Between two date , how can i take all date with one id - sql

empid startdate expirydate
18426 2018-01-01 2018-01-05
i am expecting :-
date id
2018-01-01 18426
2018-01-02 18426
2018-01-03 18426
2018-01-04 18426
2018-01-05 18426


You need cte with recursive way :
with t as (
select empid, startdate, enddate
from table
union all
select empid, dateadd(day, 1, startdate), enddate
from t
where stardate < enddate
)
select *
from t
option (maxrecursion 0);


Use a Calendar Table or a Tally Table.
You can also do it on the fly. This example should get you on the right path:
DECLARE #StartDate date, #EndDate date;
SET #StartDate = '20180101';
SET #EndDate = '20190105';
WITH N AS (
SELECT *
FROM (VALUES(NULL),(NULL),(NULL),(NULL),(NULL),(NULL),(NULL),(NULL),(NULL)) V(U)),
Tally AS(
SELECT ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) -1 AS I
FROM N N1
CROSS JOIN N N2
CROSS JOIN N N3
CROSS JOIN N N4)
SELECT DATEADD(DAY, I, #StartDate) AS [date]
FROM Tally T
WHERE DATEADD(DAY, I, #StartDate) <= #EndDate;

Related

Get an interval of dates from a range of dates

I have two dates 21/10/2019 and 26/6/2031, and from these two I need a list of dates with three months interval starting from the first date like this:
22/10/2019 | 21/01/2020
22/01/2020 | 21/04/2020
22/04/2020 | 21/07/2020
22/07/2020 | 21/10/2020
...
22/01/2031 | 21/04/2031
22/04/2031 | 26/06/2031
I've tried using ROW_NUMBER() and DENSE_RANK() and LAG() to group a complete list of dates between the two dates, but I can't seem to figure it out. I think I might need to partition this somehow, but I can't get it right.
If you don't understand, please let me know. I'm pretty new at this :)

You can use a recursive query:
with cte (dt, end_dt) as (
select #start_dt, #end_dt
union all
select dateadd(month, 3, dt), end_dt from cte where dt < end_dt
)
select dt,
case when dateadd(month, 3, dt) < end_dt
then dateadd(day, -1, dateadd(month, 3, dt))
else end_dt
end as end_dt
from cte
order by dt;
If you need to generate more than 100 quarters, you need to add option (maxrecursion 0) at the very end of the query.
Demo on DB Fiddle

This could also be done using a 'tally' or numbers based approach. The upper limit of the tally_cte is 12^5=248,832 (more than recursion can produce and it could be increased as much as needed).
declare
#start_dt datetime='2019-10-21',
#end_dt datetime='2031-06-26'
;with
n(n) as (select * from (values (1),(2),(3),(4),(5),(6),(7),(8),(9),(10),(11),(12)) v(n)),
tally_cte(n) as (
select row_number() over (order by (select null))
from n n1 cross join n n2 cross join n n3 cross join n n4 cross join n n5)
select t.*, cast(dt.dt as date) start_dt, cast(dateadd(MONTH, 3, dt.dt) as date) end_dt
from tally_cte t
cross apply (select dateadd(month, (n-1)*3, #start_dt) dt) dt
where n<datediff(month, #start_dt, #end_dt)/3;
N start_dt end_dt
1 2019-10-21 2020-01-21
2 2020-01-21 2020-04-21
3 2020-04-21 2020-07-21
...
45 2030-10-21 2031-01-21

In SQL how to calculate days in a year based on a start date and the number of days lapsed

