Searching SQL table for two consecutive missing dates - sql

Question

I want to search through a SQL table and find two consecutive missing dates.
For example, person 1 inserts 'diary' entry on day 1 and day 2, misses day 3 and day 4, and enters an entry on day 5.
I am not posting code because I am not sure of how to do this at all.
Thanks!

Answer 1

This uses a LEVEL aggregate to build the list of calendar dates from the first entry to the last, then uses LAG() to check a given date with the previous date, and then checks that neither of those dates had an associated entry to find those two-day gaps:
With diary as (
select to_date('01/01/2016','dd/mm/yyyy') entry_dt from dual union all
select to_date('02/01/2016','dd/mm/yyyy') entry_dt from dual union all
select to_date('04/01/2016','dd/mm/yyyy') entry_dt from dual union all
--leave two day gap of 5th and 6th
select to_date('07/01/2016','dd/mm/yyyy') entry_dt from dual union all
select to_date('08/01/2016','dd/mm/yyyy') entry_dt from dual union all
select to_date('10/01/2016','dd/mm/yyyy') entry_dt from dual )
select calendar_dt -1, calendar_dt
FROM (
select calendar_dt, entry_dt, lag(entry_dt) over (order by calendar_dt) prev_entry_dt
from diary
RIGHT OUTER JOIN (select min(entry_dt) + lvl as calendar_dt
FROM diary
,(select level lvl
from dual connect by level < (select max(entry_dt) - min(entry_dt)+1 from diary))
group by lvl) ON calendar_dt = entry_dt
order by calendar_dt
)
where entry_dt is null and prev_entry_dt is null
returns:
CALENDAR_DT-1, CALENDAR_DT
05/01/2016, 06/01/2016
I am only doing the calendar building to simplify building all 2-day gaps, as if a person took three days off that would be two overlapping two-day gaps (day 1-2, and days 2-3). If you want a far simpler query that outputs the start and end point of any gap of two or more days, then the following works:
With diary as (
select to_date('01/01/2016','dd/mm/yyyy') entry_dt from dual union all
select to_date('02/01/2016','dd/mm/yyyy') entry_dt from dual union all
select to_date('04/01/2016','dd/mm/yyyy') entry_dt from dual union all
select to_date('07/01/2016','dd/mm/yyyy') entry_dt from dual union all
select to_date('08/01/2016','dd/mm/yyyy') entry_dt from dual union all
select to_date('10/01/2016','dd/mm/yyyy') entry_dt from dual )
select prev_entry_dt +1 gap_start, entry_dt -1 gap_end
FROM (
select entry_dt, lag(entry_dt) over (order by entry_dt) prev_entry_dt
from diary
order by entry_dt
) where entry_dt - prev_entry_dt > 2

Answer 2

My high level approach to this problem would be to select from a dynamic table of dates, using an integer counter to add or subtract from the current DateTime to get as many dates as you require into the future or past, then LEFT join your data table to this, order by date and select the first row, or N many rows which have a NULL join.
So your data ends up being
DATE ENTRY_ID
---- -----
2016-01-01 1
2016-01-02 2
2016-01-03 NULL
2016-01-04 3
2016-01-05 4
2016-01-06 NULL
2016-01-07 NULL
2016-01-08 NULL
And you can pick all of the values you need from this dataset

Answer 3

Try this your problem looks like similar to this :-
Declare #temp Table(id int identity(1,1) not null,CDate smalldatetime ,val int)
insert into #temp select '10/2/2012',1
insert into #temp select '10/3/2012',1
insert into #temp select '10/5/2012',1
insert into #temp select '10/7/2012',2
insert into #temp select '10/9/2012',2
insert into #temp select '10/10/2012',2
insert into #temp select '10/13/2012',2
insert into #temp select '10/15/2012',2
DECLARE #startDate DATE= '10/01/2012'
DECLARE #endDate DATE= '10/15/2012'
SELECT t.Id, X.[Date],Val = COALESCE(t.val,0)
FROM
(SELECT [Date] = DATEADD(Day,Number,#startDate)
FROM master..spt_values
WHERE Type='P'
AND DATEADD(day,Number,#startDate) <= #endDate)X
LEFT JOIN #temp t
ON X.[Date] = t.CDate
Alternative you can try this :-
WITH dates AS (
SELECT CAST('2009-01-01' AS DATETIME) 'date'
UNION ALL
SELECT DATEADD(dd, 1, t.date)
FROM dates t
WHERE DATEADD(dd, 1, t.date) <= '2009-02-01')
SELECT t.eventid, d.date
FROM dates d
JOIN TABLE t ON d.date BETWEEN t.startdate AND t.enddate