I have a table with week number and day of week.
Example:
J2 = 18
G2 = Monday
How can I convert this to 2018-04-30?
I can find lots of threads for converting the other way around, meaning date to week. But I can't find anything on week + weekday to date.
Anyone know of a method?
The (optional) second argument of the WEEKDAY function determines what the first day of the week is. The two most common choices are:
1: Sunday (1) to Saturday (7)
or
2: Monday (1) to Sunday (7)
But, you can start on Wednesday, Friday, etc if you want. Pick an option now.
So, start with 1st January (of whichever unspecified year you're working with), and subtract the weekday of 1st January, according to whichever start-day you picked. This gives you a baseline for Week 1. To get to the start of Week n, you just need to add 7*(n-1) days.
Finally, you need to add the weekday back on. I'm going to recommend using MATCH on an array. Your array will be your weekdays, in order, surrounded by curly brackets. So, for Sunday-to-Saturday, you would use
MATCH(G2,{"Sunday","Monday","Tuesday","Wednesday","Thursday","Friday","Saturday"},0)
to get a number from 1 ("Sunday") to 7 ("Saturday"), or #NA if cell G2 does not contain a day-name.
Stick it all together, and you get something like this:
=DATE(YEAR(NOW()),1,1) - WEEKDAY(DATE(YEAR(NOW()),1,1),1) + 7*(J2-1) + MATCH(G2,{"Sunday","Monday","Tuesday","Wednesday","Thursday","Friday","Saturday"},0)
Try this, necessary comments in code:
Option Explicit
Sub GetDate()
Dim year As Date, day As Long
'convert day in G2 to integer
Select Case Range("G2").Value
Case "Monday"
day = 1
Case "Tuesday"
day = 2
'rest of cases
End Select
'declare what year you want
year = "01-01-2018"
'move from first week to the week specified in J2
year = DateAdd("d", (Range("J2").Value - 1) * 7, year)
'move back to first day of the week
Do While DatePart("w", year, vbMonday) <> 1
year = DateAdd("w", -1, year)
Loop
'find date of particular day
Do While DatePart("w", year, vbMonday) <> day
year = DateAdd("w", 1, year)
Loop
End Sub
The determination depends critically on your definition of "week number".
If you are using the ISO-8601 definition, where a Week starts on Monday and week 1 contains the first Thursday of the year, you can use this worksheet formula:
=DATE(YEAR(TODAY()),1,-2)-WEEKDAY(DATE(YEAR(TODAY()),1,7),14)+$J$2*7+MATCH($G$2&"*",{"Monday";"Tuesday";"Wednesday";"Thursday";"Friday";"Saturday";"Sunday"},0)-1
you may need to change the commas to semicolons depending on your regional settings
To explain:
First, find the last Monday of the previous year:
DATE(YEAR(TODAY()),1,-2)-WEEKDAY(DATE(YEAR(TODAY()),1,7),14)
Then add seven(7) times the number of desired weeks:
+$J$2*7
Then add the number of days from Monday to the desired day for that week:
+MATCH($G$2&"*",{"Monday";"Tuesday";"Wednesday";"Thursday";"Friday";"Saturday";"Sunday"},0)-1
Related
I need help with this function. I have a table where based on the posting date I need to assign which week it falls on.
Our week starts on Saturday, so this workaround that Microsoft published some years ago -- doesn't work.
Public Function WOY(MyDate As Date) As Integer ' Week Of Year
WOY = Format(MyDate, "ww", vbSaturday, vbFirstFourDays)
If WOY > 52 Then
If Format(MyDate + 7, "ww", vbSaturday, vbFirstFourDays) = 2 Then WOY = 1
End If
End Function
Would appreciate any help.
Thanks
Valglad
I don't think there is a unique solution here because part of the definition of Week is that it starts on Monday: https://learn.microsoft.com/en-us/office/troubleshoot/access/functions-return-wrong-week-number
One answer That is easy to convert is to assume that Each Saturday gets the same Week Number as the following Monday. Adapting the working solution from: https://learn.microsoft.com/en-us/office/troubleshoot/access/functions-return-wrong-week-number#workarounds
Public Function WOY(MyDate As Date) As Integer ' Week Of Year
MyDate = DateAdd("d", 2, MyDate) 'assuming Saturday gets the same weeknumber as the following Monday
WOY = Format(MyDate, "ww", vbMonday, vbFirstFourDays)
If WOY > 52 Then
If Format(MyDate + 7, "ww", vbMonday, vbFirstFourDays) = 2 Then WOY = 1 'borrowing Microsofts solution to handle all edge cases
End If
End Function
This solution is conceptually simple but then 12/28/19 starts week 1 of 2020.