What would be the SQL to calculate the number of days in each year if I had a start date and the number of days that have lapsed?
For example, the date (ymd) 2013-01-01 and the days lapsed is 1000.
I would like the result to look like this
2013 = 365
2014 = 365
2015 = 270
Can this be written as a function like datediff?
I have tried using a calendar table, but of course, linking to this just gives me 2013 = 1000
My calendar table looks like this.
DATE_ID | DATE | CALENDAR_YEAR | FINANCIAL_YEAR
-----------------------------------------------
20130101 | 2013-01-01 | 2013 | 2013/14
This is what i have tried.
SELECT
D.FISCAL_YEAR, SUM([DAYS]) AS NUMBER_OF_DAYS
FROM [dbo].[FACT] F
LEFT JOIN [dbo].[DIM_DATE] D ON D.DATE_ID = F.DATE_ID
GROUP BY
D.FISCAL_YEAR
The result for this is.
FISCAL_YEAR | NUMBER_OF_DAYS
----------------------------
2013/14 |2820
2014/15 |6635
2015/16 |2409

I would personally build a tally table to do this. Once you build that, you can easly get every date and count the number of days in each year:
DECLARE #YMD date = '20130101',
#Lapsed int = 1000;
--Build a Tally table
WITH N AS(
SELECT N
FROM (VALUES(NULL),(NULL),(NULL),(NULL),(NULL),(NULL),(NULL),(NULL),(NULL),(NULL)) N(N)),
Tally AS(
SELECT ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) -1 AS I
FROM N N1, N N2, N N3, N N4), --10,000 should be enough
--Build the dates table
Dates AS(
SELECT DATEADD(DAY, T.I, #YMD) AS CalendarDate
FROM Tally T
WHERE T.I <= #Lapsed - 1)
--And count the days
SELECT DATEPART(YEAR, CalendarDate) AS Year,
COUNT(CalendarDate) AS Days
FROM Dates D
GROUP BY DATEPART(YEAR, CalendarDate);
As a function:
CREATE FUNCTION CountDays (#YMD date, #Lapsed int)
RETURNS table
AS RETURN
--Build a Tally table
WITH N AS(
SELECT N
FROM (VALUES(NULL),(NULL),(NULL),(NULL),(NULL),(NULL),(NULL),(NULL),(NULL),(NULL)) N(N)),
Tally AS(
SELECT ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) -1 AS I
FROM N N1, N N2, N N3, N N4), --10,000 should be enough
--Build the dates table
Dates AS(
SELECT DATEADD(DAY, T.I, #YMD) AS CalendarDate
FROM Tally T
WHERE T.I <= #Lapsed - 1)
--And count the days
SELECT DATEPART(YEAR, CalendarDate) AS Year,
COUNT(CalendarDate) AS Days
FROM Dates D
GROUP BY DATEPART(YEAR, CalendarDate);
GO
SELECT *
FROM (VALUES('20130101',1000),
('20150501',755))V(YMD, Lapsed)
CROSS APPLY dbo.CountDays(V.YMD,V.Lapsed) CD;

One method is a recursive CTE:
with dates as (
select v.d, 1000 - datediff(day, v.d, dateadd(year, 1, v.d)) as days
from (values (datefromparts(2013, 1, 1))) v(d)
union all
select dateadd(year, 1, d), days - datediff(day, d, dateadd(year, 1, d))
from dates
where days > 0
)
select d,
(case when days > 0 then datediff(day, d, dateadd(year, 1, d))
else datediff(day, d, dateadd(year, 1, d)) + days
end)
from dates;
Here is a db<>fiddle.

Finding missing dates compared to date range

I have one table (A) with date ranges and another (B) with just a set date. There are missing months in B that are within the date range of A. I need to identify the missing months.
A
Person StartDate EndDate
123 1/1/2016 5/1/2016
B
Person EffectiveDate
123 1/1/2016
123 2/1/2016
123 4/1/2016
123 5/1/2016
Expected result would be
123 3/1/2016
I'm using SQL Server 2012. Any assistance would be appreciated. Thanks!

