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How to Open File Dialog in C#

By Tamara Urry
Posted On Jun 26,2009
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Category: C#
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How to Open File Dialog in C#

The OpenFileDialog object interacts with the Computer’s API (Application Programming Interface) to present available files to the user and retrieves the user’s file selection back to the program. This object is part of the System.Windows.Forms library so no additional using reference will be needed.

This example uses a button click method.

private void btnOpenTextFile_Click(object sender, EventArgs e)


//First, declare a variable to hold the user’s file selection.

private void String input = string.Empty;


//Create a new instance of the OpenFileDialog because it's an object.

private voidOpenFileDialog dialog = new OpenFileDialog();


//Now set the file type

dialog.Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*";


//Set the starting directory and the title.

dialog.InitialDirectory = "C:"; dialog.Title = "Select a text file";


//Present to the user.

if (dialog.ShowDialog() == DialogResult.OK)

     strFileName = dialog.FileName;

if (strFileName == String.Empty)

     return;//user didn't select a file


If the user selected a file, then the DialogResult property value will be “OK” and we will also have the file name and path of that file. Now we can use the StreamWriter and StreamReader to read/write to the text file.
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