How to read a file directly without including in project appln? [Resolved]

Using inputstream is correct method

Stream stread = fileupload.PostedFile.InputStream;
StreamReader reader = new StreamReader(stread);
char[] delimiters = new char[] { ',' };
string[] columns = reader.ReadLine().Split(delimiters);

Oswaldlily, if this helps please login to Mark As Answer. | Alert Moderator

Use IO class to do that , as depicted below

try

        {

            // Create an instance of StreamReader to read from a file.

            // The using statement also closes the StreamReader.

            using (StreamReader sr = new StreamReader("YouFilewithPath.txt"))

            {

                String line;

                // Read and display lines from the file until the end of

                // the file is reached.

                while ((line = sr.ReadLine()) != null)

                {

                 response.write.(line);

                }

            }

        }

        catch (Exception ex)

        {

             //trap errors here 

        }

    }

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Oswaldlily, if this helps please login to Mark As Answer. | Alert Moderator

Hi Vuyiswamb is correct, but to execute this code properly you need to upload your local file to server.
unless it's not possible.

Himanshu Manjarawala
Sr. Software Engineer@AutomationAnywhere
http://fieredotnet.wordpress.com/

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1)How to add to server thru code?

2)If added in server,whn i try to import 100 files,it is wastage of memory to be add 100 files in server.

so i tried to use File.delete(filename)

3)If above is the case,if From same server 2 or 3 people tryin to import same file,this delete function can disturb others who readin the same file..

so how to avoid all

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Here you have to be careful while giving file name.
if possible then try to keep files seperation according to user.
so when more then one user try to access same file it could be managable for us.

Himanshu Manjarawala
Sr. Software Engineer@AutomationAnywhere
http://fieredotnet.wordpress.com/

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I used in this way
string path=File.ReadAllText(fileupload.FileName);

now my error is

Could not find file
'C\Program files\Common files\Microsoft Shared\Dev Server\10.0\filename.csv

my original path is C;/Desktop/folder/filename

what is this
C\Program files\Common files\Microsoft Shared\Dev Server\10.0?????????????????

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will you give me your code??

Himanshu Manjarawala
Sr. Software Engineer@AutomationAnywhere
http://fieredotnet.wordpress.com/

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? string path=File.ReadAllText(fileupload.FileName);
String strConnection = "Data Source=Sourcename-30;Initial Catalog=catalogname;User ID=sa;Password=abc25;";
string fullPath = System.IO.Path.GetFullPath(fileupload.FileName);
string file = System.IO.Path.GetFileName(fullPath);
string dirPath = System.IO.Path.GetDirectoryName(fullPath);

string excelConnectionString = "Provider=Microsoft.ACE.OLEDB.12.0;" + "Data Source= \"" + dirPath + "\\\";" + "Extended Properties=\"text;HDR=YES;FMT=Delimited\"";

OleDbConnection excelConnection = new OleDbConnection(excelConnectionString);
excelConnection.Open();
OleDbCommand cmd = new OleDbCommand("Select * from [" + fileupload.FileName +"]", excelConnection);
OleDbDataReader dReader = cmd.ExecuteReader();

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