One approach is to generate all values between the two dates. Here is an approach using a numbers table:
with n as (
select row_number() over (order by (select null)) - 1 as n
from master.spt_values
)
select a.person, dateadd(day, n.n, a.startdate) as missingdate
from a join
n
on dateadd(day, n.n, a.startdate) <= day.enddate left join
b
on b.person = a.person and b.effectivedate = dateadd(day, n.n, a.startdate)
where b.person is null;

Try this:
CREATE TABLE #A (Person INT, StartDate DATE, EndDate DATE)
INSERT INTO #A
SELECT '123','1/1/2016', '5/1/2016'
CREATE TABLE #B(Person INT, EffectiveDate DATE)
INSERT INTO #B
SELECT 123 ,'1/1/2016' UNION ALL
SELECT 123 ,'2/1/2016' UNION ALL
SELECT 123 ,'4/1/2016' UNION ALL
SELECT 123 ,'5/1/2016'
;WITH A1
AS(
SELECT PERSON , StartDate, EndDate
FROM #A
UNION ALL
SELECT PERSON ,DATEADD(MM,1,STARTDATE), EndDate
FROM A1
WHERE DATEADD(MM,1,STARTDATE) <= EndDate
)
SELECT PERSON , StartDate
FROM A1
WHERE
NOT EXISTS
(
SELECT 1 FROM #B B1
WHERE B1.Person = A1.PERSON
AND YEAR(B1.EffectiveDate) = YEAR(A1.STARTDATE) AND MONTH(B1.EffectiveDate) = MONTH(A1.STARTDATE)
)

This should work if you are interested in getting missing months
;WITH n
AS (SELECT ROW_NUMBER() OVER(ORDER BY
(
SELECT NULL
)) - 1 AS n
FROM master.dbo.spt_values)
SELECT a.person,
DATEADD(MONTH, n.n, a.startdate) AS missingdate
FROM a a
INNER JOIN n ON DATEADD(MONTH, n.n, a.startdate) <= a.enddate
LEFT JOIN b b ON MONTH(DATEADD(MONTH, n.n, a.startdate)) = MONTH(b.effectivedate) AND YEAR(DATEADD(MONTH, n.n, a.startdate)) = YEAR(b.effectivedate)
WHERE b.person IS NULL;

Split date range to year-month rows on SQL Server 2005

I need to create an output where I got one row per year-month.
Assume the dataset is:
id | dateStart | dateEnd
1 | 2015-01-01 00:00:00.000 | 2015-03-31 00:00:00.000
2 | 2014-07-01 00:00:00.000 | 2014-08-31 00:00:00.000
...
I need the following output:
id | year-month
1 | 2015-01
1 | 2015-02
1 | 2015-03
2 | 2014-07
2 | 2014-08
The output can be any datatype since I can just change that later.
That is for 2015-01 the following is ok, "2015-01-01 00:00:00.000", "2015-01-01", "201501", "2015 | jan" ect.
Note I'm using SQL Server 2005.

Here is a method that uses recursive CTEs:
with CTE as (
select id, dateStart as dte, dateEnd
from t
union all
select id, dateadd(month, 1, dte), dateEnd
from CTE
where dateadd(month, 1, dte) < dateEnd
)
select id, dte
from CTE;
You can convert the final result into any format you like. For instance:
select id, year(dte) * 10000 + month(dte) as yyyymm_int
or
select id, cast(year(dte) * 10000 + month(dte) as varchar(255)) as yyyymm

Generate tally table(just make sure you get enough rows there). tally will contain values 0,1,2,.....n. Then you do a join with condition adding thise values as months to startDate until it is greater then endDate:
DECLARE #t TABLE
(
id INT ,
dateStart DATETIME ,
dateEnd DATETIME
)
INSERT INTO #t
VALUES ( 1, '2015-01-01 00:00:00.000', '2015-03-31 00:00:00.000' ),
( 2, '2014-07-01 00:00:00.000', '2014-08-31 00:00:00.000' )
;WITH cte AS(SELECT -1 + ROW_NUMBER() OVER(ORDER BY t1.m) m
FROM(VALUES(0),(0),(0),(0),(0),(0),(0),(0),(0),(0))t1(m) CROSS JOIN
(VALUES(0),(0),(0),(0),(0),(0),(0),(0),(0),(0))t2(m))
SELECT t.id,
DATEADD(mm, c.m, t.dateStart) AS year_month
FROM cte c
JOIN #t t ON DATEADD(mm, c.m, t.dateStart) <= t.dateEnd
ORDER BY t.id, year_month
Output:
id year_month
1 2015-01-01 00:00:00.000
1 2015-02-01 00:00:00.000
1 2015-03-01 00:00:00.000
2 2014-07-01 00:00:00.000
2 2014-08-01 00:00:00.000

