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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
4 Comments
 
LVL 15

Accepted Solution

by:
JackOfPH earned 250 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!
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LVL 32

Assisted Solution

by:Robberbaron (robr)
Robberbaron (robr) earned 250 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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