Advertisement
JannickP8

Open File c#

Apr 7th, 2022
1,040
0
Never
Not a member of Pastebin yet? Sign Up, it unlocks many cool features!
C# 0.28 KB | None | 0 0
  1. OpenFileDialog openFileDialog1 = new OpenFileDialog();
  2.             if (openFileDialog1.ShowDialog() == DialogResult.OK)
  3.             {
  4.                 openFileDialog1.Title = "Open";
  5.                 fastColoredTextBox1.Text = File.ReadAllText(openFileDialog1.FileName);
  6.             }
Advertisement
Add Comment
Please, Sign In to add comment
Advertisement