Solved

put a double in a fixed width txt file

Posted on 2004-09-24
8
171 Views
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 ?
0
Comment
Question by:weeb0
  • 2
  • 2
8 Comments
 
LVL 85

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
0
 
LVL 16

Expert Comment

by:jimbobmcgee
ID: 12143007
A double is usually 16 characters long...
0
 
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...
0
 
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
0
 
LVL 85

Accepted Solution

by:
Mike Tomlinson earned 20 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
0

Featured Post

Free Tool: Postgres Monitoring System

A PHP and Perl based system to collect and display usage statistics from PostgreSQL databases.

One of a set of tools we are providing to everyone as a way of saying thank you for being a part of the community.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Suggested Solutions

There are many ways to remove duplicate entries in an SQL or Access database. Most make you temporarily insert an ID field, make a temp table and copy data back and forth, and/or are slow. Here is an easy way in VB6 using ADO to remove duplicate row…
Since upgrading to Office 2013 or higher installing the Smart Indenter addin will fail. This article will explain how to install it so it will work regardless of the Office version installed.
Get people started with the process of using Access VBA to control Excel using automation, Microsoft Access can control other applications. An example is the ability to programmatically talk to Excel. Using automation, an Access application can laun…
Get people started with the utilization of class modules. Class modules can be a powerful tool in Microsoft Access. They allow you to create self-contained objects that encapsulate functionality. They can easily hide the complexity of a process from…

685 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question