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Multiple filters OpenFileDialog

At the moment I have three buttons on a form, each opens a different form (form2 with a textbox to display the text from the textfile, form3 with a picturebox to display the image)
What I am trying to do is put the two together for my last button so the user can filter which type to open (TXT Files or Image files). I am not sure how I can put the two together and get them to work.
The code I used to just open text files:
private void button1_Click(object sender, EventArgs e)
{
OpenFileDialog ofd = new OpenFileDialog();
ofd.InitialDirectory = #"C:\";
ofd.Filter = "TXT Files(*.txt;)|*.txt;";
if(ofd.ShowDialog() == DialogResult.OK)
{
using(StreamReader rdText = new StreamReader(ofd.FileName))
{
string info = File.ReadAllText(ofd.FileName);
TextDocumentForm newTextDocument = new TextDocumentForm();
newTextDocument.TextFileName = info;
newTextDocument.Show();
}
}
}
What I use to open my image files
private void button2_Click(object sender, EventArgs e)
{
OpenFileDialog ofdi = new OpenFileDialog();
ofdi.InitialDirectory = #"C:\";
ofdi.Filter = "Image Files(*.jpg;*.jpeg;*.bmp)|*.jpg;*.jpeg;.bmp;";
if (ofdi.ShowDialog() == DialogResult.OK)
{
Image image = Image.FromFile(ofdi.FileName);
ImgDoc newImageDoc = new ImgDocumentForm();
newImageDoc.ImageShow = image;
newImageDoc.Show();
}
}
Any help is appreciated as I am trying to develop my understanding of how OpenFileDialog still works.

Combining filters:
var openFile = new OpenFileDialog
{
InitialDirectory = #"C:\",
Filter = "TXT Files(*.txt;)|*.txt;|Image Files(*.jpg;*.jpeg;*.bmp)|*.jpg;*.jpeg;.bmp;"
};
Then use Path.GetExtension() to see which route you should take:
if (openFile.ShowDialog() == true)
{
var ext = System.IO.Path.GetExtension(openFile.FileName);
if (ext == ".txt")
{
// Open text file
}
else if (ext == ".jpg" || ext == ".jpeg" || ext == ".bmp")
{
// Open image file
}
}

Stream read selected listbox item query

I have an app that reads a directory, gets a list of the files (segy) and populates a listbox on the left side of the app with the filenames. Upon clicking an item in the listox I'd like the rich text box on the right to display the content of the file.
I have the app working if I use the openfiledialog to select one of the files in the directory, I'm having issues trying to get the stream reader to read the selected file I've clicked.
The working simple openfiledialog code below.
openFileDialog1.Filter = "All Files|*.*";
openFileDialog1.Title = "Open SEG-Y Files";
DialogResult result = openFileDialog1.ShowDialog();
StreamReader readFile = new StreamReader(openFileDialog1.FileName, ebcdic);
readFile.BaseStream.Seek(0, SeekOrigin.Begin);
readFile.Read(data, 0, 3200);
string stringData = "";
for (int i = 0; i < data.Length; i++)
{
if ((i % 80) == 0 && stringData != "")
stringData += Environment.NewLine;
stringData += data[i].ToString();
}
rtbHeader.Text = stringData;
rtb.AppendText(value);
rtb.AppendText(System.Environment.NewLine);
My code
private void txtUpdate(string value)
{
lstFiles.Items.Add(value + Environment.NewLine);
lstFiles.TopIndex = lstFiles.Items.Count - 1;
lstFiles.Update();
}
private void btnFolder_Click(object sender, EventArgs e)
{
txtPath.Text = "";
lstFiles.Items.Clear();
rtbHeader.Clear();
DialogResult result = folderBrowserDialog1.ShowDialog();
if (result == DialogResult.OK)
{
txtPath.Text = folderBrowserDialog1.SelectedPath;
}
}
private void btnFiles_Click(object sender, EventArgs e)
{
lstFiles.Items.Clear();
string path = txtPath.Text;
List<string> files = new List<string>(Directory.EnumerateFiles(txtPath.Text, "*.sgy", SearchOption.AllDirectories).Select(Path.GetFileName).OrderBy(x => x));
if (files == null || files.All(x => string.IsNullOrWhiteSpace(x)))
{
MessageBox.Show("There are no files with extension" + " sgy", "Error", MessageBoxButtons.OK, MessageBoxIcon.Exclamation);
return;
}
foreach (string file in files)
{
this.Invoke(new Action(() => txtUpdate(file)));
}
}
private void lstFiles_MouseClick(object sender, MouseEventArgs e)
{
rtbHeader.Clear();
String item = (Convert.ToString(lstFiles.SelectedItem));
//MessageBox.Show(item);
StreamReader readFile = new StreamReader(item, ebcdic);
readFile.BaseStream.Seek(0, SeekOrigin.Begin);
readFile.Read(data, 0, 3200);
string stringData = "";
for (int i = 0; i < data.Length; i++)
{
if ((i % 80) == 0 && stringData != "")
stringData += Environment.NewLine;
stringData += data[i].ToString();
}
rtbHeader.Text = stringData;
}
}
Im getting an Illegal characters in path exception on this bit.
StreamReader readFile = new StreamReader(item, ebcdic);
Thanks

