Generating filename using date

[ssh]

How to generate a filename that uses date values? For example, date value is November 21, 1999 and filename should be 21-11-99. Problem was how to separate date parts. Thanks

 

[ColinEssex]

Won't you get files with the same filename if there are more than one for a certain date?

Col

 

[Mile-O]

Dim strFileName As String

strFileName = Day(txtDate) & "-" & Month(txtDate) & "-" & Year(txtDate)

 

[ssh]

Towards the right way but I want 99 instead of 1999.

 

[Mile-O]

strFileName = Day(txtDate) & "-" & Month(txtDate) & "-" & Right(Year(txtDate),2)


Also, to add a 0 to the front of a month or day that is a single figure you could try this:

Dim strFileName As String
    
    If Day(txtDate) < 10 Then
        strFileName = "0" & Day(txtDate) & "-"
    Else
        strFileName = Day(txtDate) & "-"
    End If
    If Month(txtDate) < 10 Then
        strFileName = strFileName & "0" & Month(txtDate) & "-"
    Else
        strFileName = strFileName & Month(txtDate) & "-"
    End If
    strFileName = strFileName & right(Year(txtDate), 2)

I tried to do this line but it didn't work:

strFileName = Format(Day(txtDate),"dd") & "-" & Format(Month(txtDate) ,"mm") & "-" & Format(Year(txtDate),"yy")

When I tried that way the string returned was something like 08-04-1905 which I thought was rather bizarre.

 

[ssh]

I'll give a try, and let you know.

 

[Pat Hartman]

You can simplify to:

Format(YourDate,"mm-dd-yy")

08-04-1905 was not bizzare at all. The following statement is incorrect because, the format()"s are not being passed the date that you think you are passing. Format(YourDate,"mm") calculates the month portion of the date represented by YourDate. As I have mentioned dozens of times here (and others have as well), dates are NOT stored as text strings. They are stored as double precision numbers. The integer portion of the number represents the number of days since Dec 30, 1899 and the decimal portion represents the time of day. If you extract the month prior and pass that value to the Format() function as in Format(Month(YourDate),"mm") what you are doing is substituting some number between 1 and 12 for the actual serial number that your date represents. Today (Jan 7, 2003) is 37628. To prove this to yourself open the immediate window and type:
print cdate(37628)
print cdate(1)
print cdate(-3)
and observe the dates that are printed.

 

[ssh]

Format(YourDate,"mm-dd-yy") worked. Love simple things!