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writing to a specific column in a text file

Posted on 2009-07-15
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Last Modified: 2013-11-26
I want to write to a text file.
I would want data to be written to specific columns because i will need to read it later with a data dictionary.

eg Staff id - starts at column 1
staff name - starts at column 10
staff salary - starts at column 50.

I am using vb6

regards

anthony
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Question by:Anthony Matovu
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4 Comments
 
LVL 15

Accepted Solution

by:
JackOfPH earned 1000 total points
ID: 24867707
Here is a sample:
 Public Sub WruteToText()
 
 
        Dim Column1 As String = "Column1"
        Dim Column2 As String = "Column2"
        Dim Column3 As String = "Column3"
        Dim Column4 As String = "Column4"
        Dim Column5 As String = "Column5"
 
        Using textFile As IO.TextWriter = IO.File.CreateText("D:\MyTextFile.txt")
            'Set a delimeter for the textfile to separate the each column.
            'The delimeter for this example is ||.
            textFile.WriteLine("{0}||{1}||{2}||{3}||{5}", Column1, Column2, Column1, Column2, Column1)
            textFile.WriteLine("{0}||{1}||{2}||{3}||{5}", Column1, Column2, Column1, Column2, Column1)
 
            'Save the data to the textfile.
            textFile.Flush()
            textFile.Close()
        End Using
 
    End Sub
 
    Private Sub ReadText()
 
 
        Dim textFile As IO.TextReader = IO.File.OpenText("D:\MyTextFile.txt")
        Dim value As String
        Dim charge As String = ""
 
        Do
            'Read each line in the text file.
            value = textFile.ReadLine
 
            If value Is Nothing Then
                Exit Do
            End If
 
            'This will split the columns in the given line
            Dim Columns() As String = value.Split("||")
 
            'Read each column in the current line in the text file.
            For Each column As String In Columns
                MsgBox(column)
            Next
 
        Loop
 
        textFile.Close()
 
    End Sub

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LVL 83

Expert Comment

by:CodeCruiser
ID: 24867920
@JackofPH
"I am using vb6" is the third last line!
0
 
LVL 32

Assisted Solution

by:Robberbaron (robr)
Robberbaron (robr) earned 1000 total points
ID: 24868379
this function is from a while ago but still in my toolbox.... havent checked the method of formating strings recently.
  uses VB6 format function and pads front and back as necessary


for each line
   sOut = ""
   sOut = sOut & Fmt$(staffid,"     00000","R")   'staffid right justified in 10 char
   sOut = sOut & Fmt$(staffname,space$(40),"L")   'name left justified in 40 char
   sOut = sOut & Fmt$(salary,"###########0.00","R")   'salary in 15 char
 
print #1,sOut
 
next line
 
'-------------------------------------------
Function Fmt$(txt As String, field$, justification$)
    'formats txt in specified field  width
    Dim a As String, x As Integer, y As Integer, totlen As Integer
    
    'Determine length of field
    x = InStr(field$, ";")   'search for separator
    If x = 0 Then x = Len(field$) Else x = x - 1
    totlen = x
 
    a = Trim$(left$(Format(txt, field$), totlen))
    
    Select Case LCase$(justification$)
        Case "L"
            x = 0: y = totlen - Len(a)
        Case "R"
            x = totlen - Len(a): y = 0
        Case "C"
            x = (totlen - Len(a)) / 2
            y = totlen - Len(a) - x
        Case Else
            'use left
            x = 0: y = totlen - Len(a): y = 0
    End Select
 
    Fmt$ = Space$(x) + a + Space$(y)
 
End Function

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