 max_price,price min_price,ROW_NUMBER() OVER (PARTITION BY [user] ORDER BY DATE) RN
FROM your_table
)A WHERE RN = 1
UNION ALL
SELECT * FROM
(
SELECT [user],price max_price,0 min_price, ROW_NUMBER() OVER (PARTITION BY [user] ORDER BY DATE DESC) RN
FROM your_table
)A WHERE RN = 1
)B
GROUP BY B.[user]
Output is-
user var_price calculated_var_price
1 30/10-1 2
2 32/12-1 1
3 33/13-1 1
Is this what you want?
select user, max(price) / min(price) - 1
from t
group by user;
Your values are monotonically increasing, so max() and min() seems like the simplest solution.
EDIT:
You can use window functions:
select user, max(last_price) / max(first_price) - 1
from (select t.*,
first_value(price) over (partition by user order by date rows between unbounded preceding and current_row) as first_price,
first_value(price) over (partition by user order by date desc rows between unbounded preceding and current_row) as last_price
from t
) t
group by user;
select user
,price as first_price
,last_value(price)
over (paritition by user
order by date
rows between unbounded preceding and unbounded following) as last_price
from mytab
qualify
row_number() -- lowest date only
over (paritition by user
order by date) = 1
This returns the row with the lowest date and adds the price of the latest date
I have a table like below, I am trying to run a query in T-SQL to get the earliest and latest costs for each project_id according to the date column and calculate the percent cost increase or decrease and return the data-set show in the second table (I have simplified the table in this question).
project_id date cost
-------------------------------
123 7/1/17 5000
123 8/1/17 6000
123 9/1/17 7000
123 10/1/17 8000
123 11/1/17 9000
456 7/1/17 10000
456 8/1/17 9000
456 9/1/17 8000
876 1/1/17 8000
876 6/1/17 5000
876 8/1/17 10000
876 11/1/17 8000
Result:
(Edit: Fixed the result)
project_id "cost incr/decr pct"
------------------------------------------------
123 80% which is (9000-5000)/5000
456 -20%
876 0%
Whatever query I run I get duplicates.
This is what I tried:
select distinct
p1.Proj_ID, p1.date, p2.[cost], p3.cost,
(nullif(p2.cost, 0) / nullif(p1.cost, 0)) * 100 as 'OVER UNDER'
from
[PROJECT] p1
inner join
(select
[Proj_ID], [cost], min([date]) min_date
from
[PROJECT]
group by
[Proj_ID], [cost]) p2 on p1.Proj_ID = p2.Proj_ID
inner join
(select
[Proj_ID], [cost], max([date]) max_date
from
[PROJECT]
group by
[Proj_ID], [cost]) p3 on p1.Proj_ID = p3.Proj_ID
where
p1.date in (p2.min_date, p3.max_date)
Unfortunately, SQL Server does not have a first_value() aggregation function. It does have an analytic function, though. So, you can do:
select distinct project_id,
first_value(cost) over (partition by project_id order by date asc) as first_cost,
first_value(cost) over (partition by project_id order by date desc) as last_cost,
(first_value(cost) over (partition by project_id order by date desc) /
first_value(cost) over (partition by project_id order by date asc)
) - 1 as ratio
from project;
If cost is an integer, you may need to convert to a representation with decimal places.
You can use row_number and OUTER APPLY over top 1 ... prior to SQL 2012
select
min_.projectid,
latest_.cost - min_.cost [Calculation]
from
(select
row_number() over (partition by projectid order by date) rn
,projectid
,cost
from projectable) min_ -- get the first dates per project
outer apply (
select
top 1
cost
from projectable
where
projectid = min_.projectid -- get the latest cost for each project
order by date desc
) latest_
where min_.rn = 1
This might perform a little better
;with costs as (
select *,
ROW_NUMBER() over (PARTITION BY project_id ORDER BY date) mincost,
ROW_NUMBER() over (PARTITION BY project_id ORDER BY date desc) maxcost
from table1
)
select project_id,
min(case when mincost = 1 then cost end) as cost1,
max(case when maxcost = 1 then cost end) as cost2,
(max(case when maxcost = 1 then cost end) - min(case when mincost = 1 then cost end)) * 100 / min(case when mincost = 1 then cost end) as [OVER UNDER]
from costs a
group by project_id
I have a table including more than 5 million rows of sales transactions. I would like to find sum of date intervals between each customer three recent purchases.
Suppose my table looks like this :
CustomerID ProductID ServiceStartDate ServiceExpiryDate
A X1 2010-01-01 2010-06-01
A X2 2010-08-12 2010-12-30
B X4 2011-10-01 2012-01-15
B X3 2012-04-01 2012-06-01
B X7 2012-08-01 2013-10-01
A X5 2013-01-01 2015-06-01
The Result that I'm looking for may looks like this :
CustomerID IntervalDays
A 802
B 135
I know the query need to first retrieve 3 resent transactions of each customer (based on ServiceStartDate) and then calculate the interval between startDate and ExpiryDate of his/her transactions.
You want to calculate the difference between the previous row's ServiceExpiryDate and the current row's ServiceStartDate based on descending dates and then sum up the last two differences:
with cte as
(
select tab.*,
row_number()
over (partition by customerId
order by ServiceStartDate desc
, ServiceExpiryDate desc -- don't know if this 2nd column is necessary
) as rn
from tab
)
select t2.customerId,
sum(datediff(day, prevEnd, ServiceStartDate)) as Intervaldays
,count(*) as purchases
from cte as t2 left join cte as t1
on t1.customerId = t2.customerId
and t1.rn = t2.rn+1 -- previous and current row
where t2.rn <= 3 -- last three rows
group by t2.customerId;
Same result using LEAD:
with cte as
(
select tab.*,
row_number()
over (partition by customerId
order by ServiceStartDate desc) as rn
,lead(ServiceExpiryDate)
over (partition by customerId
order by ServiceStartDate desc
) as prevEnd
from tab
)
select customerId,
sum(datediff(day, prevEnd, ServiceStartDate)) as Intervaldays
,count(*) as purchases
from cte
where rn <= 3
group by customerId;
Both will not return the expected result unless you subtract purchases (or max(rn)) from Intervaldays. But as you only sum two differences this seems to be not correct for me either...
Additional logic must be applied based on your rules regarding:
customer has less than 3 purchases
overlapping intervals
Assuming there are no overlaps, I think you want this:
select customerId,
sum(datediff(day, ServiceStartDate, ServieEndDate) as Intervaldays
from (select t.*, row_number() over (partition by customerId
order by ServiceStartDate desc) as seqnum
from table t
) t
where seqnum <= 3
group by customerId;
Try this:
SELECT dt.CustomerID,
SUM(DATEDIFF(DAY, dt.PrevExpiry, dt.ServiceStartDate)) As IntervalDays
FROM (
SELECT *
, ROW_NUMBER() OVER (PARTITION BY CustomerID ORDER BY ServiceStartDate DESC) AS rn
, (SELECT Max(ti.ServiceExpiryDate)
FROM yourTable ti
WHERE t.CustomerID = ti.CustomerID
AND ti.ServiceStartDate < t.ServiceStartDate) As PrevExpiry
FROM yourTable t )dt
GROUP BY dt.CustomerID
Result will be:
CustomerId | IntervalDays
-----------+--------------
A | 805
B | 138