CODEWITHSUNDEEP
vbavb6
Gopal
Gopal

Reputation: 11972

How to get the days for the date

Part 1

Dim totdays as long
totdays = DateDiff("d", "01/2011", DateAdd("m", 1, "01/2011"))

The above code will return "31"

I want to get the days for that 31

Expected Output

Monday (31/01/2011)

Need VB6 Code Help

Part 2

I want to find the sundays on the particular month....

For Example, If i select the month 01/2012, The query should give the result like this

01
08
15
22
29

The above dates are sunday.

Expected Output for 01/2012 Month

01
08
15
22
29

Upvotes: 4

Views: 759

Answers (3)

Ahmad
Ahmad

Reputation: 12707

You can use DateSerial Function in VB6 to convert a string or integer variable to Date Variable

Dim d As String
Dim datevar As Date
d = "31"
datevar = DateSerial(2011,1, Val(d))   
MsgBox Format(datevar,"DDDD dd/MMM/yyyy")

Upvotes: 0

brettdj
brettdj

Reputation: 55672

something like this (tested in )

final update for the Sunday sub-query

As per request in commentd from Gopal below

    Dim strDate As String
    Dim dtStart As Date
    Dim dtEnd As Date
    Dim stEnd As Date
    Dim lngCnt As Long
    Dim strOut As String
    strDate = "01/2012"
    dtStart = DateValue(strDate)
    dtEnd = DateAdd("d", DateDiff("d", strDate, DateAdd("m", 1, strDate) - 1), dtStart)
    lngCnt = Weekday(dtStart) - 7
    Do
        lngCnt = lngCnt + 7
        strOut = strOut & Format(lngCnt, "00") & vbNewLine
    Loop While lngCnt + 7 <= dtEnd - dtStart
    MsgBox strOut

updated

Note that I needed to use lngdays-1 to add 1 day less than a month (ie 31-Jan-2011), else you would have had 01-Feb-2011 as the result

Dim strDate As String
Dim lngdays As Long
strDate = "01/2011"
lngdays = DateDiff("d", strDate, DateAdd("m", 1, strDate))
MsgBox Format(DateAdd("d", lngdays - 1, strDate), "dddd (dd/mm/yyyy)")

old

 Dim lngdays As Long
 lngdays = DateDiff("d", "01/2011", DateAdd("m", 1, "01/2011"))
 MsgBox Format(DateSerial(2011, 1, lngdays), "dddd (dd/mm/yyyy)")

Upvotes: 4

dotnetstep
dotnetstep

Reputation: 17485

Use format function like this. Here i used Now but you can pass any date and format return string of dayname

Format(Now, "dddd")

Upvotes: 0

Related Questions