Good day,
I would like to create parameters with the following conditions:
Start Date Range between 03/01/2019-03/31/2019 (done)
End day range 03/01/2019-03/31/2019
Default end day is set to the day of today but in March (so if I open the wbook tomorrow it will be 03/22/2019
It's a bit complexe and your help is much appreciated.
1- Create calc field that takes day of today and month/year
DATEPARSE("dd, MM, yyyy", str(Today()) + ", " + str(03) + str(2019))
2- In End date parameter > when opening workbook > select calculated date
Related
I have a table with records, where each record has a date column, then a start time column and end time column.
I am trying to do a datediff to get the duration in hours from start to end date with DateDiff('s',[Start Date[,[End Date])/3600.
This works perfectly for End dates that are on same day as date column, but sometimes the end date would be the next day like 12:45 AM. The date diff will give me a large negative number, how do I let it know its next day?
I dont own the data, so not much I can do with the table
Thanks!
Try something like this:
DateDiff('s',[Start Date],DateAdd('d',IIF([End date]<[Start Date],1,0),[End Date]))/3600
It can be done with pure math:
TotalHours = TimeValue(CDate([End Date] - [Start Date] + 1)) * 24
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;
Hi I am creating a table in MS Access to store the details of children in a school.
I have a field called YearGroup which needs to calculate the school year they are in based on their date of birth and whether they have been moved up or down a year.
I.e. if the expression deems they are six years old they should be placed in year 2. If they were moved down or up a year they should be in year 1 or 3 (this is based on another field in the table called YearModifier).
The code I have at the moment is this:
Year(Now()) - IIf(Month([DOB]) > 8, Year([DOB]) + 6 + [YearModifier], Year([DOB]) + 5 + [YearModifier])
My problem is that Year(Now()) is returning as invalid expression. Lots of websites have recognised using the Now() function and also I've tried Date() but nothing seems to be accepted by Access (The version is 2010).
What is going on? How can I get today's date in a calculated field expression?
Thanks
Try creating a query with all of the fields from your table, and then add an extra field YearGroup: Year(Now()) - IIf(Month([DOB]) > 8, Year([DOB]) + 6 + [YearModifier], Year([DOB]) + 5 + [YearModifier])
It appears that Date functions can't be used in calculated columns in tables.
You could use this to get the year, which might work in field expressions:
format(date(),"yyyy")
About your function (which I have re-written very slightly)
Year( Now() )
- Year([DOB])
- IIf( Month([DOB]) > 8
, 6
, 5 )
+ [YearModifier]
however!
I don't think you want to use now(). Which year they are in depend on their age at the 1st Sept at the start of the current academic year not now! Ok now will work until 31/dec/2015, so I must assume you will not be using the function after this date!
If you are you must use 2015 not Now().
Ok?
You can calculate the age of the children with a simple function:
Public Function AgeSimple( _
ByVal datDateOfBirth As Date) _
As Integer
' Returns the difference in full years from datDateOfBirth to current date.
'
' Calculates correctly for:
' leap years
' dates of 29. February
' date/time values with embedded time values
'
' DateAdd() is used for check for month end of February as it correctly
' returns Feb. 28. when adding a count of years to dates of Feb. 29.
' when the resulting year is a common year.
' After an idea of Markus G. Fischer.
'
' 2007-06-26. Cactus Data ApS, CPH.
Dim datToday As Date
Dim intAge As Integer
Dim intYears As Integer
datToday = Date
' Find difference in calendar years.
intYears = DateDiff("yyyy", datDateOfBirth, datToday)
If intYears > 0 Then
' Decrease by 1 if current date is earlier than birthday of current year
' using DateDiff to ignore a time portion of datDateOfBirth.
intAge = intYears - Abs(DateDiff("d", datToday, DateAdd("yyyy", intYears, datDateOfBirth)) > 0)
End If
AgeSimple = intAge
End Function
Then your expression would be something like this (ignoring the modifier):
ClassYear: IIf(AgeSimple([DOB]) > 6, 2, 1)
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())
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)))