I need to make my Access query always return the Monday of the current week. I have seen a few solutions on Google/StackOverflow but they are written in SQL and I am a beginner in creating Access queries (I am using the Design view to make them).
Goal: The week should be considered as M T W T F S S. Then, the query should always return the Monday of the current week. Therefore, if it is Sunday, it should still return the Monday before, NOT the next week's Monday. Can anyone explain how to do this using the Design View in Access 2010?
Keep in mind that in this context we are working with dates, so if we do Date() - 1, we will get 1 day prior to today.
Date() ~ Today's date
DatePart(
"w" - Weekday
Date() - Today's date
2 - vBMonday (Access assumes Sunday is the first day of the week, which is why this is necessary.)
1 - vbFirstJan1 - This gets into using the first week of the year. We could have omitted this, as 1 is the default.
)
-1 - Subtract 1 from the DatePart value.
Values
Date() = 4/27/2015 (at time of this writing)
DatePart("w",Date(),2,1) = 1
DatePart("w",Date(),2,1)-1 = 0
So we have Date()-0... Okay, what's so great about that? Well, let's look at a more useful scenario where today's date is a day other than Monday.
Let's act like today is 4/28/2015 (Tuesday)
Date() = 4/28/2015
DatePart("w",Date(),2,1) = 2
DatePart("w",Date(),2,1)-1 = 1
So, from the outside, in; give me the current weekday value. (1 = Monday, 2 = Tuesday, etc.), and subtract 1 from that -> that's how many days we need to subtract from the current date to get back to the weekday value of 1 (Monday).
Here's a function that will do this:
Public Function DatePrevWeekday( _
ByVal datDate As Date, _
Optional ByVal bytWeekday As VbDayOfWeek = vbMonday) _
As Date
' Returns the date of the previous weekday, as spelled in vbXxxxday, prior to datDate.
' 2000-09-06. Cactus Data ApS.
' No special error handling.
On Error Resume Next
DatePrevWeekday = DateAdd("d", 1 - Weekday(datDate, bytWeekday), datDate)
End Function
As vbMonday is 2 and your date is today, you can use the core expression in a query:
PreviousMonday: DateAdd("d",1-Weekday(Date(),2),Date())
Related
I have a variable in a SAS dataset that has a number of dates (e.g. 01APR21). What I'm looking to do is create a new variable that shows the date of the first Monday of that week. So using the above example of 01APR21, the output would be 29/03/2021 as that what was when the Monday in that week was. I'm assuming it's using intnx, but I can't get my head around it.
data test;
format date date8.;
format first_day date10.;
date = '01APR21'd;
first_day = ?;
run;
INTNX Parameters:
Interval : WEEK
Increment: 0 (same week)
Alignment: Beginning
(Sunday)
Then add 1 to get to Monday instead of Sunday. You could probably play with the SHIFT INDEX parameter as well.
Monday = intnx('week', dateVariable, 0, 'B') + 1
I'm trying to manipulate a date value to go back in time exactly 1 ISO-8601 year.
The following does not work, but best describes what I want to accomplish:
date_add(date '2018-01-03', interval -1 isoyear)
I tried string conversion as an intermediate step, but that doesn't work either:
select parse_date('%G%V%u',safe_cast(safe_cast(format_date('%G%V%u',date '2018-01-03') as int64)-1000 as string))
The error provided for the last one is "Failed to parse input string "2017013"". I don't understand why, this should always resolve to a unique date value.
Is there another way in which I can subtract an ISO year from a date?
This gives the corresponding day of the previous ISO year by subtracting the appropriate number of weeks from the date. I based the calculation on the description of weeks per year from the Wikipedia page:
CREATE TEMP FUNCTION IsLongYear(d DATE) AS (
-- Year starting on Thursday
EXTRACT(DAYOFWEEK FROM DATE_TRUNC(d, YEAR)) = 5 OR
-- Leap year starting on Wednesday
(EXTRACT(DAY FROM DATE_ADD(DATE(EXTRACT(YEAR FROM d), 2, 28), INTERVAL 1 DAY)) = 29
AND EXTRACT(DAYOFWEEK FROM DATE_TRUNC(d, YEAR)) = 4)
);
CREATE TEMP FUNCTION PreviousIsoYear(d DATE) AS (
DATE_SUB(d, INTERVAL IF(IsLongYear(d), 53, 52) WEEK)
);
SELECT PreviousIsoYear('2018-01-03');
This returns 2017-01-04, which is the third day of the 2017 ISO year. 2018-01-03 is the third day of the 2018 ISO year.
i need to calculate difference between two date excluding sunday. I have table with dates and i need to calculate number of dates of repeated days from last date.
if i have dates like that
27-05-2017
29-05-2017
30-05-2017
I use this code in script
date(max(Date)) as dateMax,
date(min(Date)) as dateMin
And i get min date = 27-05-2017 and max date = 30-05-2017 then i use in expressions
=floor(((dateMax - dateMin)+1)/7)*6 + mod((dateMax - dateMin)+1,7)
+ if(Weekday(dateMin) + mod((dateMax - dateMin)+1,7) < 7, 0, -1)
And get result 3 days. Thats OK, but the problem is if I have next dates:
10-05-2017
11-05-2017
27-05-2017
29-05-2017
30-05-2017
When use previously code I get min date = 10-05-2017 and max date = 30-05-2017 and result 18, but this is not OK.