If our week would start on Monday -- the code in my original message
would flip from 52 to 1 without any issues
For this year perhaps, but not for any year.
Thus, the answer is - as Mathieu clearly explained: There is no solution to your question.
It is for a reason, that the ISO 8601 standard was created where week numbering is an important part, as the first and last week may span calendar year boundaries.
You may use any other week numbering system, but most likely it will have shortcomings as you have seen.
I need to print the last sunday of the year from which adding 7 to it will give me all the sundays of next year.
I have the code to print all sundays for a particular year if i have a start date but i need the user to put the year so that last sunday of the previous year will be generated and 7 will be added to get first sunday of that year and so on till it reaches last sunday of next year
For example input year is 2017 it will check the last sunday of 2016 and add 7 to it to get first sunday of 2017 which is 1-1-2017 and it will go on printing all sundays till it reaches 31st december 2017
The function next_day() takes two arguments: a date and the name of a day of the week. It returns the closest "next" day (following the date argument) that matches the given day of the week. So the result is between one and seven days forward. (If you want 'Tuesday' and the input date is a Tuesday, the function returns the date seven days later.)
If you want the last Sunday of a year, it will be between Dec. 25 and Dec. 31. So if you call the next_day() function with the arguments Dec. 24 (!!) and 'Sunday' you'll get what you want.
The result will have the same time-of-day as the date argument, so if you give a date without a time-of-day, so will be the output (which is probably what you want). So:
select next_day(date '2016-12-24', 'Sunday') from dual;
NEXT_DAY(D
----------
2016-12-25
Added: If you take an input from your user, as a bind variable, you can do something like this:
select next_day(to_date(:input_year - 1 || '-12-24', 'yyyy-mm-dd'), 'Sunday') from dual;
If you provide 2017 as input (whatever mechanism your interface has for bind variables), the output will be 2016-12-25 (in DATE data type, so don't ask "in what format" - dates don't have a format!)
Please help me with an Informx SQL query to be run every Friday.
Period: 1st of the Month to the Thursday before the Friday that it is being run on. I can't just choose date ranges as it will be an auto report so the dates needs to update automatically. Is there a way to do this?
You've got two issues: the start of the month and the date of last Thursday. The simplest expression to identify the 1st day of the current month is:
MDY(MONTH(TODAY), 1, YEAR(TODAY))
The way to identify the most recent Thursday is via a CASE statement on WEEKDAY(TODAY). Something like:
CASE
WHEN WEEKDAY(TODAY) < 5 -- (Sunday (0) - Thursday (4))
THEN TODAY - WEEKDAY(TODAY) - 3 -- Calc last Sunday, back 3 days
ELSE TODAY - WEEKDAY(TODAY) + 4 -- Calc last Sunday, forward 4 days
END
Note, you will still run into issues where the 1st of current month is later than the most recent Thursday. But hopefully the examples above will point you in the right direction.
Of course, if you know this report is only ever going to be run on Fridays, then calculating the most recent Thursday is just TODAY -1.
I have a date in sql which will always fall on a Monday and I'm subtracting 13 weeks to get a weekly report. I am trying to get the same 13 week report but for last year's figures as well.
At the moment, I'm using the following:
calendar_date >= TRUNC(sysdate) - 91
which is working fine.
I need the same for last year.
However, when I split this into calendar weeks, there will also be a partially complete week as it will include 1 or 2 days from the previous week. I need only whole weeks.
e.g. the dates that will be returned for last year will be 14-Feb-2015 to 16-May-2015. I need it to start on the Monday and be 16-Feb-2015. This will change each week as I am only interested in complete weeks...
I would do this:
Get the date by substracting 91 days as you're already doing.
Get the number of the day of the week with TO_CHAR(DATE,'d')
Add the number of days until the next monday to the date.
Something like this:
SELECT TO_DATE(TO_DATE('16/05/2015','DD/MM/YYYY'),'DD/MM/YYYY')-91 + MOD(7 - TO_NUMBER(TO_CHAR(TO_DATE(TO_DATE('16/05/2015','DD/MM/YYYY'),'DD/MM/RRRR')-91,'d'))+1,7) d
FROM dual
next_day - returns date of first weekday named by char.
with dates as (select to_date('16/05/2015','DD/MM/YYYY') d from dual)
select
trunc(next_day( trunc(d-91) - interval '1' second,'MONDAY'))
from dates;
I want to get next monday from calculated date. In situation when calculated date is monday i have to move back to previous week ( -1 second).
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