Have you ever needed to remove the time part of a date variable, or remove the date part of a date variable? I recently had a requirement to do this, and my first reaction was to use the format function because it is very flexible and quite simple to use. Unfortunately, it does’t perform very well.

First, the format method for removing the time part.

Debug.Print Format(Now, "Short Date")

Debug.Print Format(Now, "Short Time")

Seems simple enough to remove the date and/or time portion of a date variable, but is it the best way? If you only want to display it, it probably is, but if you need to use the values for calculations, it is not the best way to remove the date or time portions.

Instead, you can use the Int function.

The Int function performs a simple truncate on a number. Note: this does not do rounding. Also significant is that Int will return whatever data is sent to it. If the parameter is a double, int will return a double. If the parameter is a single, Int returns a single, if the parameter is a date, Int returns a date.

Debug.Print Int(Now)

Debug.Print Now - Int(Now)

Surprised to see a number instead of a time? Me too. But… in VB6, when you subtract dates, the resulting data type is a double.

Debug.Print TypeName(Now - Int(Now))

However, this double can be converted back to a Date variable type like this.

Debug.Print cDate(Now - Int(Now))
9:25:40 AM 

Clearly, there are 2 different methods for separating the date from the time in a Date variable. Which is better? They both return the correct value, so now it’s up to performance.

I tested the performance with this code:

Option Explicit

Private Sub Command1_Click()

    Dim i As Long
    Dim dteTemp As Date
    Dim Start As Single
    Start = Timer
    For i = 1 To 1000000
        dteTemp = RemoveDate(Now)
    Call MsgBox(Format(Timer - Start, "0.00") & " micro-seconds")
    Start = Timer
    For i = 1 To 1000000
        dteTemp = RemoveDateWithFormat(Now)
    Call MsgBox(Format(Timer - Start, "0.00") & " micro-seconds")
End Sub

Private Function RemoveDateWithFormat(ByVal DateTime As Date) As Date
    RemoveDateWithFormat = Format(DateTime, "h:mm")

End Function

Private Function RemoveDate(ByVal DateTime As Date) As Date
    RemoveDate = DateTime - Int(DateTime)
End Function

The Int method runs in 0.40 micro-seconds and the Format method takes 3 micro-seconds. The Int method is approximately 7 times faster.