I have a table like this:
Id user
-------
1 A
2 B
I want to extend it and add date in a range for every row like below:
Id user date
------------------
1 A 20190101
1 A 20190102
2 B 20190101
2 B 20190102
A simple cross join with a calendar table should work here:
WITH dates AS (
SELECT '20190101' AS dt UNION ALL
SELECT '20190102'
)
SELECT
t.Id,
t.user,
d.dt AS date
FROM yourTable t
CROSS JOIN dates d;
You can use a lateral join. In SQL Server, this uses the cross apply syntax:
select t.*, v.dte
from t cross apply
(values (convert(date, '2019-01-01')) (convert(date, '2019-01-02'))
) v(dte);
Please try this:
DECLARE #FromDate DATE = '2019-01-01', #ToDate DATE = '2019-01-02';
;WITH rs AS (SELECT #FromDate dt UNION ALL SELECT DATEADD(DAY,1,dt) FROM rs WHERE dt<#ToDate)
SELECT t.Id,t.[user],rs.dt AS [date]
FROM [YourTableName] t
CROSS APPLY rs
ORDER BY t.[user],rs.dt
OPTION (MaxRecursion 0)
;
Thanks for Tim’s response and I have an idea:
WITH thedates AS
(
SELECT CAST(#startdate AS DATETIME) AS thedates
UNION ALL
SELECT DATEADD(DAY,1,thedates)
FROM thedates
WHERE DATEADD(DAY,1,thedates)<= CAST(#enddate AS DATETIME)
)
SELECT t.user_id,t.user_name,CONVERT(NVARCHAR(10),d.thedates,112)
FROM [dbo].[test] t
CROSS JOIN thedates d
Related
I've written some code to extract the latest date from multiple columns.
select (select max(LatestDate)
from (values (col1),(col2),(col3)) as updatedate(LatestDate)
) as LatestDate
from table1
However, I only want to take the date if it's before today. When I run the code for the sample dates below, it gives me the latest date as 10/04/2019 which is after today.
The date that i'd want it to extract is 14/03/2019 (col2) as it's before today, and is the latest date of all the columns whose date is before today.
Today = 27/03/2019
col1 = 02/02/2019
col2 = 14/03/2019
col3 = 10/04/2019
Can anyone advise on this? Hope it makes sense.
Many thanks
afk
You can use APPLY with WHERE clause :
select t.*, tt.LatestDate
from table1 t outer apply
( select max(LatestDate) as LatestDate
from ( values (col1),(col2),(col3) ) as updatedate(LatestDate)
where LatestDate < convert(date, GETDATE())
) tt;
You can use the below code for achieving the same.
select (select max(LatestDate)
from (values (col1),(col2),(col3)) as updatedate(LatestDate)
where updatedate < CAST(GETDATE() AS DATE)
) as LatestDate
from table1
Try this - you can take the MAX date that is less than or equal to today:
with cte as
(
select *
from (values (col1),(col2),(col3)) as updatedate
)
select (
select max(updatedate)
from cte
where updatedate <= GETDATE()
) as LatestDate
from table1
Add a where clause. I would phrase the query using cross apply:
select max(LatestDate)
from table1 t1 cross apply
(values (t1.col1), (t1.col2), (t1.col3)
) updatedate(LatestDate)
where updateddate.LatestDate < getdate();
I have a table with the following structure:
ID: StartDate: EndDate
I want to show all dates in the date range for each ID.
Eg
ID = 1: StartDate = 01/01/2018: EndDate = 03/01/2018
ID: 1 01/01/2018
ID: 1 02/01/2018
ID: 1 03/01/2018
I think i need to use a cross join but im unsure how to create this for multiple rows?
Here is the CTE for SQL Server, the syntax is somewhat different:
declare #startdate date = '2018-01-01';
declare #enddate date = '2018-03-18';
with
dates as (
select #startdate as [date]
union all
select dateadd(dd, 1, [date]) from dates where [date] < #enddate
)
select [date] from dates
So i ended up using a date table and just cross referencing that
select *
from Date d
inner join WorkingTable w
on d.Date >= w.StartDate
and d.date < w.EndDate
In standard SQL you can use a recursive CTE:
with recursive dates as (
select date '2018-01-01' as dte
union all
select dte + interval '1 day'
from dates
where dte < date '2018-01-03'
)
select dte
from dates;
The exact syntax (whether recursive is needed and date functions) differ among databases. Not all databases support this standard functionality.
