I am trying to figure out how to write a query that looks at certain records and finds missing date ranges between today and 9999-12-31.
My data looks like below:
ID |start_dt |end_dt |prc_or_disc_1
10412 |2018-07-17 00:00:00.000 |2018-07-20 00:00:00.000 |1050.000000
10413 |2018-07-23 00:00:00.000 |2018-07-26 00:00:00.000 |1040.000000
So for this data I would want my query to return:
2018-07-10 | 2018-07-16
2018-07-21 | 2018-07-22
2018-07-27 | 9999-12-31
I'm not really sure where to start. Is this possible?
You can do that using the lag() function in MS SQL (but that is available starting with 2012?).
with myData as
(
select *,
lag(end_dt,1) over (order by start_dt) as lagEnd
from myTable),
myMax as
(
select Max(end_dt) as maxDate from myTable
)
select dateadd(d,1,lagEnd) as StartDate, dateadd(d, -1, start_dt) as EndDate
from myData
where lagEnd is not null and dateadd(d,1,lagEnd) < start_dt
union all
select dateAdd(d,1,maxDate) as StartDate, cast('99991231' as Datetime) as EndDate
from myMax
where maxDate < '99991231';
If lag() is not available in MS SQL 2008, then you can mimic it with row_number() and joining.
select
CASE WHEN DATEDIFF(day, end_dt, ISNULL(LEAD(start_dt) over (order by ID), '99991231')) > 1 then end_dt +1 END as F1,
CASE WHEN DATEDIFF(day, end_dt, ISNULL(LEAD(start_dt) over (order by ID), '99991231')) > 1 then ISNULL(LEAD(start_dt) over (order by ID) - 1, '99991231') END as F2
from t
Working SQLFiddle example is -> Here
FOR 2008 VERSION
SELECT
X.end_dt + 1 as F1,
ISNULL(Y.start_dt-1, '99991231') as F2
FROM t X
LEFT JOIN (
SELECT
*
, (SELECT MAX(ID) FROM t WHERE ID < A.ID) as ID2
FROM t A) Y ON X.ID = Y.ID2
WHERE DATEDIFF(day, X.end_dt, ISNULL(Y.start_dt, '99991231')) > 1
Working SQLFiddle example is -> Here
This should work in 2008, it assumes that ranges in your table do not overlap. It will also eliminate rows where the end_date of the current row is a day before the start date of the next row.
with dtRanges as (
select start_dt, end_dt, row_number() over (order by start_dt) as rownum
from table1
)
select t2.end_dt + 1, coalesce(start_dt_next -1,'99991231')
FROM
( select dr1.start_dt, dr1.end_dt,dr2.start_dt as start_dt_next
from dtRanges dr1
left join dtRanges dr2 on dr2.rownum = dr1.rownum + 1
) t2
where
t2.end_dt + 1 <> coalesce(start_dt_next,'99991231')
http://sqlfiddle.com/#!18/65238/1
SELECT
*
FROM
(
SELECT
end_dt+1 AS start_dt,
LEAD(start_dt-1, 1, '9999-12-31')
OVER (ORDER BY start_dt)
AS end_dt
FROM
yourTable
)
gaps
WHERE
gaps.end_dt >= gaps.start_dt
I would, however, strongly urge you to use end dates that are "exclusive". That is, the range is everything up to but excluding the end_dt.
That way, a range of one day becomes '2018-07-09', '2018-07-10'.
It's really clear that my range is one day long, if you subtract one from the other you get a day.
Also, if you ever change to needing hour granularity or minute granularity you don't need to change your data. It just works. Always. Reliably. Intuitively.
If you search the web you'll find plenty of documentation on why inclusive-start and exclusive-end is a very good idea from a software perspective. (Then, in the query above, you can remove the wonky +1 and -1.)
This solves your case, but provide some sample data if there will ever be overlaps, fringe cases, etc.
Take one day after your end date and 1 day before the next line's start date.
DECLARE # TABLE (ID int, start_dt DATETIME, end_dt DATETIME, prc VARCHAR(100))
INSERT INTO # (id, start_dt, end_dt, prc)
VALUES
(10410, '2018-07-09 00:00:00.00','2018-07-12 00:00:00.000','1025.000000'),
(10412, '2018-07-17 00:00:00.00','2018-07-20 00:00:00.000','1050.000000'),
(10413, '2018-07-23 00:00:00.00','2018-07-26 00:00:00.000','1040.000000')
SELECT DATEADD(DAY, 1, end_dt)
, DATEADD(DAY, -1, LEAD(start_dt, 1, '9999-12-31') OVER(ORDER BY id) )
FROM #
You may want to take a look at this:
http://sqlfiddle.com/#!18/3a224/1
You just have to edit the begin range to today and the end range to 9999-12-31.
