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put a double in a fixed width txt file

Posted on 2004-09-24
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Last Modified: 2010-05-02
I have to put 3 double in a fixed width txt file.  How much characters I must specify in the width of a double ?
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Question by:weeb0
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8 Comments
 
LVL 86

Expert Comment

by:Mike Tomlinson
ID: 12142703
Use the format function to display the double with the number of digits you want:

Private Sub Command1_Click()
    Dim myDouble As Double
    myDouble = 1 / 3
    Debug.Print myDouble
    Debug.Print Format(myDouble, "0000.0000")
End Sub
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LVL 16

Expert Comment

by:jimbobmcgee
ID: 12143007
A double is usually 16 characters long...
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LVL 16

Expert Comment

by:jimbobmcgee
ID: 12143091
...however, if  it is less than 1, you may need an extra character to store the 0 and if it is a negative number, another extra character to store the -.  

I would use 18 characters in a fixed width file, read it as a string and then use CDbl to format as a double...

Don't forget to error handle, when writing, to account for rational decimals, such as 0.25.  Convert to string (with CStr) and then add however many spaces to make 18 chars...
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LVL 1

Author Comment

by:weeb0
ID: 12143575
is there an easy way to format the fixed width ???

I would like to have trailing spaces for a total of 30 characters in 1st field, the second one is a date... so, I know how to deal with it ...
the the 2 last are double with 2 decimals.

thank you
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LVL 86

Accepted Solution

by:
Mike Tomlinson earned 80 total points
ID: 12143972
Here is one way to pad your string with trailing spaces:

Private Sub Command1_Click()
    Dim dblValue As Double
    Dim strValue As String
   
    dblValue = 1 / 3
    strValue = Format(dblValue, "#,##0.00") ' or however you decide to format it...
    strValue = trailingPad(strValue, 30)
    Debug.Print strValue
End Sub

Private Function trailingPad(ByVal strInput As String, ByVal padLength As Integer)
    Dim i As Integer
    Dim spacesNeeded As Integer
    Dim strLength As Integer
    Dim spaces As String
   
    strLength = Len(strLength)
    If strLength > padLength Then
        trailingPad = Left(strInput, padLength)
    Else
        spacesNeeded = padLength - strLength
        spaces = Space(spacesNeeded)
        trailingPad = strInput & spaces
    End If
End Function
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