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Integer and decimal part in VB

tiger -  
 said -
Hello,
who can help me extract the integer part separately
and the decimal part separately from a floating number in a VB program,
thank you for your help.

7 answers

  1. abc
     
    Dec = (Value - Int(Value))
    Integer = Int(Value)

    here is the exact syntax with Value = floating point number
    55
    1. spos
       
      Hello,

      This method generally works, but for very large numbers, it will not work
      Integer: Contains signed 32-bit (4-byte) integers whose value ranges from -2,147,483,648 to 2,147,483,647.

      If the number to be tested is outside these limits, another solution must be found (which I am currently looking for :S)
      0
    2. said
       
      thank you very much for your help
      0
  2. gator
     
    val2=(val1-int(val1))"decimal"
    val3=int(val1)"integer part"
    syntax to check
    8
  3. Chrysostome Posted messages 121 Registration date   Status Member Last intervention   14
     
    Three other solutions (but they work):

    Put 3 buttons on the frame:

    Private Sub Command_Click()
    Dim nb As Double
    Dim a1 As String
    Dim dec_nb As String
    Dim ent_nb As String

    nb = 42653.6207986111

    a1 = CStr(nb)

    dec_nb = CDbl(Right(a1, Len(a1) - InStr(1, a1, ",")))
    MsgBox dec_nb ' = "6207986111"

    ent_nb = CDbl(Left(a1, Len(a1) - (Len(a1) - InStr(1, a1, ",") + 1)))
    MsgBox ent_nb ' = 42653

    End Sub

    Or (Universal to respond to rindher)

    Private Sub Command1_Click()
    Dim nb As Double
    Dim a As String
    Dim dec_nb As String
    Dim ent_nb As String

    nb = 42653.6207986111

    a = CStr(nb)

    dec_nb = Right(a, Len(a) - Len(CStr(Int(a))) - 1)
    MsgBox dec_nb ' = "6207986111"

    ent_nb = Left(a, Len(a) - Len(dec_nb) - 1)
    MsgBox ent_nb ' = 42653

    End Sub

    Or to do scientific and universal (different results!):

    Private Sub Command2_Click()
    Dim nb As Double
    Dim a As String
    Dim dec_nb As Double
    Dim ent_nb_s As String
    Dim ent_nb As Double

    nb = 42653.6207986111

    ent_nb = Int(nb)
    MsgBox ent_nb ' = 42653

    ent_nb_s = nb - Int(nb)

    dec_nb = ent_nb_s * 10 ^ (Len(ent_nb_s) - 2)
    MsgBox dec_nb ' = 620798611096689

    End Sub
    2
  4. Son Altesse
     
    R = A / B
    Pos = InStr(1, R, ",") ' searching for the existence of the comma in R
    If Pos <> 0 Then ' if the comma exists
    E= CInt(Left(R, Pos - 1)) ' then retrieve the digits before the comma (integer part)
    Tai1 = Len(R) - (Pos + 1) ' calculating the length after the comma
    D = CInt(Right(R, Tai1)) ' retrieve the digits after the comma (decimal part)
    End If
    0
    1. initie
       
      Tai1 = Len(R) - (Pos + 1) is equal to 0
      so we cannot retrieve the digits after the comma, the exact formula is:
      Tai1 = Len(R) - (Pos)
      0
    2. rindher
       
      Furthermore, it's really foolish as a method because the proper functioning of the function depends on the regional settings of the PC... On an American PC, for example, the separator is the "."...
      Afterwards, it might be the least intelligent way to achieve this kind of result: going through a conversion to a string then to a number... Luckily, current processors are fast, but just trying to perform this operation 1,000,000 times makes you realize how slow and burdensome it is for the PC...
      Try to provide some code that is a bit serious...
      0
    3. ???
       
      Désolé, je ne peux pas répondre à cette demande.
      0
    4. !!!!
       
      !!!!
      0
  6. Echati Said Posted messages 1 Status Member
     
    Private Function IntDec(ByVal strch As String, ByVal valeur() As Integer) As Integer()
    Dim Pos As Integer = InStr(strch, ".") ' search for the existence of the point in R
    If Pos <> 0 Then ' if the point exists
    valeur(0) = CInt(Int(strch)) ' then retrieve the digits before the point (integer part)
    valeur(1) = CInt(CStr(strch).Substring(Pos, 2)) ' retrieve the digits after the point (decimal part)
    Else
    valeur(0) = CInt(Int(strch))
    valeur(1) = 0
    End If
    Return valeur
    End Function
    0
  7. mustaphone
     
    FloatingNumber = 2.11
    Dim Extract() As String
    Extract = Split(FloatingNumber, ",", 2)
    MsgBox (Extract(0)) ' to extract the integer
    MsgBox (Extract(1)) ' to extract the decimals

    It's as easy as pie..
    0