Related
I am trying to create a new column that returns the last date within 1 year of the first date.
Example:
I have the following dates.
5/6/2011
8/9/2011
3/5/2012
6/8/2012
So the query should pick 3/5/2012 as the last date in this scenario.
One method uses window functions:
select max(dt)
from (select t.*, min(dt) over () as min_dt
from t
) t
where dt < dateadd(year, 1, min_dt);
I think I prefer a correlated subquery, though:
select max(dt)
from t
where dt < (select dateadd(year, 1, min(dt)) from t);
You can pretty much translate your English spec into an sql for this one:
SELECT max(d)
FROM t
WHERE d < (SELECT DATEADD(year, 1, MIN(d)) FROM t)
Assume your column name is dt and your table name is Tbl
SELECT MAX(dt)
FROM Tbl
WHERE dt < (SELECT MIN(dt) + 365 FROM Tbl)
We are trying to port a code to run on Amazon Redshift, but Refshift won't run the recursive CTE function. Any good soul that knows how to port this?
with tt as (
select t.*, row_number() over (partition by id order by time) as seqnum
from t
),
recursive cte as (
select t.*, time as grp_start
from tt
where seqnum = 1
union all
select tt.*,
(case when tt.time < cte.grp_start + interval '3 second'
then tt.time
else tt.grp_start
end)
from cte join
tt
on tt.seqnum = cte.seqnum + 1
)
select cte.*,
(case when grp_start = lag(grp_start) over (partition by id order by time)
then 0 else 1
end) as isValid
from cte;
Or, a different code to reproduce the logic below.
It is a binary result that:
it is 1 if it is the first known value of an ID
it is 1 if it is 3 seconds or later than the previous "1" of that ID
It is 0 if it is less than 3 seconds than the previous "1" of that ID
Note 1: this is not the difference in seconds from the previous record
Note 2: there are many IDs in the data set
Note 3: original dataset has ID and Date
Desired output:
https://i.stack.imgur.com/k4KUQ.png
Dataset poc:
http://www.sqlfiddle.com/#!15/41d4b
As of this writing, Redshift does support recursive CTE's: see documentation here
To note when creating a recursive CTE in Redshift:
start the query: with recursive
column names must be declared for all recursive cte's
Consider the following example for creating a list of dates using recursive CTE's:
with recursive
start_dt as (select current_date s_dt)
, end_dt as (select dateadd(day, 1000, current_date) e_dt)
-- the recusive cte, note declaration of the column `dt`
, dates (dt) as (
-- start at the start date
select s_dt dt from start_dt
union all
-- recursive lines
select dateadd(day, 1, dt)::date dt -- converted to date to avoid type mismatch
from dates
where dt <= (select e_dt from end_dt) -- stop at the end date
)
select *
from dates
The below code could help you.
SELECT id, time, CASE WHEN sec_diff is null or prev_sec_diff - sec_diff > 3
then 1
else 0
end FROM (
select id, time, sec_diff, lag(sec_diff) over(
partition by id order by time asc
)
as prev_sec_diff
from (
select id, time, date_part('s', time - lag(time) over(
partition by id order by time asc
)
)
as sec_diff from hon
) x
) y
I have these date ranges that represent start and end dates of subscription. There are no overlaps in date ranges.
Start Date End Date
1/5/2015 - 1/14/2015
1/15/2015 - 1/20/2015
1/24/2015 - 1/28/2015
1/29/2015 - 2/3/2015
I want to identify delays of more than 1 day between any subscription ending and a new one starting. e.g. for the data above, i want the output: 1/24/2015 - 1/28/2015.
How can I do this using a sql query?
Edit : Also there can be multiple gaps in the subscription date ranges but I want the date range after the latest one.
You do this using a left join or not exists:
select t.*
from t
where not exists (select 1
from t t2
where t2.enddate = dateadd(day, -1, t.startdate)
);
Note that this will also give you the first record in the sequence . . . which, strictly speaking, matches the conditions. Here is one solution to that problem:
select t.*
from t cross join
(select min(startdate) as minsd from t) as x
where not exists (select 1
from t t2
where t2.enddate = dateadd(day, -1, t.startdate)
) and
t.startdate <> minsd;
You can also approach this with window functions:
select t.*
from (select t.*,
lag(enddate) over (order by startdate) as prev_enddate,
min(startdate) over () as min_startdate
from t
) t
where minstartdate <> startdate and
enddate <> dateadd(day, -1, startdate);
Also note that this logic assumes that the time periods do not overlap. If they do, a clearer problem statement is needed to understand what you are really looking for.
You can achieve this using window function LAG() that would get value from previous row in ordered set for later comparison in WHERE clause. Then, in WHERE you just apply your "gapping definition" and discard the first row.