Now got this for only one id..,
create table #dateTable(id int, col1 date, col2 date)
insert into #dateTable values(1,'05-May-2018','08-May-2018') ,(2,'05-May-2018','05-May-2018')
select *from #dateTable
with cte(start, ends) as(
select start = (select top 1 col1 from #dateTable), ends = (select top 1 col2 from #dateTable)
union all
select DATEADD(dd,1,start),ends from cte where start <> ends
)select start from cte option (maxrecursion 10)
I'm still working... I update soon...!
I have the following table:
oDateTime pvalue
2017-06-01 00:00:00 70
2017-06-01 01:00:00 65
2017-06-01 02:00:00 90
ff.
2017-08-01 08:00:00 98
The oDateTime field is an hourly data which is impossible to have a duplicate value.
My question is, how can I know if the oDateTime data is correct? I meant, I need to make sure the data is not jump? It should be always 'hourly' base.
Am I missing the date? Am I missing the time?
Please advice. Thank you.
Based on this answer, you can get the missing times form your table MyLogTable it like this:
DECLARE #StartDate DATETIME = '20170601', #EndDate DATETIME = '20170801'
SELECT DATEADD(hour, nbr - 1, #StartDate)
FROM ( SELECT ROW_NUMBER() OVER ( ORDER BY c.object_id ) AS Nbr
FROM sys.columns c
) nbrs
WHERE nbr - 1 <= DATEDIFF(hour, #StartDate, #EndDate) AND
NOT EXISTS (SELECT 1 FROM MyLogTable WHERE DATEADD(hour, nbr - 1, #StartDate)= oDateTime )
If you need to check longer period, you can just add CROSS JOIN like this
FROM sys.columns c
CROSS JOIN sys.columns c1
It enables you to check much more than cca thousand records (rowcount of sys.columns table) in one query.
Since your table is not having any unique id number, use a row_number() to get the row number in the cte , then perform an self inner join with the row id and next id ,take the difference of oDateTime accordingly, this will show exactly which row do not have time difference of one hour
;with cte(oDateTime,pValue,Rid)
As
(
select *,row_number() over(order by oDateTime) from [YourTableName] t1
)
select *,datediff(HH,c1.oDateTime,c2.oDateTime) as HourDiff from cte c1
inner join cte c2
on c1.Rid=c2.Rid-1 where datediff(HH,c1.oDateTime,c2.oDateTime) >1
You could use DENSE_RANK() for numbering the hours in a day from 1 to 24. Then all you have to do is to check whether the max rank is 24 or not for a day. if there is at least one entry for each hour, then dense ranking will have max value of 24.
Use the following query to find the date when you have a oDateTime missing.
SELECT [date]
FROM
(
SELECT *
, CAST(oDateTime AS DATE) AS [date]
, DENSE_RANK() OVER(PARTITION BY CAST(oDateTime AS DATE) ORDER BY DATEPART(HOUR, oDateTime)) AS rank_num
FROM Test
) AS t
GROUP BY [date]
HAVING(MAX(rank_num) != 24);
If you need validation for each row of oDateTime, you could do self join based on rank and get the missing hour for each oDateTime.
Perhaps you are looking for this? This will return dates having count < 24 - which indicates a "jump"
;WITH datecount
AS ( SELECT CAST(oDateTime AS DATE) AS [date] ,
COUNT(CAST(oDateTime AS DATE)) AS [count]
FROM #temp
GROUP BY ( CAST(oDateTime AS DATE) )
)
SELECT *
FROM datecount
WHERE [count] < 24;
EDIT: Since you changed the requirement from "How to know if there is missing" to "What is the missing", here's an updated query.
DECLARE #calendar AS TABLE ( oDateTime DATETIME )
DECLARE #min DATETIME = (SELECT MIN([oDateTime]) FROM #yourTable)
DECLARE #max DATETIME = (SELECT MAX([oDateTime]) FROM #yourTable)
WHILE ( #min <= #max )
BEGIN
INSERT INTO #calendar
VALUES ( #min );
SET #min = DATEADD(hh, 1, #min);
END;
SELECT t1.[oDateTime]
FROM #calendar t1
LEFT JOIN #yourTable t2 ON t1.[oDateTime] = t2.[oDateTime]
GROUP BY t1.[oDateTime]
HAVING COUNT(t2.[oDateTime]) = 0;
I first created a hourly calendar based on your MAX and MIN Datetime, then compared your actual table to the calendar to find out if there is a "jump".
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];