formatting number without rounding

[pkailas]

Is there a function that will format a number such as 2.348 by dropping the 3rd digit after the decimal? I don't want this rounded up to 2.35 as if I did formatnumber(2.348, 2).

 

[DrJavaJoe]

Probably not the best way, but you can try this:

Private Sub Command2_Click()
MsgBox NoRound(2.348)
End Sub

Private Function NoRound(dblValue As Double) As Double
Dim strTemp As String
NoRound = CDbl(Fix(dblValue) & "." & Left(100 * (dblValue - Fix(dblValue)), 2))
End Function

 

[dpaulson]

Heres another


Text1 = Fix(num * 100) / 100

David Paulson

 

[LPlates]

Option Explicit

Private Sub Command1_Click()
Dim a As Double
a = TwoDecNoRound(3.248)
MsgBox a
End Sub

Private Function TwoDecNoRound(strNum) As Double
Dim char As Integer
char = InStr(1, strNum, ".&quot
Select Case char
Case 0
TwoDecNoRound = strNum
Case Else
TwoDecNoRound = Left(strNum, char + 2)
End Select
End Function

 

[CCLINT]

I use dpaulson's shown method.
You can easily put it in a proceedure and make it dynamic, passing the number of decimal places to show (Optional = 2)
and then multiplying the figure by

Function ChoppitOff (myNumber As Double, Optional iDigitsToShow As Integer = 2)
ChoppitOff = Fix(myNumber * 10 ^ iDigitsToShow)/10 ^ iDigitsToShow
End Function

 

[1127]

Hi,

Whatch out with the dot evaluation. The whole story will be wrong if the decimal separator is ","

Tibi

 

[CCLINT]

Opps, forgot to ad the return value type (otherwise it's a variant)

Function ChopItOff(myNumber As Double, Optional iDigitsAfterDec As Integer = 2) As Double
ChopItOff = Fix(myNumber * 10 ^ iDigitsAfterDec) / 10 ^ iDigitsAfterDec
End Function
´
Also, the below will accept a number or string if it can be converted to a number - then you can plug in a textbox.text property directly:

Function ChopItOff([myNumber As Variant, Optional iDigitsAfterDec As Integer = 2) As Double
If Not IsNumeric(myNumber) Then Exit Function
myNumber = CDbl(myNumber)
ChopItOff = Fix(myNumber * 10 ^ iDigitsAfterDec) / 10 ^ iDigitsAfterDec
End Function

Dim myNumber As Double
myNumber = ChopItOff(Text1.Text)
Text1.Text= CStr(ChopItOff(Text1.Text))

 

[CCLINT]

>The whole story will be wrong if the decimal separator is ","

Correct. Therefore you do not evaluate a decimal seperator, or, at most, you query the system to find out what seperator is used and sticj it into a variable, and use the variable instead.

 

[MichaelRed]

A slightly different -more flexible if more convolouted- variation on the theme:

Public Function basTrunc(MyVal As Variant, _
                         Optional DelimChr As String = ".", _
                         Optional NChar As Integer = 2)

    'Michael Red 6/5/2003 Tek-Tips thread222-567798 for pkailas

    'Sample Usage:
    '? basTrunc(3.1284)
    ' 3.12

    '? basTrunc("Michael Lee Red", " ", 0)
    'Michael


    Dim MyStr As String
    Dim MyParts() As String
    Dim rtnVal As Variant

    If (IsNumeric(MyVal)) Then
        MyStr = CStr(MyVal)
     Else
        MyStr = MyVal
    End If

    MyParts = Split(MyStr, DelimChr)

    rtnVal = MyParts(0) & DelimChr & Left(MyParts(1), NChar)

    If (IsNumeric(MyVal)) Then
        basTrunc = CDbl(rtnVal)
     Else
        basTrunc = rtnVal
    End If

End Function

Pleas also note the complete lack of any error checking!





MichaelRed
m.red@att.net

Searching for employment in all the wrong places