In an ideal world you would have a calendar table, then your query would simply be:
SELECT t.id,
c.FirstDayOfMonth
FROM YourTable AS t
INNER JOIN dbo.Calendar c
ON c.FirstDayOfMonth >= t.DateStart
AND c.FirstDayOfMonth <= t.DateEnd
AND c.DayOfMonth = 1;
Assuming that you don't have a calendar table then you can do it with a list of numbers generated on the fly (Read this article for more on this). The following will generate a list from 1-10,000:
WITH N1 AS (SELECT N FROM (VALUES (1),(1),(1),(1),(1),(1),(1),(1),(1),(1)) N (N)),
N2 (N) AS (SELECT 1 FROM N1 AS N1 CROSS JOIN N1 AS N2),
N3 (N) AS (SELECT 1 FROM N2 AS N1 CROSS JOIN N2 AS N2),
Numbers (Number) AS (SELECT ROW_NUMBER() OVER(ORDER BY N3.N) FROM N3)
SELECT * FROM Numbers;
Then you can join this to your original table:
DECLARE #T TABLE (id INT, DateStart DATE, DateEnd DATE);
INSERT #T (ID, DateStart, DateEnd)
VALUES (1, '20150101', '20150331'), (2, '20140701', '20140831');
WITH N1 AS (SELECT N FROM (VALUES (1),(1),(1),(1),(1),(1),(1),(1),(1),(1)) N (N)),
N2 (N) AS (SELECT 1 FROM N1 AS N1 CROSS JOIN N1 AS N2),
N3 (N) AS (SELECT 1 FROM N2 AS N1 CROSS JOIN N2 AS N2),
Numbers (Number) AS (SELECT ROW_NUMBER() OVER(ORDER BY N3.N) FROM N3)
SELECT t.ID,
[year-month] = DATEADD(MONTH, n.Number + DATEDIFF(MONTH, 0, t.DateStart), 0)
FROM #T AS t
INNER JOIN Numbers AS N
ON N.Number - 1 <= DATEDIFF(MONTH, t.DateStart, t.DateEnd);

How to Select continuous date in sql

Is there any function to check for continuous date. I'm having problem on working with this issue below:
My table has a datetime column with the following data:
----------
2015-03-11
2015-03-12
2015-03-13
2015-03-16
Given start date as 2015-3-11 and end date as 2015-3-17. I want the result as:
----------
2015-03-11
2015-03-12
2015-03-13
Can anyone suggest anything ?

I'm thinking this is somewhat a variation of Grouping Islands of Contiguous Dates problem. This can be done using ROW_NUMBER():
SQL Fiddle
CREATE TABLE Test(
tDate DATETIME
)
INSERT INTO Test VALUES
('20150311'), ('20150312'), ('20150313'), ('20150316');
DECLARE #startDate DATE = '20150311'
DECLARE #endDate DATE = '20150317'
;WITH Cte AS(
SELECT
*,
RN = DATEADD(DD, - (ROW_NUMBER() OVER(ORDER BY tDATE) - 1), tDate)
FROM Test
WHERE
tDate >= #startDate
AND tDate < DATEADD(DAY, 1, #endDate)
)
SELECT CAST(tDate AS DATE)
FROM CTE
WHERE RN = #startDate
RESULT
|------------|
| 2015-03-11 |
| 2015-03-12 |
| 2015-03-13 |
Here is the SQL Server 2005 version:
SQL Fiddle
DECLARE #startDate DATETIME
DECLARE #endDate DATETIME
SET #startDate = '20150311'
SET #endDate = '20150317'
;WITH Cte AS(
SELECT
*,
RN = DATEADD(DD, -(ROW_NUMBER() OVER(ORDER BY tDATE)-1), tDate)
FROM Test
WHERE
tDate >= #startDate
AND tDate < DATEADD(DAY, 1, #endDate)
)
SELECT CONVERT(VARCHAR(10), tDate, 121)
FROM CTE
WHERE RN = #startDate

