Last Monday of October

Hello All

I am curious to see if there is a cleaner way to query by the last monday in october of the previous year.  I have the below and it works but i am sure there is a cleaner way to do it that i am not thinking of.

>=
IIf(Weekday(DateSerial(Year(Date())-1,10,31),1)=1,DateSerial(Year(Date())-1,10,31-6),
IIf(Weekday(DateSerial(Year(Date())-1,10,31),1)=7,DateSerial(Year(Date())-1,10,31-5), IIf(Weekday(DateSerial(Year(Date())-1,10,31),1)=6,DateSerial(Year(Date())-1,10,31-4), IIf(Weekday(DateSerial(Year(Date())-1,10,31),1)=5,DateSerial(Year(Date())-1,10,31-3), IIf(Weekday(DateSerial(Year(Date())-1,10,31),1)=4,DateSerial(Year(Date())-1,10,31-2), IIf(Weekday(DateSerial(Year(Date())-1,10,31),1)=3,DateSerial(Year(Date())-1,10,31-1), IIf(Weekday(DateSerial(Year(Date())-1,10,31),1)=2,DateSerial(Year(Date())-1,10,31),"ERROR")))))))
garyrobbinsAsked:
Who is Participating?
 
Rgonzo1971Connect With a Mentor Commented:
HI,

pls try

=DateSerial(Year(Date())-1,11,1)-Weekday(DateSerial(Year(Date())-1,11,8-2))

Open in new window

Regards
0
 
Dale FyeCommented:
I have a table of numbers (tbl_Numbers) with a single field (lngNumber) and records from 0 to 99.

SELECT Max(DateSerial(Year(Date())-1,10,[lngNumber])) AS LastMonday
FROM qry_Numbers
WHERE (Weekday(DateSerial(Year(Date())-1,10,[lngNumber]),1)=1)
AND (qry_Numbers.lngNumber Between 1 And 31);
0
 
garyrobbinsAuthor Commented:
how would you use that as the criteria?
0
Cloud Class® Course: Amazon Web Services - Basic

Are you thinking about creating an Amazon Web Services account for your business? Not sure where to start? In this course you’ll get an overview of the history of AWS and take a tour of their user interface.

 
Dale FyeCommented:
Are you talking about a critieria in a query?

SELECT * FROM yourTable
WHERE [DateField] >= (
SELECT Max(DateSerial(Year(Date())-1,10,[lngNumber])) AS LastMonday
FROM qry_Numbers
WHERE (Weekday(DateSerial(Year(Date())-1,10,[lngNumber]),1)=1)
AND (qry_Numbers.lngNumber Between 1 And 31)
)
0
 
PatHartmanCommented:
I've attached a database full of date functions.  Look at the second tab to see this one in action.
Here is a function that will do it:
Pass in November 1st and a 2 for Monday
Function fDateDayPrevious(dtmDate As Date, bytDaySought As Byte) As Date
'   Function to return the date of the named day previous to the passed date
'   Accepts:bytDaySought (e.g.: vbSunday)
'   dtmDate as date
    If bytDaySought - WeekDay(dtmDate) <= 0 Then
        fDateDayPrevious = dtmDate + bytDaySought - WeekDay(dtmDate)
    Else
        fDateDayPrevious = dtmDate + bytDaySought - WeekDay(dtmDate) - 7
    End If
End Function

Open in new window

UsefulDateFunctions160822WorksWith64.zip
0
 
Gustav BrockCIOCommented:
You can use this generic function for finding a weekday of a month:
' Calculates the date of the occurrence of Weekday in the month of DateInMonth.
'
' If Occurrence is 0 or negative, the first occurrence of Weekday in the month is assumed.
' If Occurrence is 5 or larger, the last occurrence of Weekday in the month is assumed.
'
' If Weekday is invalid or not specified, the weekday of DateInMonth is used.
'
' 2016-06-10. Gustav Brock, Cactus Data ApS, CPH.
'
Public Function DateWeekdayInMonth( _
    ByVal DateInMonth As Date, _
    Optional ByVal Occurrence As Integer, _
    Optional ByVal Weekday As VbDayOfWeek = -1) _
    As Date
    
    Const DaysPerWeek   As Long = 7
    
    Dim Offset      As Integer
    Dim Month       As Integer
    Dim Year        As Integer
    Dim ResultDate  As Date
    
    ' Validate Weekday.
    Select Case Weekday
        Case _
            vbMonday, _
            vbTuesday, _
            vbWednesday, _
            vbThursday, _
            vbFriday, _
            vbSaturday, _
            vbSunday
        Case Else
            ' Zero, none or invalid value for VbDayOfWeek.
            Weekday = VBA.Weekday(DateInMonth)
    End Select
    
    ' Validate Occurrence.
    If Occurrence <= 0 Then
        Occurrence = 1
    ElseIf Occurrence > 5 Then
        Occurrence = 5
    End If
    
    ' Start date.
    Month = VBA.Month(DateInMonth)
    Year = VBA.Year(DateInMonth)
    ResultDate = DateSerial(Year, Month, 1)
    
    ' Find offset of Weekday from first day of month.
    Offset = DaysPerWeek * (Occurrence - 1) + (Weekday - VBA.Weekday(ResultDate) + DaysPerWeek) Mod DaysPerWeek
    ' Calculate result date.
    ResultDate = DateAdd("d", Offset, ResultDate)
    
    If Occurrence = 5 Then
        ' The latest occurrency of Weekday is requested.
        ' Check if there really is a fifth occurrence of Weekday in this month.
        If VBA.Month(ResultDate) <> Month Then
            ' There are only four occurrencies of Weekday in this month.
            ' Return the fourth as the latest.
            ResultDate = DateAdd("d", -DaysPerWeek, ResultDate)
        End If
    End If
    
    DateWeekdayInMonth = ResultDate
  
End Function

Open in new window

Then:

    >= DateWeekdayInMonth(DateSerial(Year(Date) - 1, 10, 1), 5, vbMonday)

or in a query:

    >=DateWeekdayInMonth(DateSerial(Year(Date())-1,10,1),5,2)

/gustav
0
 
xtermieCommented:
Have something similar, which checks if the date is the last monday of the month using T-SQL.
The following select will return 1 if the current date is the last monday of the month, and 0 if not
select
 case
 when datepart(dw, GETDATE()) = 2 and DATEPART(month, DATEADD(day, 7, GETDATE())) <> DATEPART(month, GETDATE())
 then 1
 else 0
 end
---
datepart(dw, GETDATE()) returns the day of the week. Monday is 2. The second part adds 7 days to the current date and checks that within 7 days the month has changed (if it does not, it is not the last monday).  Change the GETDATE()'s to any date you want to check.
0
 
xtermieCommented:
You can also try to make it into a generic function and use it with any date you like:
CREATE FUNCTION
IsLastMondayOfMonth(@dateToCheck datetime)
RETURNS bit
AS
BEGIN
DECLARE
@result bit

SELECT @result =
    CASE  
       WHEN datepart(dw, @dateToCheck) = 2 AND DATEPART(month, DATEADD(day, 7, @dateToCheck)) <> DATEPART(month, @dateToCheck)
    THEN 1
 ELSE 0
 END
 RETURN @result
END
0
 
garyrobbinsAuthor Commented:
This is exactly what I was looking for, Thank you.
0
Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.

All Courses

From novice to tech pro — start learning today.