Open a File using openfiledialog...

[helloitsme]

Hello!

I am trying to open a file that is selected by a user from the openfiledialog box, but the file does not open. Below is the part of the code that does so...

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OpenFileDialog fdlg = new OpenFileDialog();
fdlg.Filter = "Text Files(*.txt)|*.txt";
fdlg.RestoreDirectory = true;
if (fdlg.ShowDialog() == DialogResult.OK)
{
// fdlg.OpenFile(); //this does not open the selected file
textBoxOutput.Text = fdlg.FileName;
fdlg.OpenFile(); //this does not open the selected file

}
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Please help....

Thank you....

[Banfa]

fdlg.OpenFile(); //this does not open the selected file

What evidence do you have for that since you do not save the the stream returned by this method?

[helloitsme]

Thank you for replying...

According to what I understand from the description of the filedialog, the method openfile opens the file that is selected by the user in the dialog box of filedialog that appears...

Please let me know if I understood something wrong...I am not sure what is to be saved for that to happen...

Thnax..

[Banfa]

According to what I understand from the description of the filedialog, the method openfile opens the file that is selected by the user in the dialog box of filedialog that appears...

That is my understanding of the description to, however in your first post you have asserted that this doesn't happen but provided no evidence of that assertion.

stream^ instream = fdlg.OpenFile();
 

[helloitsme]

Hello..Thanx for replying...

well, I do not understand what you mean by evidence in this case....

if you mean I need a stream reader or somthing, then I am able to read the file using a streamreader and then display the contents in a richtextbox. but what I want is that the file should open as it does when you open it up from the windows explorer....

please let me know ....
thanx again !