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Help a Brazilian programmer urgent, please!

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Join Date: Oct 2009

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Reputation: Domus BRASIL is an unknown quantity at this point

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Domus BRASIL Domus BRASIL is offline

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Help a Brazilian programmer urgent, please!

-1

Oct 7th, 2009

Hello, I represent a company developing software in Brazil.

I wonder why my program (VB6)
reading a number stored in the notepad, it reads
the decimal number.

for example if the value is stored 30.5
will display the value in textbox 305

how to make it understand the point.

Thanks

Last edited by Domus BRASIL; Oct 7th, 2009 at 1:15 pm.

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Join Date: Sep 2009

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Reputation: Vineeth K is an unknown quantity at this point

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Vineeth K

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Oct 8th, 2009

I think you made mistake in declarations if any ..
Integers never load a decimal point.But a Double do.

Are you trying to open the notepad ??It is not clear .
Read from a notepad is equal to open it.Then you will get the whole data in notepad.

Who wants to be a GOD ??

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Join Date: Mar 2009

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Reputation: vb5prgrmr will become famous soon enough

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vb5prgrmr

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Posting Shark

Oct 8th, 2009

It may also depend upon your local settings on your computer. I don't know about brazil but some countries use the decimal/period character as a thousands seperator and since you have 30.5 and not 30.500 it may be stripping it.

So to perhaps solve this you may want to read the value into a string and then convert to a double.

Good Luck

If anyone has helped you solve your problem, please mark your thread as solved.

Thanks

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Join Date: Oct 2009

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Domus BRASIL Domus BRASIL is offline

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Oct 8th, 2009

I have several textbox to search a number that is in a line of a file in. txt

but if this number is decimal, it reads as if the number was over

what I do for a textbox recognize this number as a decimal?

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Join Date: Mar 2009

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Reputation: vb5prgrmr will become famous soon enough

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vb5prgrmr

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Posting Shark

Oct 8th, 2009

So, you first read the value in from a text file, right? So at that point you have "30.5". Then somewhere it gets changed, right?

 Help with Code Tags 
 Visual Basic 4 / 5 / 6 Syntax (Toggle Plain Text) 
Dim SomeVariable As Double
SomeVariable = CDbl("30.5")
Dim SomeVariable As Double
SomeVariable = CDbl("30.5")

Good Luck

If anyone has helped you solve your problem, please mark your thread as solved.

Thanks

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Join Date: May 2006

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ithelp

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Posting Virtuoso

Oct 9th, 2009

Yes correct all your variable declarations.

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dandanventura dandanventura is offline

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Oct 26th, 2009

oi...
pelo que entendi vc esta com problemas de declaracao de variaveis como os outros colegas disseram. O maior problema e se vc esta conseguindo pegar o numero 3.5 , se afirmativo tente jogar numa string, apos trocar o ponto por virgula e entao joga-lo para double.

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AndreRet

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Posting Whiz in Training

Oct 26th, 2009

You can also try your textbox format -

 Help with Code Tags 
 Visual Basic 4 / 5 / 6 Syntax (Toggle Plain Text) 
Text1.text = Format(Text1.text, "### ###.##")
'Where ## is the actual number entered into text box.
Text1.text = Format(Text1.text, "### ###.##")
'Where ## is the actual number entered into text box.

You can also try this sub I've done some while back

 Help with Code Tags 
 Visual Basic 4 / 5 / 6 Syntax (Toggle Plain Text) 
Public Function Txtnumdeci(Key As Integer, Strinfo As String) As Integer
Dim Deciprs As Boolean
Dim Ctr As Integer
Dim Decipos As Integer
 
Decipos = InStr(Strinfo, ".")
If Decipos Then
    Deciprs = True
Else
    Deciprs = False
End If
 
Select Case Key
Case vbKey0 To vbKey9
    If Deciprs = True Then
        Ctr = Len(Mid(Strinfo, Decipos + 1, Len(Strinfo) - Decipos + 1))
        If Ctr = 2 Then
            Txtnumdeci = 0
        Else
            Txtnumdeci = Key
        End If
    Else
        Txtnumdeci = Key
    End If
Case vbKeyBack
    Txtnumdeci = Key
Case vbKeyDecimal, Asc(".")
    If Deciprs = True Then
        Txtnumdeci = 0
    Else
        Txtnumdeci = Key
    End If
Case vbKeyReturn
    SendKeys "{TAB}"
Case Else
    Txtnumdeci = 0
End Select
End Function
Public Function Txtnumdeci(Key As Integer, Strinfo As String) As Integer
Dim Deciprs As Boolean
Dim Ctr As Integer
Dim Decipos As Integer

Decipos = InStr(Strinfo, ".")
If Decipos Then
    Deciprs = True
Else
    Deciprs = False
End If

Select Case Key
Case vbKey0 To vbKey9
    If Deciprs = True Then
        Ctr = Len(Mid(Strinfo, Decipos + 1, Len(Strinfo) - Decipos + 1))
        If Ctr = 2 Then
            Txtnumdeci = 0
        Else
            Txtnumdeci = Key
        End If
    Else
        Txtnumdeci = Key
    End If
Case vbKeyBack
    Txtnumdeci = Key
Case vbKeyDecimal, Asc(".")
    If Deciprs = True Then
        Txtnumdeci = 0
    Else
        Txtnumdeci = Key
    End If
Case vbKeyReturn
    SendKeys "{TAB}"
Case Else
    Txtnumdeci = 0
End Select
End Function

Hope this helps.

Please mark questions as answered when done.

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