Looking to get the "Total Consecutive Days" per row.
Shift Table Contains ShiftId, ClientID, ServiceId and ProviderID
ShiftDetails Table has the dates.
There can be multiple Shifts with the same ShiftId, ClientID, ServiceId
There can even be duplicate dates.
Would like to have multiple "Total Consecutive Days" calculations.
Consecutive Days Per ClientID
Consecutive Days Per ClientID and ServiceID
Consecutive Days Per ClientID , ServiceID and ProviderID.
SQL Fiddle
This is a gaps-and-islands problem. You can identify the islands by subtracting a sequence of integers from the date -- constant differences identify adjacent dates. Then use count(*) as a window function:
select t.*,
count(*) over (partition by clientid, serviceid, dateadd(day, -seqnum, date)) as consecutive_days
from (select t.*,
row_number() over (partition by clientid, serviceid order by date) as seqnum
from t
) t
I have a requirement to get values from a table based on an offset conditions on a date column.
Say for eg: for the below attached table, if there is any dates that comes close within 15 days based on effectivedate column I should return only the first one.
So my expected result would be as below:
Here for A1234 policy, it returns 6/18/16 entry and skipped 6/12/16 entry as the offset between these 2 dates is within 15 days and I took the latest one from the list.
If you want to group rows together that are within 15 days of each other, then you have a variant of the gaps-and-islands problem. I would recommend lag() and cumulative sum for this version:
select polno, min(effectivedate), max(expirationdate)
from (select t.*,
sum(case when prev_ed >= dateadd(day, -15, effectivedate)
then 1 else 0
end) over (partition by polno order by effectivedate) as grp
from (select t.*,
lag(expirationdate) over (partition by polno order by effectivedate) as prev_ed
from t
) t
) t
group by polno, grp;
I have a column of dates and I want to find the three maximum dates for each year I have tried the following.
select max(date, rank() over (partition by SPLIT_PART(date, '-', 1) order by date desc)
from table
;
My desired output would be
2013,2010-12-31
2013,2010-12-30
2013,2010-12-29
also there are repeats dates in the table so I would have to filter those out as well
Assuming there are no duplicate dates, you can partition by the year part of date and get the latest 3 dates per year. Use distinct (if needed) in the final query to remove the duplicates, if any.
select yr,date
from (select date_part(year,date) as yr,date
,dense_rank() over (partition by date_part(year,date) order by date desc) as rnk
from table
) t
where rnk<=3
In my postgreSQL data base, I have a table with columns of dates and prices. ('transdate' and 'price')
I would like to form a query which selects one row for each week over a date range which spans more than one year.
From another question/answer here, I implemented this code which works for date ranges of less than a year:
;with cte as
(
select *,
row_number() over (partition by Extract (week from transdate) order by transdate desc) as rn
from "tablename" where transdate between '06-01-1999' and '06-01-1999'::timestamp + `'50 week'::interval
)
select transdate, price from cte where rn = 1 order by transdate;
However, when I extend the interval greater than 50 weeks, it still only selects a max of 12 months.
How can I re-write this code to select one date/price from every week in the range?
Your problem is that week numbers wrap around at year boundaries but you want to look at the week number and the year at the same time. Lucky for you, you can PARTITION BY several things at once:
row_number() over (
partition by extract(week from transdate),
extract(year from transdate)
order by transdate desc
) as rn
I would appreciate a little expert help please.
in an SQL SELECT statement I am trying to get the last day with data per month for the last year.
Example, I am easily able to get the last day of each month and join that to my data table, but the problem is, if the last day of the month does not have data, then there is no returned data. What I need is for the SELECT to return the last day with data for the month.
This is probably easy to do, but to be honest, my brain fart is starting to hurt.
I've attached the select below that works for returning the data for only the last day of the month for the last 12 months.
Thanks in advance for your help!
