Call ShowDialog() to display an OpenFileDialog : Open File Dialog : GUI Windows Form C# Source Code

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);

        }
    }
}

 
    
   
  
   

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• Set Filter
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