Reading from a textfile and adding the contents to different listboxes

[knicksfan426]

I have a form with 2 listboxes. I've added items in each listbox. I save all the items from each listbox to a text file. What I want to do is open the text file and I want each string inside the text file to go to the appropriate listbox. How can I make my program recognize where each string belongs?

[minitech]

Use some sort of flag when saving, something that would never be in either listbox. For example:

'Save routine
Dim sw As New IO.StreamWriter(filename)
Dim i As Integer = 0
While i < Me.ListBox1.Items.Count
     sw.WriteLine(CStr(Me.ListBox1.Items(i)))
     i+=1
End While
i = 0
sw.WriteLine("[SEPARATORSTRING]")
While i < Me.ListBox2.Items.Count
     sw.WriteLine(CStr(Me.ListBox2.Items(i)))
     i+=1
End While
sw.Close()

And do the opposite to save. Ignore that one line and switch listboxes after it.

[jmcilhinney]

I would suggest using separate files.

In fact, you don't even really need files at all. You could add a StringCollection to your application settings for each ListBox and then the data will be stored in the config file for you.

[knicksfan426]

Use some sort of flag when saving, something that would never be in either listbox. For example:

'Save routine
Dim sw As New IO.StreamWriter(filename)
Dim i As Integer = 0
While i < Me.ListBox1.Items.Count
     sw.WriteLine(CStr(Me.ListBox1.Items(i)))
     i+=1
End While
i = 0
sw.WriteLine("[SEPARATORSTRING]")
While i < Me.ListBox2.Items.Count
     sw.WriteLine(CStr(Me.ListBox2.Items(i)))
     i+=1
End While
sw.Close()

And do the opposite to save. Ignore that one line and switch listboxes after it.

I switched your code around a little bit but I worked off of it. Thank you. Check out the code, I have a little problem

Save File code:

Private Sub SaveToolStripMenuItem_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles SaveToolStripMenuItem.Click
        SaveFileDialog1.Filter = ".txt (*.txt)|*.txt"
        SaveFileDialog1.DefaultExt = "txt"
 
        If SaveFileDialog1.ShowDialog = Windows.Forms.DialogResult.OK Then
 
            Using myWriter As New System.IO.StreamWriter(SaveFileDialog1.FileName)
 
                myWriter.WriteLine("|")
 
                For Each Item As String In ListBox1.Items
                    myWriter.WriteLine(Item)
                Next
 
                myWriter.WriteLine("||")
 
                For Each Item As String In ListBox2.Items
                    myWriter.WriteLine(Item)
                Next
 
 
            End Using
        End If
 
    End Sub

Open File code:

Private Sub OpenToolStripMenuItem_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles OpenToolStripMenuItem.Click
        OpenFileDialog1.Filter = ".txt (*.txt)|*.txt"
        OpenFileDialog1.DefaultExt = "txt"
 
        If OpenFileDialog1.ShowDialog = Windows.Forms.DialogResult.OK Then
 
            Dim myReader As New System.IO.StreamReader(OpenFileDialog1.FileName)
 
            If myReader.ReadLine = "|" Then
 
                Do Until myReader.ReadLine = "||"
                    ListBox1.Items.Add(myReader.ReadLine)
                Loop
 
            End If
 
        End If
 
    End Sub

As you can see, I've used the separator "|" for the listbox1 items and "||" for listbox2 items. In my open file coding, whenever I run the program, VB gives me an error on line 12 saying "Value cannot be null. Parameter name: item." What do I do about this? I can't find any way to properly work around it.

[minitech]

One of the lines will be blank if you use .WriteLine(). Just add "On Error Resume Next" at the beginning of your Sub.

[knicksfan426]

Alright, I did what you said and the code now looks like this

Open File code:

Private Sub OpenToolStripMenuItem_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles OpenToolStripMenuItem.Click
        On Error Resume Next
 
        OpenFileDialog1.Filter = ".txt (*.txt)|*.txt"
        OpenFileDialog1.DefaultExt = "txt"
 
        If OpenFileDialog1.ShowDialog = Windows.Forms.DialogResult.OK Then
 
            Dim myReader As New System.IO.StreamReader(OpenFileDialog1.FileName)
 
            If myReader.ReadLine = "|" Then
 
                Do Until myReader.ReadLine = "||"
                    ListBox1.Items.Add(myReader.ReadLine)
                Loop
 
            End If
 
        End If
 
    End Sub

I still get an error in "Immediate Window" that says "A first chance exception of type 'System.ArgumentNullException' occurred in System.Windows.Forms.dll" What do I do?

[knicksfan426]

bump

[jmcilhinney]

That's not an error, or at least not one you have to worry about. A message about a "first chance exception" means that an exception has been thrown, which is perfectly legal. It when an exception isn't caught that you have to worry. Does the application keep running and do what it's supposed to do?

[minitech]

First chance exceptions don't really seem to mean much. Just ignore them if you see that in the Debug window.

[knicksfan426]

That's not an error, or at least not one you have to worry about. A message about a "first chance exception" means that an exception has been thrown, which is perfectly legal. It when an exception isn't caught that you have to worry. Does the application keep running and do what it's supposed to do?

No, the application freezes.

[jmcilhinney]

I just noticed that you have On Error Resume Next in your code. Get rid of that and never use it again. Now your code will crash and we can see what the actual issue. Then we can fix it.

[minitech]

1. Try taking out the OpenFileDialog1.DefaultExt line.
2. Make sure the program has permissions to the file.
3. Take out On Error Resume Next if it still doesn't work.

[knicksfan426]

1. Try taking out the OpenFileDialog1.DefaultExt line.
2. Make sure the program has permissions to the file.
3. Take out On Error Resume Next if it still doesn't work.

I did everything you said, my application still freezes. To everyone, here's the open file dialog code:

Note: Please remember the "|" is the separator I have made that comes before all listbox1 items and "||" is the separator that comes before all listbox2 items.

Private Sub OpenToolStripMenuItem_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles OpenToolStripMenuItem.Click
        OpenFileDialog1.Filter = ".txt (*.txt)|*.txt"
 
        If OpenFileDialog1.ShowDialog = Windows.Forms.DialogResult.OK Then
 
            Dim myReader As New System.IO.StreamReader(OpenFileDialog1.FileName)
 
            If myReader.ReadLine = "|" Then
 
                Do Until myReader.ReadLine = "||"
                    ListBox1.Items.Add(myReader.ReadLine)
                Loop
 
            End If
 
        End If
 
    End Sub
End Class

At the line of code where it says "ListBox1.Items.Add(myReader.ReadLine)", VB says there is an error there: "Value cannot be null.Parameter name: item" What do I do to fix this?

[knicksfan426]

bump

[minitech]

Change Do Until myReader.ReadLine = "||" To Do Until (myReader.Peek() = -1) OrElse (myReader.ReadLine = "||").
See if that fixes the problem.

[knicksfan426]

Change Do Until myReader.ReadLine = "||" To Do Until (myReader.Peek() = -1) OrElse (myReader.ReadLine = "||").
See if that fixes the problem.

There is improvement but it is minimal. The file is opened with out any errors now but only part of the file is being opened, not all text is being distributed to the proper listboxes.

[minitech]

That's because you haven't added a loop for the second listbox.
After your first loop, add:

While myReader.Peek > -1
     Me.ListBox2.Items.Add(myReader.ReadLine())
End While

By the way, if you haven't already, check your file; It needs to look like this:

|
List Box 1 Item 1
List Box 1 Item 2
||
List Box 2 Item 1
List Box 2 Item 2
List Box 3 Item 3

If it doesn't, I have no idea what the problem is.