I am loading data into a table. I don't have any info on how frequent or when the source data is loaded, all I know is I need data from the source to run my script.
Here's the issue, if I run max(date) I get the latest date from the source, but I don't know if the data is still loading. I've ran into cases where I've only gotten a percentage of the data. Thus, I need the next business day after max date.
I want to know is there a way to get the second latest date in the system. I know I can get max(date) - 1, but that give me literally the day after. I don't need the literal day after.
Example, if I run the script on Tuesday, max(date) will be Monday, but since weekend are not in the source system, I need to get Friday instead of Monday.
DATE
---------
2017-04-29
2017-04-25
2017-04-21
2017-04-19
2017-04-18
2017-04-15
2017-04-10
max(date) = 2017-04-29
how do I get 2017-04-25?
Depending on your version of SQL Server, you can use a windowing function like row_number:
select [Date]
from
(
select [Date],
rn = row_number() over(order by [Date] desc)
from #yourtable
) d
where rn = 2
Here is a demo.
Should you have multiple of the same date, you can perform a distinct first:
;with cte as
(
select distinct [date]
from #yourtable
)
select [date]
from
(
select [date],
rn = row_number() over(order by [date] desc)
from cte
) x
where rn = 2;
You can use row_number and get second as below
select * from ( select *, Rown= row_number() over (order by date desc) from yourtable ) a
where a.RowN = 2
More recent SQL Server versions support FETCH FIRST:
select date
from tablename
order by date desc
offset 1 fetch first 1 row only
OFFSET 1 means skip one row. (The 2017-04-29 row.)
;With cte([DATE])
AS
(
SELECT '2017-04-29' union all
SELECT '2017-04-25' union all
SELECT '2017-04-21' union all
SELECT '2017-04-19' union all
SELECT '2017-04-18' union all
SELECT '2017-04-15' union all
SELECT '2017-04-10'
)
SELECT [DATE] FROM
(
SELECT *,ROW_NUMBER()OVER(ORDER BY Seq)-1 As Rno FROM
(
SELECT *,MAX([DATE])OVER(ORDER BY (SELECT NULL))Seq FROM cte
)dt
)Final
WHERE Final.Rno=1
OutPut
DATE
-----
2017-04-25
You can also use FIRST_VALUE with a dynamic date something like DATEADD(DD, -1, GETDATE()). The example below has the date hard coded.
SELECT DISTINCT
FIRST_VALUE([date]) OVER(ORDER BY [date] DESC) AS FirstDate
FROM CTE
WHERE [date] < '2017-04-25'
Another way
DECLARE #T TABLE ([DATE] DATE)
INSERT INTO #T VALUES
('2017-04-29'),
('2017-04-25'),
('2017-04-21'),
('2017-04-19'),
('2017-04-18'),
('2017-04-15'),
('2017-04-10');
SELECT
MAX([DATE]) AS [DATE]
FROM #T
WHERE DATENAME(DW,[DATE]) NOT IN ('Saturday','Sunday')
Another way of doing it, just for example sake...
SELECT MIN(A.date)
FROM
(
SELECT TOP 2 DISTINCT date
FROM YourTable AS C
ORDER BY date DESC
) AS A
Related
I am explain problem in short.
select distinct DATE from #Table where DATE >='2016-01-01'
Output :
Date
2016-11-23
2016-11-22
2016-11-21
2016-11-19
2016-11-18
Now i need to find out missing date a compare to our calender dates from year '2016'
i.e. Here date '2016-11-20' is missing.
I want list of missing dates.
Thanks for reading this. Have nice day.
You need to generate dates and you have to find missing ones. Below with recursive cte i have done it
;WITH CTE AS
(
SELECT CONVERT(DATE,'2016-01-01') AS DATE1
UNION ALL
SELECT DATEADD(DD,1,DATE1) FROM CTE WHERE DATE1<'2016-12-31'
)
SELECT DATE1 MISSING_ONE FROM CTE
EXCEPT
SELECT * FROM #TABLE1
option(maxrecursion 0)
Using CTE and get all dates in CTE table then compare with your table.
