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Get complete file path from file click in File Explorer

Using Visual Studio 2017 and Windows 10 I want to be able to open a file explorer and navigate to a file outside of the program. Once my file is collected I want to get the file path and the complete file name for the file explorer.
OpenFileDialog openFileDialog1 = new OpenFileDialog();
openFileDialog1.Filter = "All files (*.*)|*.*";
if (openFileDialog1.ShowDialog() == DialogResult.OK)
textBoxFolderpath.Text= openFileDialog1.FileName;
if (textBoxFolderpath.Text.Contains('\\'))
textBoxFolderpath.Text = textBoxFolderpath.Text.Substring(0, textBoxFolderpath.Text.LastIndexOf('\\'));
This is what I have tried so far. The textBoxFolderpath is being used to hold the values.
What am I missing or doing incorrectly?

When working with filenames and paths I'd highly recommend using built-in classes to handle this, like the Path class:
Path.GetDirectoryName(openFileDialog1.FileName)
This returns the path, without the filename. It's also cross-platform compatible.

Thanks Zer0 - below is what I ended up using.
private void btnCurrentFolder_Click(object sender, EventArgs e)
{
OpenFileDialog openDialog1 = new OpenFileDialog();
// Determine starting directory
if (chkSetToRoot.Checked)
{
openDialog1.InitialDirectory = #"K:\RESULTS";
}
openDialog1.Title = "Select A Test File";
openDialog1.Filter = "All Files (*.*)|*.*";
if (openDialog1.ShowDialog() == DialogResult.OK)
{
textBoxFolderpath.Text = Path.GetDirectoryName(openDialog1.FileName);
textBoxFileName.Text = Path.GetFileName(openDialog1.FileName);
}
}

C# winform File.Copy from a SaveDialog

I am attempting to copy a .txt from my application directory or some kind an export feature to users desire path and filename using savedialog on C# my code is below.
private void button2_Click(object sender, EventArgs e)
{
string directory = AppDomain.CurrentDomain.BaseDirectory + "output.txt";
using (SaveFileDialog dialog = new SaveFileDialog())
{
dialog.Filter = "txt files (*.txt);
dialog.FilterIndex = 2;
dialog.RestoreDirectory = true;
if (dialog.ShowDialog() == DialogResult.OK)
{
File.Copy(directory, Path.GetDirectoryName(dialog.FileName) + dialog.FileName);
}
}
}
But I am getting an error
The given path's format is not supported.
I am new with C# and want to understand this error and in addition, I want to set the file name extention default as .txt also, any suggestion would be great.

There are a couple of things you need to change.
First, of course, is your copy call. This line makes no sense
File.Copy(directory, Path.GetDirectoryName(dialog.FileName) + dialog.FileName);
dialog.FileName contains already the full file name of your destination file. So there is no need to extract the directory and then add all the path again. Write just
File.Copy(directory, dialog.FileName);
But this creates a possible error. What if your user doesn't change the destination folder to another directory? You end up writing on the same file you want to read.
So I would add a sanity check like this
if(directory == dialog.FileName)
MessageBox.Show("Copy","Choose a different output folder");
else
File.Copy(directory, dialog.FileName);
Finally, if you want to force the output file to have always the .TXT extension you could add this line to the SaveDialog configuration
// Fix also your filter property. The one you have is invalid
dialog.Filter = "txt files (*.txt)|*.txt";
dialog.FilterIndex = 0; // 2 ?? There is no index 2 in your filter string
dialog.RestoreDirectory = true;
// Force the .TXT extension
dialog.AddExtension = true;

How to use openfiledialog to open any file as text in C#?

I am writing a winforms program in C# that uses the openfiledialog. I would like it to be able to take the file that the user selected and open it as text, regardless of the file type.
I tried it like this:
if (openFileDialog1.ShowDialog() == DialogResult.OK)
{
textBox1.Text = Process.Start("notepad.exe", openFileDialog1.ToString()).ToString();
}
However, that didn't work and I'm not sure if I"m even on the right track.