I need to count only dates from
27-05-2017
29-05-2017
30-05-2017
I need to get max date and go throw loop repeated dates and if have brake to see is that date sunday if yes then step that date and continue to count repeated dates and if i again have break and if not sunday than close loop and remember number of days.
In my case instead of 18 days i need to get 3 days.
Any idea?
I'd recommend you creating a master calendar in the script where you can apply weights or any other rule to your days. Then in your table or app you can just loop through the dates or perform operations and sum their weights (0: if sunday, 1: if not). Let's see an example:
// In this case I'll do a master calendar of the present year
LET vMinDate = Num(MakeDate(year(today()),1,1));
LET vMaxDate = Num(MakeDate(year(today()),12,31));
Calendar_tmp:
LOAD
$(vMinDate) + Iterno() - 1 as Num,
Date($(vMinDate) + Iterno() - 1) as Date_tmp
AUTOGENERATE 1 WHILE $(vMinDate) + Iterno() - 1 <= $(vMaxDate);
Master_Calendar:
LOAD
Date_tmp AS Date,
Week(Date_tmp) as Week,
Year(Date_tmp) as Year,
Capitalize(Month(Date_tmp)) as Month,
Day(Date_tmp) as Day,
WeekDay(Date_tmp) as WeekDay,
if(WeekDay = '7',0,1) as DayWeight //HERE IS WHERE YOU COULD DEFINE A VARIABLE TO DIRECTLY COUNT THE DAY IF IT IS NOT SUNDAY
'T' & ceil(num(Month(Date_tmp))/3) as Quarter,
'T' & ceil(num(Month(Date_tmp))/3) & '-' & right(year(Date_tmp),2) as QuarterYear,
date(monthStart(Date_tmp),'MMMM-YYYY') as MonthYear,
date(monthstart(Date_tmp),'MMM-YY') as MonthYear2
RESIDENT Calendar_tmp
ORDER BY Date_tmp ASC;
DROP Table Calendar_tmp;
I have a spreadsheet that asks people to enter in a day of the month when we need to send out a bill. What I want to do is create a calendar event based on that. So, essentially what I need is an event that starts at the current month, day from the spreadsheet, and continues to a specified point in time.
var monthlyDate = row[6]; // Seventh column, monthly date of payment
var curDate = new Date();
var curMonth = curDate.getMonth();
var curYear = curDate.getYear();
curDate.setDate(curMonth, monthlyDate, curYear);
Logger.log("Day of month: %s", monthlyDate);
Logger.log("Current Date: %s", curDate);
Logger.log("Current Date: %s", Date());
What I'm seeing is that the monthly date is coming in as a float "6.0" for example, and no matter what I enter in for monthlyDate in the setDate line, it keeps setting the date to 10/9/15 (Today is 10/15/15). I've hard-coded that value to many different numbers, but for some reason it's just not working.
How can I create a date (in any format) that follows the scheme "Current Month / Day from Speadsheet / Current Year" ?
The getMonth() method returns a "zero-indexed" number. So, it returns the number 9 for the 10th month. setDate() doesn't set the date, it sets the "Day of the Month". The name of that method is misleading.
Documentation - setDate()
So, the last two parameters that you are using in setDate() are doing nothing. You are setting the day of the month to 9.
If you want to set multiple date parameters at the same time, you need to use the new Date() method:
var d = new Date(year, month, day, hours, minutes, seconds, milliseconds);
The month parameter accept values from 0 to 11, 0 is Jan and 11 is Dec
Date Reference
In my quarterly report Im trying to validate the two parameters StartDate and EndDate.
I first check if the difference between the dates is 2 months:
Switch(DateDiff(
DateInterval.Month, Parameters!StartDate.Value, Parameters!EndDate.Value) <> 2,
"Error message")
Then I try to add whether the StartDate is the first day of month AND EndDate is last day of month:
And (Day(Parameters!StartDate.Value) <> 1
And Day(DATEADD(DateInterval.Day,1,Parameters!EndDate.Value)))
So the whole expression looks like this:
Switch(DateDiff(DateInterval.Month, Parameters!StartDate.Value, Parameters!EndDate.Value) <> 2
And
Parameters!IsQuarterly.Value = true
And
Day(Parameters!StartDate.Value) <> 1
And
Day(DATEADD(DateInterval.Day,1,Parameters!EndDate.Value))<>1),
"Error: Quarterly report must include 3 months")
But It works wrong when the difference between dates is still 2 months, but StartDate and EndDate are not first and last day of the whole period.
I'd appreciate any help :)
I would say just change the implementation Add another two Parameter With Quarter and Year
Quarter like Q1,Q2,Q3 & Q4 with Value 1,2,3 & 4 respectively and year 2012,2013,2014 & so on
Now based on the parameter selected Qtr & Year set Default value of start & End Date
=DateSerial(Parameters!Year.Value), (3*Parameters!Qtr.Value)-2, 1) --First day of Quarter
=DateAdd("d",-1,DateAdd("q",1,Parameters!Year.Value, (3*Parameters!Qtr.Value)-2, 1))) --Last day of quarter
Doing this no need to do any validation bcz its always get the correct Date Difference.
Other Reference
First day of current quarter
=DateSerial(Year(Now()), (3*DatePart("q",Now()))-2, 1)
Last day of current quarter
=DateAdd("d",-1,DateAdd("q",1,DateSerial(Year(Now()), (3*DatePart("q",Now()))-2, 1)))