The code example really should be simpler. Much simpler. And the problem description more specific. Much more specific. See https://stackoverflow.com/help/mcve and https://stackoverflow.com/help/how-to-ask
That said, I believe that if you change this statement in the txtUpdate() method:
lstFiles.Items.Add(value + Environment.NewLine);
to this:
lstFiles.Items.Add(value);
It will work. The exception is most likely caused by the fact that you have newline characters in your strings. Not only does that make the filename not the one you want, it's not a valid character in Windows paths.
Also note that the items in the ListBox you've added are already strings. You don't need to call Convert.ToString() on them. You can just cast them back to a string:
String item = (string)lstFiles.SelectedItem;

How to save filename only to the gridview but I need path while downloading the file

I got finally, how to download the file using the path. But I was wondering how can I keep the file name only on the grid view. While I need the full path for downloading.
On debugging I came to see that I cannot keep file name only on file upload. Since it is carried to the downloading section. If I keep file name, then file name is carried to the downloading part and the file is not downloaded.
Can anyone help me
Codes
private void UploadAttachment(DataGridViewCell dgvCell)
{
using (OpenFileDialog fileDialog = new OpenFileDialog())
{
//Set File dialog properties
fileDialog.CheckFileExists = true;
fileDialog.CheckPathExists = true;
fileDialog.Filter = "All Files|*.*";
fileDialog.Title = "Select a file";
fileDialog.Multiselect = true;
if (fileDialog.ShowDialog() == System.Windows.Forms.DialogResult.OK)
{
string strfilename = fileDialog.FileName;
cncInfoDataGridView.Rows[dgvCell.RowIndex].Cells[1].Value = strfilename;
}
}
}
/// <summary>
/// Download Attachment from the provided DataGridViewCell
/// </summary>
/// <param name="dgvCell"></param>
private void DownloadAttachment(DataGridViewCell dgvCell)
{
string fileName = Convert.ToString(dgvCell.Value);
if (!string.IsNullOrEmpty(fileName))
{
byte[] objData;
FileInfo fileInfo = new FileInfo(fileName);
string fileExtension = fileInfo.Extension;
//show save as dialog
using (SaveFileDialog saveFileDialog1 = new SaveFileDialog())
{
//Set Save dialog properties
saveFileDialog1.Filter = "Files (*" + fileExtension + ")|*" + fileExtension;
saveFileDialog1.Title = "Save File as";
saveFileDialog1.CheckPathExists = true;
saveFileDialog1.FileName = fileName;
if (saveFileDialog1.ShowDialog() == DialogResult.OK)
{
string s = cncInfoDataGridView.Rows[dgvCell.RowIndex].Cells[1].Value.ToString();
objData = File.ReadAllBytes(s);
File.WriteAllBytes(saveFileDialog1.FileName, objData);
}
}
}
}
}

Your question is not clear. But if i get you right. Why don't you use use a column for the name and a hidden/0 width column for the file path. Its a grid and therefore you may have many columns. Plus i am thinking the code below should return the full path , unless you trim and can't see where you trim .
string strfilename = fileDialog.FileName;
To get the file name only you can use the Path
string filenameOnly= System.IO.Path.GetFileName(strfilename);
The above should return your file name only you can check here for better understanding.
Add another column and set the width to 0 . Example below
DataGridViewColumn column = dataGridView.Columns[0];
column.Width = 0;
cncInfoDataGridView.Columns.Add(column);
Save your file path in the new column width a 0 width and retrieve during download.
A similar question was asked here.
How to extract file name from file path name?

