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

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C# / C Sharp » GUI Windows Form » Open File Dialog Screenshots

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

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