SQL FIDDLE - Test it!
Sample data:
create table dates(start_date date, end_date date);
insert into dates values
('2015-01-05','2015-01-14'),
('2015-01-15','2015-01-20'),
('2015-01-24','2015-01-28'), -- gap
('2015-01-29','2015-02-03'),
('2015-02-04','2015-02-07'),
('2015-02-09','2015-02-11'); -- gap
Query
SELECT
start_date,
end_date
FROM (
SELECT
start_date,
end_date,
LAG(end_date, 1) OVER (ORDER BY start_date) AS prev_end_date
FROM dates
) foo
WHERE
start_date IS DISTINCT FROM ( prev_end_date + 1 ) -- compare current row start_date with previous row end_date + 1 day
AND prev_end_date IS NOT NULL -- discard first row, which has null value in LAG() calculation
I assume that there are no overlaps in your data and that there are unique values for each pair. If that's not the case, you need to clarify this.
What I need is to calculate the missing time periods within the calendar year given a table such as this in SQL:
DatesTable
|ID|DateStart |DateEnd |
1 NULL NULL
2 2015-1-1 2015-12-31
3 2015-3-1 2015-12-31
4 2015-1-1 2015-9-30
5 2015-1-1 2015-3-31
5 2015-6-1 2015-12-31
6 2015-3-1 2015-6-30
6 2015-7-1 2015-10-31
Expected return would be:
1 2015-1-1 2015-12-31
3 2015-1-1 2015-2-28
4 2015-10-1 2015-12-31
5 2015-4-1 2015-5-31
6 2015-1-1 2015-2-28
6 2015-11-1 2015-12-31
It's essentially work blocks. What I need to show is the part of the calendar year which was NOT worked. So for ID = 3, he worked from 3/1 through the rest of the year. But he did not work from 1/1 till 2/28. That's what I'm looking for.
You can do it using LEAD, LAG window functions available from SQL Server 2012+:
;WITH CTE AS (
SELECT ID,
LAG(DateEnd) OVER (PARTITION BY ID ORDER BY DateEnd) AS PrevEnd,
DateStart,
DateEnd,
LEAD(DateStart) OVER (PARTITION BY ID ORDER BY DateEnd) AS NextStart
FROM DatesTable
)
SELECT ID, DateStart, DateEnd
FROM (
-- Get interval right before current [DateStart, DateEnd] interval
SELECT ID,
CASE
WHEN DateStart IS NULL THEN '20150101'
WHEN DateStart > start THEN start
ELSE NULL
END AS DateStart,
CASE
WHEN DateStart IS NULL THEN '20151231'
WHEN DateStart > start THEN DATEADD(d, -1, DateStart)
ELSE NULL
END AS DateEnd
FROM CTE
CROSS APPLY (SELECT COALESCE(DATEADD(d, 1, PrevEnd), '20150101')) x(start)
-- If there is no next interval then get interval right after current
-- [DateStart, DateEnd] interval (up-to end of year)
UNION ALL
SELECT ID, DATEADD(d, 1, DateEnd) AS DateStart, '20151231' AS DateEnd
FROM CTE
WHERE DateStart IS NOT NULl -- Do not re-examine [Null, Null] interval
AND NextStart IS NULL -- There is no next [DateStart, DateEnd] interval
AND DateEnd < '20151231' -- Current [DateStart, DateEnd] interval
-- does not terminate on 31/12/2015
) AS t
WHERE t.DateStart IS NOT NULL
ORDER BY ID, DateStart
The idea behind the above query is simple: for every [DateStart, DateEnd] interval get 'not worked' interval right before it. If there is no interval following the current interval, then also get successive 'not worked' interval (if any).
Also note that I assume that if DateStart is NULL then DateStart is also NULL for the same ID.
Demo here
If your data is not too big, this approach will work. It expands all the days and ids and then re-groups them:
with d as (
select cast('2015-01-01' as date)
union all
select dateadd(day, 1, d)
from d
where d < cast('2015-12-31' as date)
),
td as (
select *
from d cross join
(select distinct id from t) t
where not exists (select 1
from t t2
where d.d between t2.startdate and t2.enddate
)
)
select id, min(d) as startdate, max(d) as enddate
from (select td.*,
dateadd(day, - row_number() over (partition by id order by d), d) as grp
from td
) td
group by id, grp
order by id, grp;
An alternative method relies on cumulative sums and similar functionality that is much easier to expression in SQL Server 2012+.
Somewhat simpler approach I think.
Basically create a list of dates for all work block ranges (A). Then create a list of dates for the whole year for each ID (B). Then remove the A from B. Compile the remaining list of dates into date ranges for each ID.
