about OpenFileDialog problem

Question

Hi:

I meet a problem when using OpenFileDialog. when I writing a function to call it, and choose the file. And then , if I try to 

write data and save to another file, the operation will fail. It will pass the compiler, but, will not create any file 

if I perform the function. Is there any solution to solve the problem, thanks.

Answer 1

Hi,

Check out below sample

Create a test.txt file and select it using filedialog. and write below code

OpenFileDialog ofd = new OpenFileDialog();
            string text=string.Empty;
            if (ofd.ShowDialog() == true)
            {
                string filePath = ofd.FileName;
                string safeFilePath = ofd.SafeFileName;
                StreamReader streamReader = new StreamReader(filePath);
                text = streamReader.ReadToEnd();
                streamReader.Close();
            }
            text = text + " Rajwadi";
            using (StreamWriter outfile =  new StreamWriter(@"E:\AllTxtFiles.txt"))
            {
                outfile.Write(text);
            }

Thanks,

Rajnikant

Answer 2

Thanks a lot.