C# file open dialog
file open and the name return
OpenFileDialog openFile = new OpenFileDialog();
openFile.DefaultExt = "jpg";
openFile.Filter = "Images Files(*.jpg; *.jpeg; *.gif; *.bmp; *.png)|*.jpg;*.jpeg;*.gif;*.bmp;*.png";
openFile.ShowDialog();
if (openFile.FileNames.Length > 0)
{
foreach (string filename in openFile.FileNames)
{
this.textBox1.Text = filename;
}
}
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