sql server - Filling in missing ranges in a set -


this question on filing missing ranges of data. specifically, have result set each row contains startdate , enddate value. let's have:

start     end 1/15      1/20 1/12      3/15

i need query produces adds following row data:

1/21      2/11

most other related questions filling gaps know set (like list of dates). case i'm looking start/end of missing data.

assuming meant write 1/21 instead of 2/21, here's way it:

with dates (     select '2016-01-15' dtstart, '2016-01-20' dtend union     select '2016-02-12', '2016-03-15' union     select '2016-03-21', '2016-04-11' ),  calcs (     select          dateadd(day, 1, dtend) rangestart,         (select dateadd(day, -1, min(dtstart))              dates d2 d2.dtstart > d.dtend) rangeend     dates d )  select * calcs c c.rangeend >= c.rangestart

the table dates 3 rows of sample dates. in thecalcs table, rangestart column next day after each dtend. rangeend column takes next dtstart , subtracts day. finally, last row have null because there missing range after last dtend, ignore row null rangeend value.

edit: if you're unfamiliar with statement in code, it's cte. used here quick way create table sample data in (dates) , place store calculations (calcs).

edit2: since mentioned using join in comment, here's way that:

with dates (     select '2016-01-15' dtstart, '2016-01-20' dtend union     select '2016-02-12', '2016-03-15' union     select '2016-03-21', '2016-04-11' ),  calcs (     select         dateadd(day, 1, d1.dtend) rangestart,         dateadd(day, -1, min(d2.dtstart)) rangeend     dates d1     join dates d2 on d1.dtend < d2.dtstart     group d1.dtend )  select * calcs datediff(day, rangestart, rangeend) >= 0

edit3: updated datediff inequality include ranges of 1 day


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