Question

Below is my code, i tried to open my excel file in my c# application but the program give’s me an error message “Cannot Access “my excel.xls”. But when I specify the file path in my string path variable it works, the problem is I need to get the file path from an openFileDialog.

using System;
using System.IO;
using System.Collections.Generic;
using System.Text;
using System.Data;
using System.Windows.Forms;
using System.Data.OleDb;
using System.Reflection;
using MOIE = Microsoft.Office.Interop.Excel;
using OFFICE = Microsoft.Office.Core;

namespace EmpUploader
{
   public class ExcelCon
    {
       private OleDbDataReader reader = null;
       private OleDbCommand excelCommand = new OleDbCommand();
       private OleDbDataAdapter adapter = new OleDbDataAdapter();
       private DataTable excelData = new DataTable();
       private MOIE.ApplicationClass objExcel = new MOIE.ApplicationClass();
       private MOIE.Workbook wb = null;
       private string myConn = "";
       private string strSQL = "";
       private string err = "";
       private string path2 = "";
       private int sheetCount = 0;
       private OleDbConnection Con = new OleDbConnection("");

 #region "excel interop prarameters"
       private static object xl_missing = Type.Missing;
       private static object xl_true = true;
       private static object xl_false = false;


       private object xl_update_links = xl_missing;
       private object xl_read_only = xl_missing;
       private object xl_format = xl_missing;
       private object xl_password = xl_missing;
       private object xl_write_res_password = xl_missing;
       private object xl_ignore_read_only = xl_missing;
       private object xl_origin = xl_missing;
       private object xl_delimiter = xl_missing;
       private object xl_editable = xl_missing;
       private object xl_notify = xl_missing;
       private object xl_converter = xl_missing;
       private object xl_add_to_mru = xl_missing;
       private object xl_local = xl_missing;
       private object xl_corrupt_load = xl_missing;
#endregion
  }
//MY CODE FOR OPENING THE EXCEL
//note that my file path came from an openfiledialog
  public void InitializeConnection(string path)
       {


               //connection string for excel
               myConn = @"Provider=Microsoft.Jet.OLEDB.4.0; Data Source=" + path + ";      Extended Properties =Excel 8.0";
               Con.ConnectionString = myConn;
               Con.Open();

               //this is the sample specified path that worked when i test my application
               //path = @"C:\shinetsu p5 emp list.xls";
               objExcel.Visible = false;
               wb = objExcel.Workbooks.Open(path, xl_update_links, xl_read_only, xl_format, xl_password, xl_write_res_password, xl_ignore_read_only, xl_origin, xl_delimiter, xl_editable, xl_notify, xl_converter, xl_add_to_mru, xl_local, xl_corrupt_load);

           sheetCount = wb.Worksheets.Count;



       }
}

Answer 1

The OpenFileDialog class has a property “FileName” which contains the full name of the file (Path + Filename). So just use that property as your “path” variable.

http://msdn.microsoft.com/en-us/library/system.windows.forms.openfiledialog_members.aspx