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Open File Dialog

C# - Jan Welker Februar 27, 2006
4 Ratings 3.25/5

Sprache: C#

Open File Dialog mit Filter
string Pfad = string.Empty;

OpenFileDialog openFileDialog1 = new OpenFileDialog();
openFileDialog1.Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*";
if(openFileDialog1.ShowDialog() == DialogResult.OK)
    Pfad = openFileDialog1.FileName;
string Pfad = string.Empty;

OpenFileDialog openFileDialog1 = new OpenFileDialog();
openFileDialog1.Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*";
if(openFileDialog1.ShowDialog() == DialogResult.OK)
    Pfad = openFileDialog1.FileName;
Abgelegt unter DialogResult FileName Filter openFileDialog

Alte URL: /snippet/open-file-dialog/15

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