CREATE TABLE #yourTable(_Values DATE)
INSERT INTO #yourTable(_Values)
SELECT '2016-11-23' UNION ALL
SELECT '2016-11-22' UNION ALL
SELECT '2016-11-21' UNION ALL
SELECT '2016-11-19' UNION ALL
SELECT '2016-11-18'
DECLARE #DATE DATE = '2016-11-01'
;WITH CTEYear (_Date) AS
(
SELECT #DATE
UNION ALL
SELECT DATEADD(DAY,1,_Date)
FROM CTEYear
WHERE _Date < EOMONTH(#DATE,0)
)
SELECT * FROM CTEYear
WHERE NOT EXISTS(SELECT 1 FROM #yourTable WHERE _Date = _Values)
OPTION(maxrecursion 0)
You need to generate the dates and then find the missing ones. A recursive CTE is one way to generate a handful of dates. Another way is to use master..spt_values as a list of numbers:
with n as (
select row_number() over (order by (select null)) - 1 as n
from master..spt_values
),
d as (
select dateadd(day, n.n, cast('2016-01-01' as date)) as dte
from n
where n <= 365
)
select d.date
from d left join
#table t
on d.dte = t.date
where t.date is null;
If you are happy enough with ranges of missing dates, you don't need a list of dates at all:
select date, (datediff(day, date, next_date) - 1) as num_missing
from (select t.*, lead(t.date) over (order by t.date) as next_date
from #table t
where t.date >= '2016-01-01'
) t
where next_date <> dateadd(day, 1, date);
I have an sql table like that:
Id Date Price
1 21.09.09 25
2 31.08.09 16
1 23.09.09 21
2 03.09.09 12
So what I need is to get min and max date for each id and dif in days between them. It is kind of easy. Using SQLlite syntax:
SELECT id,
min(date),
max(date),
julianday(max(date)) - julianday(min(date)) as dif
from table group by id
Then the tricky one: how can I receive the price per day during this difference period. I mean something like this:
ID Date PricePerDay
1 21.09.09 25
1 22.09.09 0
1 23.09.09 21
2 31.08.09 16
2 01.09.09 0
2 02.09.09 0
2 03.09.09 12
I create a cte as you mentioned with calendar but dont know how to get the desired result:
WITH RECURSIVE
cnt(x) AS (
SELECT 0
UNION ALL
SELECT x+1 FROM cnt
LIMIT (SELECT ((julianday('2015-12-31') - julianday('2015-01-01')) + 1)))
SELECT date(julianday('2015-01-01'), '+' || x || ' days') as date FROM cnt
p.s. If it will be in sqllite syntax-would be awesome!
You can use a recursive CTE to calculate all the days between the min date and max date. The rest is just a left join and some logic:
with recursive cte as (
select t.id, min(date) as thedate, max(date) as maxdate
from t
group by id
union all
select cte.id, date(thedate, '+1 day') as thedate, cte.maxdate
from cte
where cte.thedate < cte.maxdate
)
select cte.id, cte.date,
coalesce(t.price, 0) as PricePerDay
from cte left join
t
on cte.id = t.id and cte.thedate = t.date;
One method is using a tally table.
To build a list of dates and join that with the table.
The date stamps in the DD.MM.YY format are first changed to the YYYY-MM-DD date format.
To make it possible to actually use them as a date in the SQL.
At the final select they are formatted back to the DD.MM.YY format.
First some test data:
create table testtable (Id int, [Date] varchar(8), Price int);
insert into testtable (Id,[Date],Price) values (1,'21.09.09',25);
insert into testtable (Id,[Date],Price) values (1,'23.09.09',21);
insert into testtable (Id,[Date],Price) values (2,'31.08.09',16);
insert into testtable (Id,[Date],Price) values (2,'03.09.09',12);
The SQL:
with Digits as (
select 0 as n
union all select 1
union all select 2
union all select 3
union all select 4
union all select 5
union all select 6
union all select 7
union all select 8
union all select 9
),
t as (
select Id,
('20'||substr([Date],7,2)||'-'||substr([Date],4,2)||'-'||substr([Date],1,2)) as [Date],
Price
from testtable
),
Dates as (
select Id, date(MinDate,'+'||(d2.n*10+d1.n)||' days') as [Date]
from (
select Id, min([Date]) as MinDate, max([Date]) as MaxDate
from t
group by Id
) q
join Digits d1
join Digits d2
where date(MinDate,'+'||(d2.n*10+d1.n)||' days') <= MaxDate
)
select d.Id,
(substr(d.[Date],9,2)||'.'||substr(d.[Date],6,2)||'.'||substr(d.[Date],3,2)) as [Date],
coalesce(t.Price,0) as Price
from Dates d
left join t on (d.Id = t.Id and d.[Date] = t.[Date])
order by d.Id, d.[Date];
The recursive SQL below was totally inspired by the excellent answer from Gordon Linoff.