For MSSQL 2012. This will return MAX continuous groups:
DECLARE #t TABLE(d DATE)
INSERT INTO #t VALUES
('20150311'),
('20150312'),
('20150313'),
('20150316')
;WITH
c1 AS(SELECT d, IIF(DATEDIFF(dd,LAG(d, 1, DATEADD(dd, -1, d)) OVER(ORDER BY d), d) = 1, 0, 1) AS n FROM #t),
c2 AS(SELECT d, SUM(n) OVER(ORDER BY d) AS n FROM c1)
SELECT TOP 1 WITH TIES MIN(d) AS StartDate, MAX(d) AS EndDate, COUNT(*) AS DayCount
FROM c2
GROUP BY n
ORDER BY DayCount desc
Output:
StartDate EndDate DayCount
2015-03-11 2015-03-13 3
For
('20150311'),
('20150312'),
('20150313'),
('20150316'),
('20150317'),
('20150318'),
('20150319'),
('20150320')
Output:
StartDate EndDate DayCount
2015-03-16 2015-03-20 5
Apply filtering in c1 CTE:
c1 AS(SELECT d, IIF(DATEDIFF(dd,LAG(d, 1, DATEADD(dd, -1, d)) OVER(ORDER BY d), d) = 1, 0, 1) AS n FROM #t WHERE d BETWEEN '20150311' AND '20150320'),
For MSSQL 2008:
;WITH
c1 AS(SELECT d, (SELECT MAX(d) FROM #t it WHERE it.d < ot.d) AS pd FROM #t ot),
c2 AS(SELECT d, CASE WHEN DATEDIFF(dd,ISNULL(pd, DATEADD(dd, -1, d)), d) = 1 THEN 0 ELSE 1 END AS n FROM c1),
c3 AS(SELECT d, (SELECT SUM(n) FROM c2 ci WHERE ci.d <= co.d) AS n FROM c2 co)
SELECT TOP 1 WITH TIES MIN(d) AS StartDate, MAX(d) AS EndDate, COUNT(*) AS DayCount
FROM c3
GROUP BY n
ORDER BY DayCount desc

you don't need to declare any start date or end date as other answers says, you need a row_num with datediff function:
create table DateFragTest (cDate date);
insert into DateFragTest
values ('2015-3-11'),
('2015-3-12'),
('2015-3-13'),
('2015-3-16')
with cte as
(select
cDate,
row_number() over (order by cDate ) as rn
from
DateFragTest)
select cDate
from cte t1
where datediff(day,
(select cDate from cte t2 where t2.rn=t1.rn+1),
t1.cDate)<>1
Output:
cDate
2015-03-11
2015-03-12
2015-03-13
SQLFIDDLE DEMO

For sql server 2012-
WITH cte
AS
(
SELECT [datex]
, lead([datex]) OVER ( ORDER BY [datex]) lead_datex
, datediff(dd,[datex],lead([datex]) OVER ( ORDER BY [datex]) ) AS diff
FROM [dbo].[datex]
)
SELECT c.[datex]
FROM [cte] AS c
WHERE diff >=1

Use BETWEEN
The query will go like this:
SELECT *
FROM your_table_name
WHERE your_date_column_name BETWEEN '2015-3-11' AND '2015-3-13'

(dt between x and y) or just (dt >= x and dt <= y).