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") 4/1/2009 Debug.Print Format(Now, "Short Time") 09:16
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) 4/1/2009 Debug.Print Now - Int(Now) 0.390729166669189
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)) Double
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) Next Call MsgBox(Format(Timer - Start, "0.00") & " micro-seconds") Start = Timer For i = 1 To 1000000 dteTemp = RemoveDateWithFormat(Now) Next 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.