Break apart a date in Access. Example 7/1/2015

[harvella]

Hi,



I'm trying to figure out how to take apart a date when I can only pull a full date.

Example is that my data can only give me 7/1/2015 but I need it to have specific columns for "7", "1", and "2015". I have searched everywhere I can but I'm not sure if I am searching with the correct keywords.

Thanks for the help.

[orange]

You should try using Google

MSAccess date functions

See http://www.techonthenet.com/access/functions/

[ItsMe]

Here is some more info and an example.
https://msdn.microsoft.com/EN-US/lib.../gg251733.aspx

Dim intMonth As Integer
Dim dtMyDate As Date
dtMyDate = "01/01/2010"
intMonth = DatePart("m", dtMyDate)
MsgBox intMonth

[harvella]

I did find this site but I am fairly new to access and I'm not understanding the way the expression is written. I get that I need to use a datepart expression but I'm not sure if I put the expression category in first or second.

This is how I am writing based on what I understand but I know its in the wrong order. but I can't figure out exactly how to write it.

datepart( [Date (Calendar)], "D")

You should try using Google

MSAccess date functions

See http://www.techonthenet.com/access/functions/

[June7]

What date part are you trying to extract? The example at bottom of link referenced by ItsMe explicitly shows correct syntax.

Date is a function that has no argument - it simply returns the current system date.

Is Calendar a field in table?

DatePart("d", [Calendar])

[orange]

Did you see and try the examples

Example

Let's look at how to use the DatePart function in MS Access:
DatePart ("yyyy", #15/10/1998#)
Result: 1998

DatePart ("m", #15/10/2003#)
Result: 10

DatePart ("d", #15/10/2003#)
Result: 15 Example in VBA Code

The DatePart function can be used in VBA code in Microsoft Access.
For example:
Dim LValue As Integer

LValue = DatePart ("d", #15/10/2003#) In this example, the variable called LValue would now contain the value of 15.

[IrogSinta]

While you're learning the very DatePart useful function, might as well learn some other useful Date Functions:

Month(#7/1/2015#)	'this returns 1
Day(#7/1/2015#)	'this returns 7
Year(#7/1/2015#)	'this returns 2015

Ron