OleDb through open file dialog in Visual Basic Express 2010

Question

I am using Visual Basic Express 2010. Form 3 has code to import a csv file into a DataGridView while creating a DataSet. This event happens when I open Form 3. The path to the file is in the code. I would like to have a button on Form 1 that opens an OpenFileDialog so the user can browse for the csv file. Once the user selects the file, the DataGridView and DataSet on Form 3 initiate. The code I am currently using is below. Is there a way to edit the code to have an open file dialog from a button on Form 1 and not auto load through a pathway? Any assistance would be appreciated.

Public Class Form3

    Private Sub Form3_Load(ByVal sender As Object, ByVal e As EventArgs) Handles MyBase.Load

        Dim file As String = "test.csv"
        Dim path As String = "C:\Users\laptop\Desktop\"
        Dim ds As New DataSet

        End If

        Try
            If IO.File.Exists(IO.Path.Combine(path, file)) Then
                Dim ConStr As String = _
                "Provider=Microsoft.Jet.OLEDB.4.0;Data Source=" & _
                path & ";Extended Properties=""Text;HDR=Yes;IMEX=1;FMT=CSVDelimited\"""
                Dim conn As New OleDb.OleDbConnection(ConStr)
                Dim da As New OleDb.OleDbDataAdapter("Select * from " & _
                file, conn)
                da.Fill(ds, "TextFile")

            End If

        Catch ex As Exception
            MessageBox.Show(ex.ToString)

        End Try

        DataGridView1.DataSource = ds.Tables(0)

    End Sub

Answer 1

Change your Form3 class adding a couple of global vars and force the caller of Form3 to pass an initialization value. Then use the values passed in the opening of the connection

Public Class Form3
    Dim mFileCSV As String
    Dim mPathCSV As String

    Public Sub New(ByVal fileCSV as String, ByVal pathCSV as String)
        mFileCSV = fileCSV
        mPathCSV = pathCSV
    End Sub

    Private Sub Form3_Load(ByVal sender As Object, ByVal e As EventArgs) Handles MyBase.Load
       Dim fullPath = Path.Combine(mPathCSV, mFileCSV)
       if File.Exist(fullPath)  then
          .... open the connection ....


    End Sub

after these changes your Form1 when need to open the Form3 will call

   Dim of = new OpenFileDialog()
   if of.ShowDialog() = DialogResult.OK Then

       .... extract file and path and pass to form3 constructor

       Dim f3 As Form3 = New Form3(file, path)
       f3.Show()

Answer 2

I would like to have a button on Form 1 that opens an OpenFileDialog so the user can browse for the csv file. Once the user selects the file, the DataGridView and DataSet on Form 3 initiate

In order to solve this you need to do two things

1. Use an OpenFileDialog

This is easy. Just use the OpenFileDialog The below snippet adapted from the MSDN sample Opens a File Dialog box and if the

Dim openFileDialog1 As New OpenFileDialog()
Dim fileLocation As String
openFileDialog1.InitialDirectory = "c:\"
openFileDialog1.Filter = "txt files (*.csv)|*.csv|All files (*.*)|*.*"
openFileDialog1.FilterIndex = 1
openFileDialog1.RestoreDirectory = True 

If openFileDialog1.ShowDialog() = System.Windows.Forms.DialogResult.OK Then 

    fileLocation = openFileDialog1.FileName

End If

2. Pass the fileLocation to Form3

Forms are just objects. So all the ways that you pass information from one object to another is at your disposal.

For example you could create a public property FileLocation on Form3 and then set it before you Show Form3. Also instead of using Form.Load you'd need to use Form.Shown instead.

Another option would be to overload the Constructor for Form3 that has a FileLocation parameter.