Dictionary<int, byte[]> _myAttachments;
private void btnUpload_Click(object sender, EventArgs e)
{
try
{
//Throw error if attachment cell is not selected.
//make sure user select only single cell
if (dataGridView1.SelectedCells.Count == 1 && dataGridView1.SelectedCells[0].ColumnIndex == 1)
{
UploadAttachment(dataGridView1.SelectedCells[0]);
}
else
MessageBox.Show("Select a single cell from Attachment column", "Error uploading file", MessageBoxButtons.OK, MessageBoxIcon.Error);
}
catch (Exception ex)
{
MessageBox.Show(ex.Message, "Error uploading file", MessageBoxButtons.OK, MessageBoxIcon.Error);
}
}
private void btnDownload_Click(object sender, EventArgs e)
{
//Throw error if attachment cell is not selected.
//make sure user select only single cell
//and the cell have a value in it
if (dataGridView1.SelectedCells.Count == 1 && dataGridView1.SelectedCells[0].ColumnIndex == 1 && dataGridView1.SelectedCells[0].Value != null)
{
DownloadAttachment(dataGridView1.SelectedCells[0]);
}
else
MessageBox.Show("Select a single cell from Attachment column", "Error uploading file", MessageBoxButtons.OK, MessageBoxIcon.Error);
}
private void dataGridView1_CellDoubleClick(object sender, DataGridViewCellEventArgs e)
{
//Throw error if attachment cell is not selected.
//make sure user select only single cell
//and the cell have a value in it
if (dataGridView1.SelectedCells.Count == 1 && dataGridView1.SelectedCells[0].ColumnIndex == 1 && dataGridView1.SelectedCells[0].Value != null)
{
DownloadAttachment(dataGridView1.SelectedCells[0]);
}
else
MessageBox.Show("Select a single cell from Attachment column", "Error uploading file", MessageBoxButtons.OK, MessageBoxIcon.Error);
}
private void UploadAttachment(DataGridViewCell dgvCell)
{
using (OpenFileDialog fileDialog = new OpenFileDialog())
{
//Set File dialog properties
fileDialog.CheckFileExists = true;
fileDialog.CheckPathExists = true;
fileDialog.Filter = "All Files|*.*";
fileDialog.Title = "Select a file";
fileDialog.Multiselect = false;
if (fileDialog.ShowDialog() == DialogResult.OK)
{
FileInfo fileInfo = new FileInfo(fileDialog.FileName);
byte[] binaryData = File.ReadAllBytes(fileDialog.FileName);
dataGridView1.Rows[dgvCell.RowIndex].Cells[1].Value = fileInfo.Name;
if (_myAttachments.ContainsKey(dgvCell.RowIndex))
_myAttachments[dgvCell.RowIndex] = binaryData;
else
_myAttachments.Add(dgvCell.RowIndex, binaryData);
}
}
}
private void DownloadAttachment(DataGridViewCell dgvCell)
{
string fileName = Convert.ToString(dgvCell.Value);
//Return if the cell is empty
if (fileName == string.Empty)
return;
FileInfo fileInfo = new FileInfo(fileName);
string fileExtension = fileInfo.Extension;
byte[] byteData = null;
//show save as dialog
using (SaveFileDialog saveFileDialog1 = new SaveFileDialog())
{
//Set Save dialog properties
saveFileDialog1.Filter = "Files (*" + fileExtension + ")|*" + fileExtension;
saveFileDialog1.Title = "Save File as";
saveFileDialog1.CheckPathExists = true;
saveFileDialog1.FileName = fileName;
if (saveFileDialog1.ShowDialog() == DialogResult.OK)
{
byteData = _myAttachments[dgvCell.RowIndex];
File.WriteAllBytes(saveFileDialog1.FileName, byteData);
}
}
}

how to remember the last path when insert files into Listbox

i have application with Listbox and files, each time i press on Add button the default C drive open and i want the application to remember the last path i used
private void btnAdd_Click(object sender, EventArgs e)
{
System.IO.Stream myStream;
OpenFileDialog thisDialog = new OpenFileDialog();
thisDialog.InitialDirectory = "c:\\";
thisDialog.Filter = "(*.snoop, *.pcap, *.cap, *.net)|*.snoop; *.pcap; *.cap; *.net|" + "All files (*.*)|*.*";
thisDialog.FilterIndex = 1;
thisDialog.RestoreDirectory = false;
thisDialog.Multiselect = true; // Allow the user to select multiple files
thisDialog.Title = "Please Select Source File";
thisDialog.FileName = lastPath;
List<string> list = new List<string>();
if (thisDialog.ShowDialog() == DialogResult.OK)
{
foreach (String file in thisDialog.FileNames)
{
try
{
if ((myStream = thisDialog.OpenFile()) != null)
{
using (myStream)
{
listBoxFiles.Items.Add(file);
}
}
}
catch (Exception ex)
{
MessageBox.Show("Error: Could not read file from disk. Original error: " + ex.Message);
}
}
}
}

