How to open existing excel sheet from floder

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1.00/5 (1 vote)

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C#

C#4.0

Want to open existing workbook from folder using c#.net web application.

Posted 23-Sep-15 21:12pm

Member 11838322

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Tadit Dash (ତଡିତ୍ କୁମାର ଦାଶ) 24-Sep-15 3:20am

So, what is the issue?

Member 11838322 24-Sep-15 13:04pm

here my code:
string extn = Path.GetExtension(FileUpload18.PostedFile.FileName);
Microsoft.Office.Interop.Excel.Application xlApp;
Microsoft.Office.Interop.Excel.Workbook xlWorkBook;
Microsoft.Office.Interop.Excel.Worksheet xlWorkSheet;
object misValue = System.Reflection.Missing.Value;
xlApp = new Microsoft.Office.Interop.Excel.ApplicationClass();
string constr = Server.MapPath("~\\Input_MRDB_Cost_Upload\\MRDB_Cost rates_Template.xlsx");
//xlWorkBook = xlApp.Workbooks.Open(constr, 0, false, 5, "", "", false, Excel.XlPlatform.xlWindows, "",true, false, 0, true, false, false);
xlWorkBook = xlApp.Workbooks.Open(Filename: @constr, ReadOnly: false);
xlWorkSheet = (Microsoft.Office.Interop.Excel.Worksheet)xlWorkBook.Worksheets.get_Item("MRDB_Cost rates_Template");
Microsoft.Office.Interop.Excel.Range formatRange;
formatRange = xlWorkSheet.get_Range("A3").EntireColumn;
formatRange.NumberFormat = "@";
string constr1 = Server.MapPath("~/Input_MRDB_Cost_Upload/") + Guid.NewGuid() + extn;
xlApp.ActiveWorkbook.SaveAs(constr1, Microsoft.Office.Interop.Excel.XlFileFormat.xlWorkbookDefault, Type.Missing, Type.Missing, false, false, Microsoft.Office.Interop.Excel.XlSaveAsAccessMode.xlNoChange, Type.Missing, Type.Missing, Type.Missing, Type.Missing, Type.Missing);
xlApp.ActiveWorkbook.Saved = true;
// xlWorkBook.SaveAs(constr, Microsoft.Office.Interop.Excel.XlFileFormat.xlWorkbookNormal, null, null, false, false, Microsoft.Office.Interop.Excel.XlSaveAsAccessMode.xlShared, false, false, null, null, null);
xlWorkBook.Close();
xlApp.Quit();

Sorry, we couldn't find C:\Users\goutam.x.pal\Desktop\NEWDFM\Input_MRDB_Cost_Upload\MRDB_Cost rates_Template.xlsx. Is it possible it was moved, renamed or deleted?

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Sisir Patro · Answer 1 · 2015-09-23T21:33:00

Solution 1

Hi,

you can use process to open any file.

System.Diagnostics.Process.Start(@"file path");

Thanks,
Sisir Patro

Posted 23-Sep-15 21:33pm

Sisir Patro