Develop Reference

How to close a text file that is opened through openDialogBox?

I have created a button that opens a text file through openDialogBox1 and display the content in a listBox. Then I wanted to create another button to close this file, but I couldn't figure out what code to use. Can anyone help me please? Thank you.
Here is a sample of my openFile button code :
private void openButton_Click(object sender, EventArgs e)
{
try
{
StreamReader inputFile;
string File;
OpenFileDialog openFileDialog1 = new OpenFileDialog();
openFileDialog1.DefaultExt = "txt";
openFileDialog1.Filter = "Text Files (*txt)|*txt";
if (openFileDialog1.ShowDialog() == System.Windows.Forms.DialogResult.OK)
{
inputFile = File.OpenText(openFileDialog1.FileName);
fileListBox.Items.Clear();
while (!inputFile.EndOfStream)
{
File = inputFile.ReadLine();
fileListBox.Items.Add(File);
}
inputFile.Close();
}
}
catch
{
MessageBox.Show("Invalid File!");
}
}

Declare StreamReader inputFile; globally so it is available in the second button click.
However, the file should be opened and closed as soon as you are done with it to release resources. Plus, you don't know, the second button may never get clicked.

Open an existing text file in WPF and C#

I have a program that enables a user to search text files in an open file dialog. The user is then able to open an existing text file they choose and edit it. However, my problem is that when they file opens it appears blank. What am I missing?
private void Open_Click(object sender, RoutedEventArgs e)
{
TextBox openText = new TextBox();
var OpenFile = new Microsoft.Win32.OpenFileDialog();
Nullable<bool> Success = OpenFile.ShowDialog();
OpenFile.DefaultExt = ".txt";
OpenFile.Filter = "Text documents (.txt)|*.txt";
if (Success.HasValue && Success.Value)
{
openText.Text = OpenFile.FileName;
}
else
{
//cannot open file
}
}

Replace this:
openText.Text = OpenFile.FileName;
with this:
openText.Text = System.IO.File.ReadAllText(OpenFile.FileName);

Use File.ReadAllText()
openText.Text = File.ReadAllText(OpenFile.FileName);

How to make a duplicate copy of the file selected in the OpenFileDialog control

// Browses file with OpenFileDialog control
private void btnFileOpen_Click(object sender, EventArgs e)
{
OpenFileDialog openFileDialogCSV = new OpenFileDialog();
openFileDialogCSV.InitialDirectory = Application.ExecutablePath.ToString();
openFileDialogCSV.Filter = "CSV files (*.csv)|*.csv|All files (*.*)|*.*";
openFileDialogCSV.FilterIndex = 1;
openFileDialogCSV.RestoreDirectory = true;
if (openFileDialogCSV.ShowDialog() == DialogResult.OK)
{
this.txtFileToImport.Text = openFileDialogCSV.FileName.ToString();
}
}
In the code above, i browse for a file to open. What I want to do is, browse for a file, select it and then press ok. On clicking ok, i want to make a copy of the seleted file and give that duplicate file a .txt extension. I need help on achieving this.
Thanks

if (openFileDialogCSV.ShowDialog() == DialogResult.OK)
{
var fileName = openFileDialogCSV.FileName;
System.IO.File.Copy( fileName ,Path.Combine(Path.GetDirectoryName(fileName), Path.GetFileNameWithoutExtension(fileName)+".txt"));
}
Above code will copy selected file as txt with same name and in to same directory.
if you need to overwrite existing file with same name add another parameter to Copy method as true.
System.IO.File.Copy(source, destination, true);

You use File.Copy as follows,
File.Copy(openFileDialogCSV.FileName., openFileDialogCSV.FileName + ".txt");

Try this
private void btnFileOpen_Click(object sender, EventArgs e)
{
OpenFileDialog openFileDialogCSV = new OpenFileDialog();
openFileDialogCSV.InitialDirectory = Application.ExecutablePath.ToString();
openFileDialogCSV.Filter = "CSV files (*.csv)|*.csv|All files (*.*)|*.*";
openFileDialogCSV.FilterIndex = 1;
openFileDialogCSV.RestoreDirectory = true;
if (openFileDialogCSV.ShowDialog() == DialogResult.OK)
{
this.txtFileToImport.Text = openFileDialogCSV.FileName.ToString();
System.IO.File.Copy(this.txtFileToImport.Text,"C://123.txt")
}
}
123 can be changed by any file name that you want.

Filtering text files in C#

How do I just open files with the .txt extension, I want to my program to pop up an error message if the file is not a .txt file I want a code that can modify this code below
private void button1_Click(object sender, EventArgs e)
{
OpenFileDialog of = new OpenFileDialog();
of.ShowDialog();
textBox1.Text = of.FileName;
}
Can someone help let's say I want to put this loop
if fileextension is .txt then
OpenFileDialog of = new OpenFileDialog();
of.ShowDialog();
textBox1.Text = of.FileName;
else show error message(like can not open this file)

As I understood correctly, you want to see only txt files in your dialog?
If so, use Filter property.
OpenFileDialog of = new OpenFileDialog();
of.Filter = "Text files (*.txt)|*.txt";

Can use Path.GetExtension method for this
OpenFileDialog of = new OpenFileDialog();
if(of.ShowDialog() == DialogResult.OK)
{
if(Path.GetExtension(of.FileName).Equals("txt",
StringComparison.InvariantCultureIgnoreCase))
textBox1.Text = of.FileName;
}

You shouldn't allow all extensions if you only allow txt extension.
of.Filter = "Text Files|*.txt";
Will make the OpenFileDialog accept only txt extension files.

OpenfileDialog - Window is not popping out

I am working on form . I want small window to pop up when I click button and to will select XML file of my choice from various folder.
I guess, this OpenFileDialog will help me.
private void button3_Click(object sender, EventArgs e)
{
/
OpenFileDialog OpenFileDialog1 = new OpenFileDialog();
openFileDialog1.Filter = " XML Files|*.xml";
openFileDialog1.InitialDirectory = #"D:\";
if (OpenFileDialog1.ShowDialog() == DialogResult.OK)
{
MessageBox.Show(filed.FileName.ToString());
}
}
I tried using following code but when I click on the button there window doesn't pop up.
I am not geting what mistake I have made.
What is the problem with that?
Thanks!

You cant just open the file dialog from a console app. You will have to workaround it with some setting to single thread apartment (STA).
[STAThread]
static void Main(string[] args)
{
MessageBox.Show("Test");
}
--EDIT--
Following works on click event:
OpenFileDialog f = new OpenFileDialog();
f.Filter = "XML Files|*.xml";
f.InitialDirectory = "D:\\";
if(f.ShowDialog() == DialogResult.OK)
{
MessageBox.Show(f.FileName);
}

You cant open file fialog in console app.
You say I have button, so this must be Win app, use
openFileDialog1.ShowDialog(); is missing
private void button3_Click(object sender, EventArgs e)
{
OpenFileDialog OpenFileDialog1 = new OpenFileDialog();
openFileDialog1.Filter = " XML Files|*.xml";
openFileDialog1.InitialDirectory = #"D:\";
openFileDialog1.ShowDialog();
// Get file name and use OpenFileDialog1.FileName or something like that
}