Call ShowDialog() to display an OpenFileDialog

 

using System;
using System.Windows.Forms;

class MainClass {
    static void Main(string[] args) {

        OpenFileDialog dlg = new OpenFileDialog();

        if (dlg.ShowDialog() == DialogResult.OK) {
            Console.WriteLine(dlg.FileName);

        }
    }
}

Related examples in the same category

1.	Set InitialDirectory
2.	Set CheckFileExists to true
3.	Set Filter
4.	FileOk Action
5.	shows browsing for a set of files
6.	shows browsing for a file
7.	Open File Dialog with file types
8.	Demonstrates using an OpenFileDialog to prompt for a file name, and to open a file
9.	Show the use of some of the OpenFile dialog box

Call ShowDialog() to display an OpenFileDialog : Open File Dialog « GUI Windows Form « C# / C Sharp

Related examples in the same category