how to send directory of a file from form to the class

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Let say in a form,I select which file from the desktop is to be read through menu strip.but how to send directory of that file to the class so that the class know which file is to be read?The class already have all the readtext code.thanks.

Posted 19-Jul-12 23:37pm

Esmond90

Updated 19-Jul-12 23:38pm

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dimpledevani 20-Jul-12 5:38am

If you are working on web then use FileUpload Control or you must use FileDialog box control to select and specify your required file

Esmond90 20-Jul-12 5:47am

Thanks for reply.Ya,i use FileDialog box.But how the code should be written at directory path for the class?

eg. class: System.IO.StreamReader sr = new System.IO.StreamReader(DIRECTORY PATH)

Form: DialogResult result = this.openFileDialog1.ShowDialog();

if (result == DialogResult.OK) { string file = openFileDialog1.FileName;

}

Sunny_Kumar_ 20-Jul-12 5:52am

Initialize the streamreader with a filestream like this:
StreamReader sr = new StreamReader(new FileStream("filepath", FileMode.Open));

As I know, StreamReader never takes a Directory as an argument.
replace "filepath" with the actual filename which you're selecting using OpenFileDialog.

Esmond90 20-Jul-12 6:00am

Ok.Thanks for the information.I am trying on it,let you guys know when I know what actually the problem is.Thanks!

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Sunny_Kumar_ · Answer 1 · 2012-07-19T23:49:00

Hi Esmond,

Well, first I would like to let you know that if you're using OpenFileDialog to select a file, it returns the full path of selected file or files, So you won't need at all to indicate the drive to search a file. Just pass the complete file name which is absolute. If you're using any other mean to select a file then please specify that in your question using "Improve question" link in your post so that we can get a better picture of your problem.

You can parse contents from a text file like this:

string DataInTextFile = "";

            OpenFileDialog ofd = new OpenFileDialog();
            ofd.Filter = "Text Files|*.txt";
            DialogResult dr = ofd.ShowDialog();
            if (dr == DialogResult.OK)
            {
                StreamReader SR = new StreamReader(ofd.FileName);
                DataInTextFile = SR.ReadToEnd();
                SR.Close();
                SR.Dispose();
            }

            MessageBox.Show(DataInTextFile);

Go ahead with the manipulation on text file contents using "DataInTextFile" string object.

Hope this helps.

Happy Coding :)