I have this formula in my column to calculate the start date of a given week :
dateadd(week,[Week]-(1),
dateadd(day,(-1),
dateadd(week,
datediff(week,(0),
CONVERT([varchar](4),[Year],(0))+'-01-01'),
(1))))
Where Week and Year are other fields like 38 and 2012
Problem is, it calculates the start date of week 38/2012 as a monday (17th Sept), I would like it to be a sunday instead (16th Sept) is this possible?
Many thanks.
This will return you the first day of the week, given a week number and a year, assuming that the first day of the week is a Sunday.
Standard exclusions apply, e.g. don't try year 1499.
declare #tbl table ([Week] int, [Year] int)
insert #tbl select 38,2012
union all select 1,2012
union all select 0,2012
union all select 1,2013
select DATEADD(
Week,
[Week]-1,
DATEADD(
Day,
(8-##datefirst)-DATEPART(dw, CAST([Year]*10000+101 AS VARCHAR(8))),
CAST([Year]*10000+101 AS VARCHAR(8))))
from #TBL
Result
2012-09-16 00:00:00.000
2012-01-01 00:00:00.000
2011-12-25 00:00:00.000
2012-12-30 00:00:00.000
Note that the Week number starts from 1, and if the week doesn't start on a Sunday, then the first day of that week could end up in an earlier year (row #4). Because the Weeks are relative, you can use Week 0, -1 etc and it will still give you a result (row #3), rightly or wrongly.
You may also notice I used a different method to create a date out of the year, just as an alternative.
The (8-##datefirst) portion of the query makes it robust regardless of your DATEFIRST setting.
If you want the first day of the week to be Sunday you could change your database to make it so, using the DATEFIRST setting.
function getFirstDateOfThisWeek(d)
{
var TempDate = new Date(d || new Date());
TempDate.setDate(TempDate.getDate() + (#Config.WeekStartOn - 1 - TempDate.getDay() - 7) % 7);
return TempDate;
}
var StartDate = getFirstDateOfThisWeek(new Date()); //1st date of this week
Related
Consider the date "2022-07-02"
For the month July first week only have 3 days in it.
I need to find the number of days in the week for the given date.
In above date the week has 3 days where "2022-07-02" day reside.
Example 2 :
For month June in 2022 first week has 5 days in the week
Therefore if i declare a date as "2022-06-03" it should pass the number of days in the week as 5
I need a query to find the number of days for the specific week.
set datefirst 1 -- assumes Monday is set as start of week
declare #myDate date = getdate();
-- calculate the next last day of week
declare #nextSunday date = dateadd(day, 7 - datepart(weekday, #myDate), #myDate);
select case
-- advancing into the next month implies a partial week
-- datediff(month, #myDate, nextSunday) = 1 would be equivalent
when day(#nextSunday) < day(#myDate) then 7 - day(#nextSunday)
-- else see if still within first week
when day(#nextSunday) < 7 then day(#nextSunday)
else 7 end;
Within a query you might use it this way:
select case
when day(nextSunday) < day(dateColumn) then 7 - day(nextSunday)
when day(nextSunday) < 7 then day(nextSunday)
else 7 end
from
myData cross apply (
values (dateadd(day, 7 - datepart(weekday, dateColumn), dateColumn))
) v(nextSunday);
https://dbfiddle.uk/?rdbms=sqlserver_2019&fiddle=ee5bfb52dabe31dd619cfd136689db59
If you don't want the shorthand form then just replace every instance of nextSunday in the final step with its full expression.
There's nothing in the logic that prevents this from working with another first day of week. I just chose a variable name that helped ellucidate this particular problem.
I have two event tables with timestamped data: Registered, Signed_In.
Both have rows such as: original_timestamp, user_id
I am trying to find out users who haven't signed in within 30 days after registering. I have used the following query but I cannot add a WHERE clause to it.
I tried a query but I am getting hourly difference, whereas I wanted days difference which is unsupported in BigQuery.
SELECT Signed_In.user_id, TIMESTAMP_DIFF(Registered.original_timestamp, Signed_In.original_timestamp, HOUR) AS days_difference
FROM `test_db.Signed_In` signed_in
JOIN `test_db.Registered` registered
ON Signed_In.user_id = Registered.user_id
GROUP BY 1,2
ORDER BY 2 DESC
WHERE days_difference > '30'
I am getting two columns: user_id, days_difference but the days_difference shows hours and my WHERE clause is rejected when I use it.
You can try this below code-
Note: Using Ordinal Position for GROUP BY and ORDER BY is not a good practice. Its always safe and standard to use the column names directly.
SELECT Signed_In.user_id,
TIMESTAMP_DIFF(Registered.original_timestamp, Signed_In.original_timestamp, HOUR) AS days_difference
FROM `test_db.Signed_In` signed_in
JOIN `test_db.Registered` registered
ON Signed_In.user_id = Registered.user_id
WHERE DATE_DIFF(Registered.original_timestamp, Signed_In.original_timestamp, Day) > '30'
GROUP BY 1,2
ORDER BY 2 DESC
Just replace HOUR to DAY in your query:
SELECT Signed_In.user_id, TIMESTAMP_DIFF(Registered.original_timestamp, Signed_In.original_timestamp, DAY) AS days_difference
Correct values are:
MICROSECOND
MILLISECOND
SECOND
MINUTE
HOUR
DAYOFWEEK
DAY
DAYOFYEAR
WEEK: Returns the week number of the date in the range [0, 53]. Weeks begin with Sunday, and dates prior to the first Sunday of the year are in week 0.
