Oracle select sum by time window - sql

Lets assume that we have the ORACLE table of the following format and data:
TIMESTAMP MESSAGENO ORGMESSAGE
------------------------- ---------------------- -------------------------------------
27.04.13 1 START PERIOD
27.04.13 3 10
27.04.13 4 5
28.04.13 5 6
28.04.13 3 20
29.04.13 4 25
29.04.13 5 26
30.04.13 2 END PERIOD
30.04.13 1 START PERIOD
01.05.13 3 10
02.05.13 4 15
02.05.13 5 16
03.05.13 3 30
03.05.13 4 35
04.05.13 5 36
05.05.13 2 END PERIOD
I want to select sum of all the ORGMESSAGE for all the period (window between START PERIOD and END PERIOD) grouped by MESSAGENO.
Exapmle output would be:
PERIOD START PERIOD END MESSAGENO SUM
------------ ------------- -------- ----
27.04.13 30.04.13 3 25
27.04.13 30.04.13 4 30
27.04.13 30.04.13 5 32
30.04.13 05.05.13 3 45
30.04.13 05.05.13 4 50
30.04.13 05.05.13 5 52
I am guessing that use of ORACLE Analityc function woulde be suitable but really dont know how and where to start.
Thanks in advance for any help.

If we assume that the period starts and ends match, then a simple way to find the matching messages is to count the preceding number of starts. This is a cumulative sum and it is easy in Oracle. The rest is just aggregation:
select min(timestamp) as periodstart, max(timestamp) as periodend, messageno, count(*)
from (select om.*,
sum(case when messageno = 1 then 1 else 0 end) over (order by timestamp) as grp
from orgmessages om
) om
where messageno not in (1, 2)
group by grp, messageno;
Note that this method (as with the others) really wants the timestamp to be unique on each record. In the data presented, these solutions will work. But if you have multiple starts and ends on the same day, none of them will work assuming that timestamp only has the date.

First find all period ends per period start. Then join with your table to group and sum.
select
dates.start_date,
dates.end_date,
messageno,
sum(to_number(orgmessage)) as period_sum
from mytable
join
(
select start_dates.timestmp as start_date, min(end_dates.timestmp) as end_date
from (select * from mytable where orgmessage = 'START PERIOD') start_dates
join (select * from mytable where orgmessage = 'END PERIOD') end_dates
on start_dates.timestmp < end_dates.timestmp
group by start_dates.timestmp
) dates on mytable.timestmp between dates.start_date and dates.end_date
where mytable.orgmessage not like '%PERIOD%'
group by dates.start_date, dates.end_date, messageno
order by dates.start_date, dates.end_date, messageno;
SQL fiddle: http://www.sqlfiddle.com/#!4/365de/15.

please, try this one, replace rrr with your table name
select periodstart, periodend, messageno, sum(to_number(orgmessage)) s
from (select TIMESTAMP periodstart,
(select min (TIMESTAMP) from rrr r2 where orgmessage = 'END PERIOD' and r2.TIMESTAMP > r.TIMESTAMP) periodend
from rrr r
where orgmessage = 'START PERIOD'
) borders, rrr r
where r.TIMESTAMP between borders.periodstart and borders.periodend
and r.orgmessage not in ('END PERIOD', 'START PERIOD')
group by periodstart, periodend, messageno
order by periodstart, periodend, messageno

Related

How to create a start and end date with no gaps from one date column and to sum a value within the dates

