how to retrieve file data(.txt) to respective text box?

Question

in my code i am writing text boxes data to a text file using save file dialog, which will save my text box data to a specified text file.and my problem is i need to retrieve back file data to respective text boxes when ever user required ...how can i do it?

  private void SaveData_Click(object sender, EventArgs e)
    {
       // Stream myStream;
        SaveFileDialog saveFileDialog1 = new SaveFileDialog();

        saveFileDialog1.InitialDirectory = "c:\\";
        saveFileDialog1.Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*";
        saveFileDialog1.FilterIndex = 2;
        saveFileDialog1.RestoreDirectory = true;

        if (saveFileDialog1.ShowDialog() == DialogResult.OK)
        {
            if ((myStream = saveFileDialog1.OpenFile()) != null)
            {
                // Code to write the stream goes here.
                using (StreamWriter objWriter = new StreamWriter(myStream))
                {
                    objWriter.Write(textBox1.Text);
                    objWriter.Write(","); 
                    objWriter.Write(textBox2.Text);
                    objWriter.Write(",");
                    objWriter.Write(textBox3.Text);
                    objWriter.Write(",");
                    objWriter.Write(textBox4.Text);

                    MessageBox.Show("Details have been saved");
                }


                myStream.Close();
            }
        }
    }

    private void Retrieve_Click(object sender, EventArgs e)
    {
        //Stream myStream = null;
        OpenFileDialog openFileDialog1 = new OpenFileDialog();

        openFileDialog1.InitialDirectory = "c:\\";
        openFileDialog1.Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*";
        openFileDialog1.FilterIndex = 2;
        openFileDialog1.RestoreDirectory = true;

        if (openFileDialog1.ShowDialog() == DialogResult.OK)
        {
            try
            {
                if ((myStream = openFileDialog1.OpenFile()) != null)
                {
                    using (myStream)
                    {
                        // Insert code to read the stream here.
                        textBox1.Text = (myStream).ToString();
                      textBox2.Text = ().ToString();
                      textBox3.Text = ().Tostring();
                      textBox4.text = ().Tostring();

                    }
                }
            }
            catch (Exception ex)
            {
                MessageBox.Show("Error: Could not read file from disk. Original error: " + ex.Message);
            }
        }
    }

Answer 1

You need to use an OpenFileDialog Control and pass it to the ReadAllText Method ..

Here is an Example :

    myAmazingTextBox.Text = File.ReadAllText(openFileDialog1.FileName);

Answer 2

Save the location where user saves data.\ Next time you should read the path first. You can save the saveFileDialog1.FileName.

Answer 3

Use the code below to retrieve the text saved in the file (code changed within try block)

private void Retrieve_Click(object sender, EventArgs e)
{
    //Stream myStream = null;
    OpenFileDialog openFileDialog1 = new OpenFileDialog();

    openFileDialog1.InitialDirectory = "c:\\";
    openFileDialog1.Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*";
    openFileDialog1.FilterIndex = 2;
    openFileDialog1.RestoreDirectory = true;

    if (openFileDialog1.ShowDialog() == DialogResult.OK)
    {
        try
        {
            // Read all text stored in the file
            string fileData = File.ReadAllText(openFileDialog1.FileName);

            // As you are appending textbox data using comma as separator, 
            // so split the text read from file on comma separator
            string[] parts = fileData.Split(','); 

            // as there were 4 textboxes, so after split, the 'parts' array should contain 4 elements, otherwise, the file/data is invalid
            if(parts.Length != 4)
            {
                 MessageBox.Show("Invalid source file.");
                 return;
            }

            // set the respective values into the textboxes 
            textBox1.Text = parts[0];
            textBox2.Text = parts[1];
            textBox3.Text = parts[2];
            textBox4.text = parts[3];
        }
        catch (Exception ex)
        {
            MessageBox.Show("Error: Could not read file from disk. Original error: " + ex.Message);
        }
    }

I would like to suggest that you don't use comma as a separator as the user may enter text which contains comma in the content. I would suggest that you encode the text into Base64 string (which contains only A-Za-z0-9 with additional two characters which does not contain comma. So, you can separate the base64 encoded string with comma separator as then you will be 100% sure that comma is only the separator and not the part of content.

While reading, decode the base64 string and show into textboxes.