And a recursive SQL is probably more performant for this anyway.
(He should get the 15 points for the accepted answer).
The difference in this version is that the datestamps are first formatted to YYYY-MM-DD.
with t as (
select Id,
('20'||substr([Date],7,2)||'-'||substr([Date],4,2)||'-'||substr([Date],1,2)) as [Date],
Price
from testtable
),
cte as (
select Id, min([Date]) as [Date], max([Date]) as MaxDate from t
group by Id
union all
select Id, date([Date], '+1 day'), MaxDate from cte
where [Date] < MaxDate
)
select cte.Id,
(substr(cte.[Date],9,2)||'.'||substr(cte.[Date],6,2)||'.'||substr(cte.[Date],3,2)) as [Date],
coalesce(t.Price, 0) as PricePerDay
from cte
left join t
on (cte.Id = t.Id and cte.[Date] = t.[Date])
order by cte.Id, cte.[Date];
I have a query as following:
SELECT [Date] FROM [TableX] ORDER BY [Date]
The result is:
2016-06-01
2016-06-03
2016-06-10
2016-06-11
How can I get following pairs?
From To
2016-06-01 2016-06-03
2016-06-03 2016-06-10
2016-06-10 2016-06-11
If you're using SQL Server 2012 or later, you can use the LEAD method.
Accesses data from a subsequent row in the same result set without the use of a self-join in SQL Server 2016. LEAD provides access to a row at a given physical offset that follows the current row.
I think it would look like this for you:
SELECT [Date] AS [From], LEAD([Date], 1) OVER (ORDER BY [Date]) AS [To]
FROM TableX
ORDER BY [Date]
Note that on the last row, the [To] field will be NULL. If you wanted to remove that row, you could put it in an inner query:
SELECT *
FROM
(
SELECT [Date] AS [From], LEAD([Date], 1) OVER (ORDER BY [Date]) AS [To]
FROM TableX
) x
WHERE [To] IS NOT NULL
All you need to do is add a row number for each date.
Then unite all these rows by the next row (except the last row)
WITH cteDates AS
(
SELECT [Date],
ROW_NUMBER() OVER (ORDER BY (SELECT [Date])) As RowNum
FROM TableX
)
SELECT TOP(SELECT COUNT(*) - 1 FROM cteDates)
[Date] [From],
(SELECT [Date] FROM cteDates WHERE RowNum = d.RowNum + 1) [To]
FROM cteDates d
A little tricky solution for SQL 2008.
declare #tbl table(dt datetime)
insert #tbl values
('2016-06-01'),
('2016-06-03'),
('2016-06-10'),
('2016-06-11')
;with cte as (
select dt, ROW_NUMBER() over(order by dt) rn --add number
from #tbl
),
newTbl as (
select t1.dt start, t2.dt [end]
from cte t1 inner join cte t2 on t1.rn+1=t2.rn
)
select *
from newTbl
The result is what you wish.
Since there are never any gaps as you stated, you can just used DATEADD()
SELECT DISTINCT
[Date] as [FROM],
DATEADD(DAY,1,[Date]) as [TO]
FROM TableX
ORDER BY [Date] DESC
Below is the data I have
date value
30/03/2014 625949
23/03/2014 624549
16/03/2014 623149
09/03/2014 621549
02/03/2014 619749
23/02/2014 617749
16/02/2014 616149
09/02/2014 614549
02/02/2014 612949
19/01/2014 609749
12/01/2014 608149
06/01/2014 606749
if I want to get only the last value in a month, for example output like below, how I can get it?
date value
30/03/2014 625949
23/02/2014 617749
19/01/2014 609749
;with Cte as(Select tDate,value, C=ROW_NUMBER()
over(PARTITION by convert(varchar(6), tdate, 112) order by tdate desc)
From #Temp )
Select *
from cte
where C=1
Fiddle Sample
SELECT date,value FROM table1 WHERE date in (
SELECT MAX(date) FROM table1 GROUP BY MONTH(date)
)
I have the below.
Name Date
A 2011-01-01 01:00:00.000
A 2011-02-01 02:00:00.000
A 2011-03-01 03:00:00.000
B 2011-04-01 04:00:00.000
A 2011-05-01 07:00:00.000
The desired output is
Name StartDate EndDate
-------------------------------------------------------------------
A 2011-01-01 01:00:00.000 2011-04-01 04:00:00.000
B 2011-04-01 04:00:00.000 2011-05-01 07:00:00.000
A 2011-05-01 07:00:00.000 NULL
How to achieve the same using TSQL in a set based approach.
