How to read binary text file into datagridview ?

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hi all

I want to know how to write the code to read the binary text file "MyText.txt" which has few line of data like this, that has two values I want to get it back by separate each values by using ";" and through or using data we get to add into datagrid view line by line, i want to know how to write that code

youtube.com/watch?v=apY6IIFJfd0;youtube.com
youtube.com/watch?v=shlPQgJhG0g;youtube.com
youtube.com/watch?v=apY6IIFJfd0;youtube.com

Result i want to get in datagridview

-------------------------------------------------
|Website URL                     | Website Name |
-------------------------------------------------
|youtube.com/watch?v=apY6IIFJfd0 | youtube.com
|youtube.com/watch?v=shlPQgJhG0g | youtube.com
|youtube.com/watch?v=apY6IIFJfd0 | youtube.com

What I have tried:

I haven't try to code yet being find the solution

Posted 12-May-21 16:25pm

Mr.Kim2050

Updated 12-May-21 20:07pm

RickZeeland

v2

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OriginalGriff · Answer 1 · 2021-05-12T19:22:00

Solution 1

Use File.ReadAllLines[^] To read the data.
Process each line into separate sections using the string.Split[^] method on teh semicolon.
Then you have a choice: either add the items to existing columns of a DGV, or create a "holding class" to hold the two values in properties, and build a collection of those to provide the DGV with a DataSource.

VB

Dim lines As String() = File.ReadAllLines("d:\Test Data\yt.txt")
Dim items As List(Of [MyClass]) = lines.[Select](Function(l) New [MyClass](l.Split(";"c))).ToList()
myDataGridView.DataSource = items

VB.NET

Public Class [MyClass]
    Public Property URL As String
    Public Property Name As String

    Public Sub New(ByVal arr As String())
        URL = arr(0)
        Name = arr(1)
    End Sub
End Class

Your choice!

Posted 12-May-21 19:22pm

OriginalGriff

Comments

Maciej Los 13-May-21 2:01am

5ed!

Mr.Kim2050 13-May-21 8:09am

hi OriginalGriff

thanks, with your reply I will try the code you send to me soon but i have some more question relate to this code below which i have search from internet how to read binary text file into DataGridview, i got 3 kinds of code which i dont know why or where error with logic, please help to edit or correct the mistake for me if you know the answer why? thanks in advance

Code1:

If My.Computer.FileSystem.FileExists(MyFileName) Then
' Need to write load text file
Dim ReadStream As FileStream
Dim AcceptValue As String

Try
' Here is the error part
ReadStream = New FileStream(MyFileName, FileMode.Open)
Dim ReadBinaryVal As New BinaryReader(ReadStream)
AcceptValue = ReadBinaryVal.ReadString()
Dim Row As String() = New String() {"Model 1", AcceptValue, "Online"}
DataGridView1.Rows.Add(Row)
ReadStream.Close()
Else
Try
TextBinaryWrite = New BinaryWriter(New FileStream(MyFileName, FileMode.Create))
txtStatuse.Text = "Statuse : " & MyFileName & " is created."
TextBinaryWrite.Close()
Catch ex As Exception
txtStatuse.Text = ex.Message + "Cannot create file."
Return
End Try
End If

Mr.Kim2050 13-May-21 8:10am

Error code 2:

If My.Computer.FileSystem.FileExists(MyFileName) Then
Dim Br As BinaryReader
Dim AcceptValue As String

Try
Using BinReader As New BinaryReader(File.Open(MyFileName, FileMode.Open))
Dim Pos As Integer = 0
Dim Length As Integer = BinReader.BaseStream.Length

While Pos < Length
Dim AcceptValue As String = BinReader.ReadString
Dim Row As String() = New String() {Pos, AcceptValue, "Online"}
DataGridView1.Rows.Add(Row)
Pos += 1
End While
End Using

Else
Try
TextBinaryWrite = New BinaryWriter(New FileStream(MyFileName, FileMode.Create))
txtStatuse.Text = "Statuse : " & MyFileName & " is created."
TextBinaryWrite.Close()
Catch ex As Exception
txtStatuse.Text = ex.Message + "Cannot create file."
Return
End Try
End If

Mr.Kim2050 13-May-21 8:10am

Error code 3:

Try
Br = New BinaryReader(New FileStream(MyFileName, FileMode.Open))
AcceptValue = Br.ReadString()
Dim Row As String() = New String() {"Pos1", AcceptValue, "Online"}
DataGridView1.Rows.Add(Row)
Catch ex As Exception
txtStatuse.Text = ex.Message & "Error Load File"
Return
End Try
Br.Close()
Else
Try
TextBinaryWrite = New BinaryWriter(New FileStream(MyFileName, FileMode.Create))
txtStatuse.Text = "Statuse : " & MyFileName & " is created."
TextBinaryWrite.Close()
Catch ex As Exception
txtStatuse.Text = ex.Message + "Cannot create file."
Return
End Try