I am new SQL coding using in SQL developer.
I have a table that has 4 columns: Patient ID (ptid), service date (dt), insurance payment amount (insr_amt), out of pocket payment amount (op_amt). (see table 1 below)
What I would like to do is (1) create two columns "start_dt" and "end_dt" using the "dt" column where if there are no gaps in the date by the patient ID then populate the start and end date with the first and last date by patient ID, however if there is a gap in service date within the patient ID then to create the separate start and end date rows per patient ID, along with (2) summing the two payment amounts by patient ID with in the one set of start and end date visits (see table 2 below).
What would be the way to run this using SQL code in SQL developer?
Thank you!
Table 1:
Ptid
dt
insr_amt
op_amt
A
1/1/2021
30
20
A
1/2/2021
30
10
A
1/3/2021
30
10
A
1/4/2021
30
30
B
1/6/2021
10
10
B
1/7/2021
20
10
C
2/1/2021
15
30
C
2/2/2021
15
30
C
2/6/2021
60
30
Table 2:
Ptid
start_dt
end_dt
total_insr_amt
total_op_amt
A
1/1/2021
1/4/2021
120
70
B
1/6/2021
1/7/2021
30
20
C
2/1/2021
2/2/2021
30
60
C
2/6/2021
2/6/2021
60
30
You didn't mention the specific database so this solution works in PostgreSQL. You can do:
select
ptid,
min(dt) as start_dt,
max(dt) as end_dt,
sum(insr_amt) as total_insr_amt,
sum(op_amt) as total_op_amt
from (
select *,
sum(inc) over(partition by ptid order by dt) as grp
from (
select *,
case when dt - interval '1 day' = lag(dt) over(partition by ptid order by dt)
then 0 else 1 end as inc
from t
) x
) y
group by ptid, grp
order by ptid, grp
Result:
ptid start_dt end_dt total_insr_amt total_op_amt
----- ---------- ---------- -------------- -----------
A 2021-01-01 2021-01-04 120 70
B 2021-01-06 2021-01-07 30 20
C 2021-02-01 2021-02-02 30 60
C 2021-02-06 2021-02-06 60 30
See running example at DB Fiddle 1.
EDIT for Oracle
As requested, the modified query that works in Oracle is:
select
ptid,
min(dt) as start_dt,
max(dt) as end_dt,
sum(insr_amt) as total_insr_amt,
sum(op_amt) as total_op_amt
from (
select x.*,
sum(inc) over(partition by ptid order by dt) as grp
from (
select t.*,
case when dt - 1 = lag(dt) over(partition by ptid order by dt)
then 0 else 1 end as inc
from t
) x
) y
group by ptid, grp
order by ptid, grp
See running example at db<>fiddle 2.

sum values based on 7-day cycle in SQL Oracle

I have dates and some value, I would like to sum values within 7-day cycle starting from the first date.
date value
01-01-2021 1
02-01-2021 1
05-01-2021 1
07-01-2021 1
10-01-2021 1
12-01-2021 1
13-01-2021 1
16-01-2021 1
18-01-2021 1
22-01-2021 1
23-01-2021 1
30-01-2021 1
this is my input data with 4 groups to see what groups will create the 7-day cycle.
It should start with first date and sum all values within 7 days after first date included.
then start a new group with next day plus anothe 7 days, 10-01 till 17-01 and then again new group from 18-01 till 25-01 and so on.
so the output will be
group1 4
group2 4
group3 3
group4 1
with match_recognize would be easy current_day < first_day + 7 as a condition for the pattern but please don't use match_recognize clause as solution !!!
One approach is a recursive CTE:
with tt as (
select dte, value, row_number() over (order by dte) as seqnum
from t
),
cte (dte, value, seqnum, firstdte) as (
select tt.dte, tt.value, tt.seqnum, tt.dte
from tt
where seqnum = 1
union all
select tt.dte, tt.value, tt.seqnum,
(case when tt.dte < cte.firstdte + interval '7' day then cte.firstdte else tt.dte end)
from cte join
tt
on tt.seqnum = cte.seqnum + 1
)
select firstdte, sum(value)
from cte
group by firstdte
order by firstdte;
This identifies the groups by the first date. You can use row_number() over (order by firstdte) if you want a number.
Here is a db<>fiddle.

