not sure if my below query script right to execute as I am trying to find just one query in Oracle SQL query
select distinct (master_id) , last_name from (
select q2.* , max (count_a) over (partition by master_id)
count_b from ( select q1.* , count (*) over (partition by
master_id order by purchased_date desc ) count_a from profile q1)
q2) where count_b > 2
I am trying to minimise timer to execute get data by reducing sub query
for example above it has two subqueries
max (count_a) over (partition by master_id) count_b
count (*) over (partition by master_id order by purchased_date desc ) count_a
so I played around until this query
max (count (*)) over (partition by master_id) count
SQL query script;
select * from profile a
join( select * from (
select master_id, max (count(*)) over (partition by
master_id) count from profile) where count >2) b
ON a. master_id = b. master_id
Thank you in advance for your help
Related
Quick question, I'm trying to update a column only when there are duplicates(partition column > 1) in the table and have selected it based on partition concept, But the current query updates the whole table! please check the query below, Any leads would be greatly appreciated :)
UPDATE public.database_tag
SET deleted_at= '2022-04-25 19:33:29.087133+00'
FROM (
SELECT *,
row_number() over (partition by title order by created_at) as RN
FROM public.database_tag
ORDER BY RN DESC) X
WHERE X.RN > 1
Thanks very much!
Assuming that every row have unique ID it can be done like below.
UPDATE database_tag
SET deleted_at= '2022-04-25 19:33:29.087133+00'
WHERE <some_unique_id> in (
select <some_unique_id> from (
SELECT <some_unique_id>,
row_number() over (partition by title order by created_at) as RN
FROM public.database_tag
) X
WHERE X.RN > 1
)
Or we can reverse query to update all but set of ID's
UPDATE database_tag
SET deleted_at= '2022-04-25 19:33:29.087133+00'
WHERE <some_unique_id> not in (
select distinct on (title)
<some_unique_id> from database_tag
order by title, created_at
)
I want to get the number of consecutive days from the current date using Postgres SQL.
enter image description here
Above is the scenario in which I have highlighted consecutive days count should be like this.
Below is the SQL query which I have created but it's not returning the expected result
with grouped_dates as (
select user_id, created_at::timestamp::date,
(created_at::timestamp::date - (row_number() over (partition by user_id order by created_at::timestamp::date) || ' days')::interval)::date as grouping_date
from watch_history
)
select * , dense_rank() over (partition by grouping_date order by created_at::timestamp::date) as in_streak
from grouped_dates where user_id = 702
order by created_at::timestamp::date
Can anyone please help me to resolve this issue?
If anyhow we can able to apply distinct for created_at field to below query then I will get solutions for my issue.
WITH list AS
(
SELECT user_id,
(created_at::timestamp::date - (row_number() over (partition by user_id order by created_at::timestamp::date) || ' days')::interval)::date as next_day
FROM watch_history
)
SELECT user_id, count(*) AS number_of_consecutive_days
FROM list
WHERE next_day IS NOT NULL
GROUP BY user_id
Does anyone have an idea how to apply distinct to created_at for the above mentioned query ?
To get the "number of consecutive days" for the same user_id :
WITH list AS
(
SELECT user_id
, array_agg(created_at) OVER (PARTITION BY user_id ORDER BY created_at RANGE BETWEEN CURRENT ROW AND '1 day' FOLLOWING) AS consecutive_days
FROM watch_history
)
SELECT user_id, count(DISTINCT d.day) AS number_of_consecutive_days
FROM list
CROSS JOIN LATERAL unnest(consecutive_days) AS d(day)
WHERE array_length(consecutive_days, 1) > 1
GROUP BY user_id
To get the list of "consecutive days" for the same user_id :
WITH list AS
(
SELECT user_id
, array_agg(created_at) OVER (PARTITION BY user_id ORDER BY created_at RANGE BETWEEN CURRENT ROW AND '1 day' FOLLOWING) AS consecutive_days
FROM watch_history
)
SELECT user_id
, array_agg(DISTINCT d.day ORDER BY d.day) AS list_of_consecutive_days
FROM list
CROSS JOIN LATERAL unnest(consecutive_days) AS d(day)
WHERE array_length(consecutive_days, 1) > 1
GROUP BY user_id
full example & result in dbfiddle
I just started learning Postgres, and I'm trying to make an aggregation table that has the columns:
user_id
booking_sequence
booking_created_time
booking_paid_time
booking_price_amount
total_spent
All columns are provided, except for the booking_sequence column. I need to make a query that shows the first five flights of each user that has at least x purchases and has spent more than a certain amount of money, then sort it by the amount of money spent by the user, and then sort it by the booking sequence column.
I've tried :
select user_id,
row_number() over(partition by user_id order by user_id) as booking_sequence,
booking_created_time as booking_created_date,
booking_price_amount,
sum(booking_price_amount) as total_booking_price_amount
from fact_flight_sales
group by user_id, booking_created_time, booking_price_amount
having count(user_id) > 5
and total_booking_price_amount > 1000
order by total_booking_price_amount;
I got 0 when I added count(user_id) > 5, and total_booking_price_amount is not found when I add the second condition in the HAVING clause.
Edit:
I managed to make the code function correctly, for those who are curious:
select x.user_id, row_number() over(partition by x.user_id)
as booking_sequence, x.booking_created_time::date as booking_created_date, x.booking_price_amount,
sum(y.booking_price_amount) as total_booking_price_amount from
(
select user_id, booking_created_time, booking_price_amount from fact_flight_sales
group by user_id, booking_created_time, booking_price_amount
) as x
join
(
select user_id, booking_price_amount
from fact_flight_sales group by user_id, booking_price_amount
) as y
on x.user_id = y.user_id
group by x.user_id, x.booking_created_time, x.booking_price_amount
having count(x.user_id) >= 1 and sum(y.booking_price_amount) >250000
order by total_booking_price_amount desc, booking_sequence asc;
Big thanks to Laurenz for the help!
About count(user_id) > 5:
HAVING is calculated before window functions are evaluated, So result rows excluded by the HAVING clause will not be used to calculate the window function.
About total_booking_price_amount in HAVING:
You cannot use aliases from the SELECT list in the HAVING clause. You will have to repeat the expression (or use a subquery).
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;