You should use this code:
First add this namespace :
using System.IO;
Then add this codes to your function:
OpenFileDialog openFileDialog = new OpenFileDialog();
openFileDialog.Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*";
if (openFileDialog.ShowDialog()== DialogResult.OK)
{
textBox1.Text = File.ReadAllText(openFileDialog.FileName);
}

To open the file using notepad, you need to pass the file name as second parameter of Start method. For example:
using (var ofd = new OpenFileDialog())
{
if(ofd.ShowDialog()== DialogResult.OK)
{
System.Diagnostics.Process.Start("notepad.exe", ofd.FileName);
}
}
Also if for any reason while knowing that not all file contents are text, you are going to read the file content yourself:
using (var ofd = new OpenFileDialog())
{
if(ofd.ShowDialog()== DialogResult.OK)
{
var txt = System.IO.File.ReadAllText(ofd.FileName);
}
}

What you are doing at the moment is starting a Process with the argument openFileDialog1.ToString(), calling ToString() on the process and setting this as text in the TextBox. If the path was valid, the result would probably be something like "System.Diagnostics.Process". But since you use openFileDialog1.ToString() as a path, your application probably crashes with a file not found error.
To get the selected file of an OpenFileDialog, use openFileDialog1.FileName. (See the docs here)
What I think you actually want to do, is read from the file and write its contents as text to the TextBox. To do this, you need a StreamReader, like so:
if (openFileDialog1.ShowDialog() == DialogResult.OK)
{
using(var reader = new StreamReader(openFileDialog1.FileName))
{
textBox1.Text = reader.ReadToEnd();
}
}
This way, you open the file with the StreamReader, read its contents and then assign them to the text box.
The using statement is there because a StreamReader needs to be disposed of after you're done with it so that the file is no longer in use and all resources are released. The using statement does this for you automatically.

Win Form: SaveFileDialog

I added the following piece of code to a save button:
if (saveFileDialog1.ShowDialog() == DialogResult.OK)
{
FileStream fs = new FileStream(saveFileDialog1.FileName, FileMode.Create);
StreamWriter writer = new StreamWriter(fs);
writer.Write(twexit.Text); // twexit is previously created
writer.Close();
fs.Close();
}
When I type the name of the file and click save, it says that file does not exist. I know it does not exist but I set FileMode.Create. So, shouldnt it create file if it does not exist?

There is an option CheckFileExists in SaveFileDialog which will cause the dialog to show that message if the selected file doesn't exist. You should leave this set to false (this is the default value).

You can simply use this:
File.WriteAllText(saveFileDialog1.FileName, twexit.Text);
instead of lot of code with stream. It create new file or overwrite it.
File is class of System.Io . If you want to say if file exist, use
File.Exist(filePath)
Bye

Use like this:
SaveFileDialog dlg = new SaveFileDialog();
dlg.Filter = "csv files (*.csv)|*.csv";
dlg.Title = "Export in CSV format";
//decide whether we need to check file exists
//dlg.CheckFileExists = true;
//this is the default behaviour
dlg.CheckPathExists = true;
//If InitialDirectory is not specified, the default path is My Documents
//dlg.InitialDirectory = Application.StartupPath;
dlg.ShowDialog();
// If the file name is not an empty string open it for saving.
if (dlg.FileName != "")
//alternative if you prefer this
//if (dlg.ShowDialog() == System.Windows.Forms.DialogResult.OK
//&& dlg.FileName.Length > 0)
{
StreamWriter streamWriter = new StreamWriter(dlg.FileName);
streamWriter.Write("My CSV file\r\n");
streamWriter.Write(DateTime.Now.ToString());
//Note streamWriter.NewLine is same as "\r\n"
streamWriter.Write(streamWriter.NewLine);
streamWriter.Write("\r\n");
streamWriter.Write("Column1, Column2\r\n");
//…
streamWriter.Close();
}
//if no longer needed
//dlg.Dispose();

File Open/Save Dialog

I am using my own Custom View to show the files and folders and also using a search box to jump to a specific folder. In that case How to send a message to File Open/Save dialog to enforce it to change the current displayed folder.
e.g. If the dialog shows files and folders of current displaying folder "C:\", I want an API (or any piece of code) to enforce to change the current folder to "D:\"

You can have the dialog open at a specific directory using InitialDirectory.
If you want to control what the dialog does at runtime, that's a bit more complex.

Set SaveFileDialog.InitialDirectory after you create it, but before you open it.
For example:
Stream myStream = null;
SaveFileDialog saveFileDialog1 = new SaveFileDialog();
saveFileDialog1 .InitialDirectory = "d:\\" ;
saveFileDialog1 .Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*" ;
saveFileDialog1 .FilterIndex = 2 ;
saveFileDialog1 .RestoreDirectory = true ;
if(saveFileDialog1 .ShowDialog() == DialogResult.OK)
{
try
{
if ((myStream = saveFileDialog1 .OpenFile()) != null)
{
// Code to write the stream goes here.
myStream.Close();
}
}
catch (Exception ex)
{
MessageBox.Show("Error: Could not save file to disk. Original error: " + ex.Message);
}
}

set InitialDirectory property to any path