SQL question - how to output using iterative date logic in SQL Server

I have the following sample table (provided with single ID for simplicity - need to perform the same logic across all IDs)
ID Visit_date
-----------------
ABC 8/7/2019
ABC 9/10/2019
ABC 9/12/2019
ABC 10/1/2019
ABC 10/1/2019
ABC 10/8/2019
ABC 10/15/2019
ABC 10/17/2019
ABC 10/24/2019
Here is what I need to get the sample output
Mark the first visit as 1 in the "new_visit" column
Compare the subsequent dates with the 1st date until it exceeds 21 days condition. Example Sep 10 is compared to Aug 7 and it doesn’t fall within 21 days of Aug 7, therefore this is considered as another new_visit, so mark new_visit as 1
Then we compare Sep 10 with the subsequent dates with 21 days criteria and mark all of them as follow_up of Sep 10 visit. Eg. Sep 12, Oct 1 are within 21 days of Sep 10; hence they are considered as follow up visits, so mark "follow_up" as 1
When the subsequent date exceeds 21 days criteria of the previous new visit (e.g. Oct 8 compared to Sep 10) then Oct 8 will be considered a new visit & mark "New_visit" as 1 and the subsequent dates will be compared against Oct 8
Sample Output :
Dates New_Visit Follow_up
-----------------------------
8/7/2019 1
9/10/2019 1
9/12/2019 1
10/1/2019 1
10/1/2019 1
10/8/2019 1
10/15/2019 1
10/17/2019 1
10/24/2019 1
You need a recursive query for this.
You would enumerate the rows, then walk through the dataset by ascending date, while keeping track of the first visit date of each group; when the interval since the last first visit exceeds 21 days, the date of the first visit resets, and a new group starts.
with
data as (
select t.*, row_number() over(partition by id order by date) rn
from mtytable t
),
cte as (
select id, visit_date, visit_date first_visit_date
from data
where rn = 1
union all
select c.id, d.visit_date, case when d.visit_date > datead(day, 21, c.first_visit_date) then d.visit_date else c.first_visit_date end
from cte c
inner join data d on d.id = c.id and d.rn = c.rn + 1
)
select
id,
date,
case when visit_date = first_visit_date then 1 else 0 end as is_new
case when visit_date = first_visit_date then 0 else 1 end as is_follow_up
from cte
If a patient may have more than 100 visits, then you need to add option (maxrecursion 0) at the very end of the query.
You need a recursive CTE to handle this. This is the idea, although the exact syntax might vary by database:
with recursive t as (
select id, date,
row_number() over (partition by id order by date) as seqnum
from yourtable
),
recursive cte as (
select id, date, visit_start as date, 1 as is_new_visit
from t
where id = 1
union all
select cte.id, t.date,
(case when t.date < visit_start + interval '21 day'
then cte.visit_start else t.date
end) as visit_start,
(case when t.date < cte.visit_start + interval '21 say'
then 0 else 1
end) as is_new_visit
from cte join
t
on t.id = cte.id and t.seqnum = cte.seqnum + 1
)
select *
from cte
where is_new_visit = 1;

SQL not returning a value if no row exist for time queried

I'm writing this SQL query which returns the number of records created in an hour in last 24 hours. I'm getting the result for only those hours that have a non zero value. If no records were created, it doesn't return anything at all.
Here's my query:
SELECT HOUR(timeStamp) as hour, COUNT(*) as count
FROM `events`
WHERE timeStamp > DATE_SUB(NOW(), INTERVAL 24 HOUR)
GROUP BY HOUR(timeStamp)
ORDER BY HOUR(timeStamp)
The output of current Query:
+-----------------+----------+
| hour | count |
+-----------------+----------+
| 14 | 6 |
| 15 | 5 |
+-----------------+----------+
But i'm expecting 0 for hours in which no records were created. Where am I going wrong?
One solution is to generate a table of numbers from 0 to 23 and left join it with your original table.
Here is a query that uses a recursive query to generate the list of hours (if you are running MySQL, this requires version 8.0):
with hours as (
select 0 hr
union all select hr + 1 where h < 23
)
select h.hr, count(e.eventID) as cnt
from hours h
left join events e
on e.timestamp > now() - interval 1 day
and hour(e.timestamp) = h.hr
group by h.hr
If your RDBMS does not support recursive CTEs, then one option is to use an explicit derived table:
select h.hr, count(e.eventID) as cnt
from (
select 0 hr union all select 1 union all select 2 ... union all select 23
) h
left join events e
on e.timestamp > now() - interval 1 day
and hour(e.timestamp) = h.hr
group by h.hr

SQL Count Numbers of Projects started Each Day Between Two Dates

I have this table that I need count how many projects (job) i have started each day.
job start end
1 01-01-2013 04-01-2013
2 01-01-2013 02-01-2013
3 01-01-2013 03-01-2013
4 03-01-2013 04-01-2013
5 03-01-2013 04-01-2013
6 03-01-2013 04-01-2013
...
i want count how many job's i have started each day.. / i mean how many job's are open each day..
date count
01-01-2013 3
02-01-2013 3
03-01-2013 5
04-01-2013 4
05-01-2013 0
...
select start, count(*) as jobs_per_day
from your_table
group by start
But this will not return a record for dates where you did not create any job.
The following works for me in postgresql
with dates as (
select aday::date
from generate_series((select min(start) from your_table),
(select max(end) from your_table),
'1 day'::interval) aday
), flat as (
select *
from dates, your_table
where dates.aday between your_table.start and your_table.end
)
select
aday,
count(*) as count
from flat
group by aday
order by aday
;
The first CTE generates a series of dates, which might have to be done differently in another RDBMS.
select start as date, count(*) as count from table_name
where start_date>="your start date" and end_date<="your end date"
group by start;