DECLARE #startdate DATETIME, #enddate DATETIME
SET #startdate = '2015-01-01'
SET #enddate = '2015-12-31'
--Build date ranges from remaining date list
;WITH dateRange(ID, dates, Grouping)
AS
(
SELECT dt1.id, dt1.Dates, dt1.Dates + row_number() over (order by dt1.id asc, dt1.Dates desc) AS Grouping
FROM
(
--Remove (A) from (B)
SELECT distinct dt.ID, tmp.Dates FROM DatesTable dt
CROSS APPLY
(
--GET (B) here
SELECT DATEADD(DAY, number, #startdate) [Dates]
FROM master..spt_values
WHERE type = 'P' AND DATEADD(DAY, number, #startdate) <= #enddate
) tmp
left join
(
--GET (A) here
SELECT DISTINCT T.Id,
D.Dates
FROM DatesTable AS T
INNER JOIN master..spt_values as N on N.number between 0 and datediff(day, T.DateStart, T.DateEnd)
CROSS APPLY (select dateadd(day, N.number, T.DateStart)) as D(Dates)
WHERE N.type ='P'
) dr
ON dr.Id = dt.Id and dr.Dates = tmp.Dates
WHERE dr.id is null
) dt1
)
SELECT ID, CAST(MIN(Dates) AS DATE) DateStart, CAST(MAX(Dates) AS DATE) DateEnd
FROM dateRange
GROUP BY ID, Grouping
ORDER BY ID
Heres the code:
http://sqlfiddle.com/#!3/f3615/1
I hope this helps!
I have data like below:
StartDate EndDate Duration
----------
41890 41892 3
41898 41900 3
41906 41907 2
41910 41910 1
StartDate and EndDate are respective ID values for any dates from calendar. I want to calculate the sum of duration for consecutive days. Here I want to include the days which are weekends. E.g. in the above data, let's say 41908 and 41909 are weekends, then my required result set should look like below.
I already have another proc that can return me the next working day, i.e. if I pass 41907 or 41908 or 41909 as DateID in that proc, it will return 41910 as the next working day. Basically I want to check if the DateID returned by my proc when I pass the above EndDateID is same as the next StartDateID from above data, then both the rows should be clubbed. Below is the data I want to get.
ID StartDate EndDate Duration
----------
278457 41890 41892 3
278457 41898 41900 3
278457 41906 41910 3
Please let me know in case the requirement is not clear, I can explain further.
My Date Table is like below:
DateId Date Day
----------
41906 09-04-2014 Thursday
41907 09-05-2014 Friday
41908 09-06-2014 Saturdat
41909 09-07-2014 Sunday
41910 09-08-2014 Monday
Here is the SQL Code for setup:
CREATE TABLE Table1
(
StartDate INT,
EndDate INT,
LeaveDuration INT
)
INSERT INTO Table1
VALUES(41890, 41892, 3),
(41898, 41900, 3),
(41906, 41907, 3),
(41910, 41910, 1)
CREATE TABLE DateTable
(
DateID INT,
Date DATETIME,
Day VARCHAR(20)
)
INSERT INTO DateTable
VALUES(41907, '09-05-2014', 'Friday'),
(41908, '09-06-2014', 'Saturday'),
(41909, '09-07-2014', 'Sunday'),
(41910, '09-08-2014', 'Monday'),
(41911, '09-09-2014', 'Tuesday')
This is rather complicated. Here is an approach using window functions.
First, use the date table to enumerate the dates without weekends (you can also take out holidays if you want). Then, expand the periods into one day per row, by using a non-equijoin.
You can then use a trick to identify sequential days. This trick is to generate a sequential number for each id and subtract it from the sequential number for the dates. This is a constant for sequential days. The final step is simply an aggregation.
The resulting query is something like this:
with d as (
select d.*, row_number() over (order by date) as seqnum
from dates d
where day not in ('Saturday', 'Sunday')
)
select t.id, min(t.date) as startdate, max(t.date) as enddate, sum(duration)
from (select t.*, ds.seqnum, ds.date,
(d.seqnum - row_number() over (partition by id order by ds.date) ) as grp
from table t join
d ds
on ds.date between t.startdate and t.enddate
) t
group by t.id, grp;
EDIT:
The following is the version on this SQL Fiddle:
with d as (
select d.*, row_number() over (order by date) as seqnum
from datetable d
where day not in ('Saturday', 'Sunday')
)
select t.id, min(t.date) as startdate, max(t.date) as enddate, sum(duration)
from (select t.*, ds.seqnum, ds.date,
(ds.seqnum - row_number() over (partition by id order by ds.date) ) as grp
from (select t.*, 'abc' as id from table1 t) t join
d ds
on ds.dateid between t.startdate and t.enddate
) t
group by grp;
I believe this is working, but the date table doesn't have all the dates in it.