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Last Monday of October

Posted on 2016-08-03
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Last Modified: 2016-08-05
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")))))))
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Question by:garyrobbins
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9 Comments
 
LVL 48

Expert Comment

by:Dale Fye
ID: 41741291
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
 

Author Comment

by:garyrobbins
ID: 41741308
how would you use that as the criteria?
0
 
LVL 48

Expert Comment

by:Dale Fye
ID: 41741319
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
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LVL 38

Expert Comment

by:PatHartman
ID: 41741415
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
 
LVL 52

Accepted Solution

by:
Rgonzo1971 earned 2000 total points
ID: 41741807
HI,

pls try

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

Open in new window

Regards
0
 
LVL 51

Expert Comment

by:Gustav Brock
ID: 41741882
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
 
LVL 18

Expert Comment

by:xtermie
ID: 41741949
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
 
LVL 18

Expert Comment

by:xtermie
ID: 41741952
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
 

Author Closing Comment

by:garyrobbins
ID: 41744340
This is exactly what I was looking for, Thank you.
0

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