WEEK(<WEEKDAY>): Returns the week number of timestamp_expression in the range [0, 53]. Weeks begin on WEEKDAY. datetimes prior to the first WEEKDAY of the year are in week 0. Valid values for WEEKDAY are SUNDAY, MONDAY, TUESDAY, WEDNESDAY, THURSDAY, FRIDAY, and SATURDAY.
ISOWEEK: Returns the ISO 8601 week number of the datetime_expression. ISOWEEKs begin on Monday. Return values are in the range [1, 53]. The first ISOWEEK of each ISO year begins on the Monday before the first Thursday of the Gregorian calendar year.
MONTH
QUARTER
YEAR
ISOYEAR: Returns the ISO 8601 week-numbering year, which is the Gregorian calendar year containing the Thursday of the week to which date_expression belongs.
DATE
DATETIME
TIME
I tried using this
dateadd(month, 0, dateadd(day,(30-datepart(dd,'2015-02-28')),'2015-02-28'))
to get the required output and instead of getting '2015-02-28' i get '2015-03-02'. How is it possible to change day of date in SQL and set last day of month if the day does not exist in the month ?
====Update with sample data =============
Note: Goal is not to get the last day of the month
If i want to change the day to 30 and if it's a month which has only 28 days. it should be the end of the month date. If not date should be 30th.
Changing the day to 30th
If Feb it should be - '2015-02-28'
If march it should be - '2015-03-30'
If April it should be - '2015-04-30'
There exists a function for this if your sql server version is 2012 or higher:
SELECT EOMONTH('2015-02-15')
returns 2015-02-28
-- Replace Day portion of #OrigDate with #NewDay.
-- If the result would be beyond the end of the month, return the last day of the month
create function ReplaceDay ( #OrigDate date, #NewDay int )
returns date
as
begin
declare #NewDate date
-- Deal with extreme cases, in case someone passes #NewDay = 7777 or #NewDay = -7777
if #NewDay < 1
set #NewDay = 1
if #NewDay > 31
set #NewDay = 31
-- Subtract the DAY part of the original date from the new day.
-- Add that many days to the original date
-- Example: if the original date is 2018-02-08 and you want to replace 08 with 17, then add 17-08=9 days
set #NewDate = DateAdd ( d, #NewDay-Day(#OrigDate), #OrigDate )
-- Have we ended up in the next month?
-- If so subtract days so that we end up back in the original month.
-- The number of days to subtract is just the Day portion of the new date.
-- Example, if the result is 2018-03-02, then subtract 2 days
if Month(#NewDate)<>Month(#OrigDate)
set #NewDate = DateAdd ( d, -Day(#NewDate), #NewDate )
return #NewDate
end
go
select dbo.ReplaceDay ( '2017-02-08', 17 ) -- Returns 2017-02-17
select dbo.ReplaceDay ( '2017-02-08', 30 ) -- Returns 2017-02-28
for sql server 2012 or higher versions please check HoneyBadger's answer.
for older versions:
SELECT DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,#mydate)+1,0))
I have to get/create date (date of first week) from the user input of day name
Examples-
if input saturday then date should be 7 (for current month 1st saturday)
if input sunday then date 1 (current month 1st sunday)
I am having to use lot of logic to get the date but couldn't get exact output
any suggestions on how to come up with the SQL query for such a function ?
If you are using SQL*plus in Oracle then the code will be as :
select next_day(sysdate,'&d')-7 from dual;
If any update required please do inform.
okay i have a link hereby where you can understand this and try to do
Your answer!!!
Try this.
SELECT DATEPART(dw,DATEADD(m, DATEDIFF(m, 0, GETDATE()), 0))
input: if getdate() returns date in jan 2017
output: 1
input:if getdate() returns date in jan 2017
output: 4
Getting the month for the current date is, obviously, straight forward, but I'm needing to get the month name with a different end date.
I need to get the month name with the start date of the month being the first Thursday after the first Wednesday of the month and the end date of the month being the first Wednesday of the following month. It's for an accounting thing, so I'm not going to argue with the spec!
e.g. for 2014, January would run from 9th Jan - 5th Feb, February would run from 6th February - 5th March, March would run from 6th March - 2nd April.
I would suggest that you create a table with your 'accounting months' in it, having a start date, end date and month name columns.
You could then query this to find the row where your date is between the start and end dates and return the month name. Putting this into a scalar function would then allow it to be reusable and relatively easily updated for next years months as well.
I think, as per Paddy's answer, a lookup table is the simplest thing to do. Here's one way to generate the rows for it:
; With Numbers(n) as (
select 4 union all select 5
), Months as (
select CONVERT(date,'20010104') as StartDt,CONVERT(date,'20010207') as EndDt,
DATENAME(month,'20010103') as Month
union all
select DATEADD(week,n1.n,StartDt),DATEADD(week,n2.n,EndDt),
DATENAME(month,DATEADD(week,n1.n,StartDt))
from Months,Numbers n1,Numbers n2 --Old-skool join, just for once
where DATEPART(day,DATEADD(week,n1.n,StartDt)) between 2 and 8 and
DATEPART(day,DATEADD(week,n2.n,EndDt)) between 1 and 7 and
StartDt < '21000101'
)
select * from Months option (maxrecursion 0)
(CW since this is effectively just an extension to Paddy's answer but I don't want to edit their answer, nor is it suitable for a comment