Save the last directory used in a global variable like this:
private string _lastPath = string.Empty;
then after the file selection initialize it:
if(thisDialog.Filenames.Length > 0)
_lastPath = Path.GetDirectoryName(thisDialog.Filenames[0]);
when you reopen the dialog set the InitialDirectory with this check:
thisDialog.InitialDirectory = (_lastPath.Length > 0 ? _lastPath: "c:\\");
and remove the thisDialog.FileName = lastPath;
EDIT --- UPDATE OF YOUR CODE ---
// This at the global level of your form
private string _lastPath = string.Empty;**
private void btnAdd_Click(object sender, EventArgs e)
{
System.IO.Stream myStream;
OpenFileDialog thisDialog = new OpenFileDialog();
thisDialog.InitialDirectory = (_lastPath.Length > 0 ? _lastPath: "c:\\");
thisDialog.Filter = "(*.snoop, *.pcap, *.cap, *.net)|*.snoop; *.pcap; *.cap; *.net|" + "All files (*.*)|*.*";
thisDialog.FilterIndex = 1;
thisDialog.RestoreDirectory = false;
thisDialog.Multiselect = true; // Allow the user to select multiple files
thisDialog.Title = "Please Select Source File";
thisDialog.FileName = lastPath;
List<string> list = new List<string>();
if (thisDialog.ShowDialog() == DialogResult.OK)
{
if(thisDialog.Filenames.Length > 0)
_lastPath = Path.GetDirectoryName(thisDialog.Filenames[0]);
foreach (String file in thisDialog.FileNames)
{
try
{
if ((myStream = thisDialog.OpenFile()) != null)
{
using (myStream)
{
listBoxFiles.Items.Add(file);
}
}
}
catch (Exception ex)
{
MessageBox.Show("Error: Could not read file from disk. Original error: " + ex.Message);
}
}
}
}

You can use a Visual Studio have the last path value for every execution of the application.
Only have to go to Project Properties->Configuration and add a value descriptor.
Example:
Name = LastPath; Type = string; Scope = User; Value = "Default path";
And then after you rebuild yout application, you can set this property this way:
Settings.Default.LastPath = LastPathSelected;
later, you can retrieve the value with:
thisDialog.InitialDirectory = Settings.Default.LastPath;

thisDialog.InitialDirectory = Path.GetDirectoryName(lastPath);

Yes, you can use the OpenFileDialog.InitialDirectory property. Note: you are setting the directory and not the file. So be sure to remove the filename from the path.
more info here

remove this line and you have the last path
thisDialog.InitialDirectory = "c:\\";

how to save data loaded from external file into the program?

Based on the function below, it is used to load data from file .dat. The problem is every time i load a new file, the previous file will be overwritten. how to store the data from previous file inside the program so that when loading a new file, the new data will be added to the previous one?
private void btnLoad_Click(object sender, EventArgs e)
{
try
{
OpenFileDialog openFileDialog1 = new OpenFileDialog();
openFileDialog1.Filter = "Data File (*.dat)|*.dat";
if (openFileDialog1.ShowDialog() == DialogResult.OK)
{
TextReader f = new StreamReader(openFileDialog1.FileName);
String line;
this.letterData.Clear();
this.letters.Items.Clear();
while ((line = f.ReadLine()) != null)
{
int sampleSize = Form1.DOWNSAMPLE_HEIGHT * Form1.DOWNSAMPLE_WIDTH;
char ch = char.ToUpper(line[0]);
bool[] sample = new bool[sampleSize];
int idx = 2;
for (int i = 0; i < sampleSize; i++)
{
if (line[idx++] == '1')
sample[i] = true;
else
sample[i] = false;
}
this.letterData.Add(ch, sample);
this.letters.Items.Add("" + ch);
}
f.Close();
}
MessageBox.Show(this, "File Loaded");
}
catch (Exception ex)
{
MessageBox.Show("Error: " + ex.Message);
}
}

Remove the following lines?
this.letterData.Clear();
this.letters.Items.Clear();
EDIT:
Or change to te following in order to have unique keys in your dictionary
this.letterData.Add(string.Format("{0}_{1}", openFileDialog1.FileName, ch), sample);
this.letters.Items.Add(ch.toString());