DDL is as under
DECLARE #t TABLE(PersonName VARCHAR(32), [Date] DATETIME)
INSERT INTO #t VALUES('A', '2011-01-01 01:00:00')
INSERT INTO #t VALUES('A', '2011-01-02 02:00:00')
INSERT INTO #t VALUES('A', '2011-01-03 03:00:00')
INSERT INTO #t VALUES('B', '2011-01-04 04:00:00')
INSERT INTO #t VALUES('A', '2011-01-05 07:00:00')
Select * from #t
;WITH cte1
AS (SELECT *,
ROW_NUMBER() OVER (ORDER BY Date) -
ROW_NUMBER() OVER (PARTITION BY PersonName
ORDER BY Date) AS G
FROM #t),
cte2
AS (SELECT PersonName,
MIN([Date]) StartDate,
ROW_NUMBER() OVER (ORDER BY MIN([Date])) AS rn
FROM cte1
GROUP BY PersonName,
G)
SELECT a.PersonName,
a.StartDate,
b.StartDate AS EndDate
FROM cte2 a
LEFT JOIN cte2 b
ON a.rn + 1 = b.rn
Because the result of CTEs are not generally materialised however
you may well find you get better performance if you materialize the
intermediate result yourself as below.
DECLARE #t2 TABLE (
rn INT IDENTITY(1, 1) PRIMARY KEY,
PersonName VARCHAR(32),
StartDate DATETIME );
INSERT INTO #t2
SELECT PersonName,
MIN([Date]) StartDate
FROM (SELECT *,
ROW_NUMBER() OVER (ORDER BY Date) -
ROW_NUMBER() OVER (PARTITION BY PersonName
ORDER BY Date) AS G
FROM #t) t
GROUP BY PersonName,
G
ORDER BY StartDate
SELECT a.PersonName,
a.StartDate,
b.StartDate AS EndDate
FROM #t2 a
LEFT JOIN #t2 b
ON a.rn + 1 = b.rn
SELECT
PersonName,
StartDate = MIN(Date),
EndDate
FROM (
SELECT
PersonName,
Date,
EndDate = (
/* get the earliest date after current date
associated with a different person */
SELECT MIN(t1.Date)
FROM #t AS t1
WHERE t1.Date > t.Date
AND t1.PersonName <> t.PersonName
)
FROM #t AS t
) s
GROUP BY PersonName, EndDate
ORDER BY 2
Basically, for every Date we find the nearest date after it such that is associated with a different PersonName. That gives us EndDate, which now distinguishes for us consecutive groups of dates for the same person.
Now we only need to group the data by PersonName & EndDate and get the minimal Date in every group as StartDate. And yes, sort the data by StartDate, of course.
Get a row number so you will know where the previous record is. Then, take a record and the next record after it. When the state changes we have a candidate row.
select
state,
min(start_timestamp),
max(end_timestamp)
from
(
select
first.state,
first.timestamp_ as start_timestamp,
second.timestamp_ as end_timestamp
from
(
select
*, row_number() over (order by timestamp_) as id
from test
) as first
left outer join
(
select
*, row_number() over (order by timestamp_) as id
from test
) as second
on
first.id = second.id - 1
and first.state != second.state
) as agg
group by state
having max(end_timestamp) is not null
union
-- last row wont have a ending row
--(select state, timestamp_, null from test order by timestamp_ desc limit 1)
-- I think it something like this for sql server
(select top state, timestamp_, null from test order by timestamp_ desc)
order by 2
;
Tested with PostgreSQL but should work with SQL Server as well
The other answer with the cte is a good one. Another option would be to iterate over the collection in any case. It's not set based, but it is another way to do it.
You will need to iterate to either A. assign a unique id to each record that corresponds to its transaction, or B. to actually get your output.
TSQL is not ideal for iterating over records, especially if you have a lot, and so I would recommend some other way of doing it, a small .net program or something that is better at iterating.
There's a very quick way to do this using a bit of Gaps and Islands theory:
WITH CTE as (SELECT PersonName, [Date]
, Row_Number() over (ORDER BY [Date])
- Row_Number() over (ORDER BY PersonName, [Date]) as Island
FROM #t)
Select PersonName, Min([Date]), Max([Date])
from CTE
GROUP BY Island, PersonName
ORDER BY Min([Date])