SELECT fd.cust_id,fd.server_name,fd.instance_name,
TRUNC(fd.coll_date) AS coll_date,fd.column_name
FROM super_table fd,
(SELECT TRUNC(daterange,'MM')-1 first_of_month
FROM (
select TRUNC(sysdate-365,'MM') + level as DateRange
from dual
connect by level<=365)
GROUP BY TRUNC(daterange,'MM')) fom
WHERE fd.cust_id = :CUST_ID
AND fd.coll_date > SYSDATE-400
AND TRUNC(fd.coll_date) = fom.first_of_month
GROUP BY fd.cust_id,fd.server_name,fd.instance_name,
TRUNC(fd.coll_date),fd.column_name
ORDER BY fd.server_name,fd.instance_name,TRUNC(fd.coll_date)
You probably need to group your data so that each month's data is in the group, and then within the group select the maximum date present. The sub-query might be:
SELECT MAX(coll_date) AS last_day_of_month
FROM Super_Table AS fd
GROUP BY YEAR(coll_date) * 100 + MONTH(coll_date);
This presumes that the functions YEAR() and MONTH() exist to extract the year and month from a date as an integer value. Clearly, this doesn't constrain the range of dates - you can do that, too. If you don't have the functions in Oracle, then you do some sort of manipulation to get the equivalent result.
Using information from Rhose (thanks):
SELECT MAX(coll_date) AS last_day_of_month
FROM Super_Table AS fd
GROUP BY TO_CHAR(coll_date, 'YYYYMM');
This achieves the same net result, putting all dates from the same calendar month into a group and then determining the maximum value present within that group.
Here's another approach, if ANSI row_number() is supported:
with RevDayRanked(itemDate,rn) as (
select
cast(coll_date as date),
row_number() over (
partition by datediff(month,coll_date,'2000-01-01') -- rewrite datediff as needed for your platform
order by coll_date desc
)
from super_table
)
select itemDate
from RevDayRanked
where rn = 1;
Rows numbered 1 will be nondeterministically chosen among rows on the last active date of the month, so you don't need distinct. If you want information out of the table for all rows on these dates, use rank() over days instead of row_number() over coll_date values, so a value of 1 appears for any row on the last active date of the month, and select the additional columns you need:
with RevDayRanked(cust_id, server_name, coll_date, rk) as (
select
cust_id, server_name, coll_date,
rank() over (
partition by datediff(month,coll_date,'2000-01-01')
order by cast(coll_date as date) desc
)
from super_table
)
select cust_id, server_name, coll_date
from RevDayRanked
where rk = 1;
If row_number() and rank() aren't supported, another approach is this (for the second query above). Select all rows from your table for which there's no row in the table from a later day in the same month.
select
cust_id, server_name, coll_date
from super_table as ST1
where not exists (
select *
from super_table as ST2
where datediff(month,ST1.coll_date,ST2.coll_date) = 0
and cast(ST2.coll_date as date) > cast(ST1.coll_date as date)
)
If you have to do this kind of thing a lot, see if you can create an index over computed columns that hold cast(coll_date as date) and a month indicator like datediff(month,'2001-01-01',coll_date). That'll make more of the predicates SARGs.
Putting the above pieces together, would something like this work for you?
SELECT fd.cust_id,
fd.server_name,
fd.instance_name,
TRUNC(fd.coll_date) AS coll_date,
fd.column_name
FROM super_table fd,
WHERE fd.cust_id = :CUST_ID
AND TRUNC(fd.coll_date) IN (
SELECT MAX(TRUNC(coll_date))
FROM super_table
WHERE coll_date > SYSDATE - 400
AND cust_id = :CUST_ID
GROUP BY TO_CHAR(coll_date,'YYYYMM')
)
GROUP BY fd.cust_id,fd.server_name,fd.instance_name,TRUNC(fd.coll_date),fd.column_name
ORDER BY fd.server_name,fd.instance_name,TRUNC(fd.coll_date)