How to open an Excel file in C#?

Question

I am trying to convert some VBA code to C#. I am new to C#. Currently I am trying to open an Excel file from a folder and if it does not exist then create it. I am trying something like the following. How can I make it work?

Excel.Application objexcel;
Excel.Workbook wbexcel;
bool wbexists;
Excel.Worksheet objsht;
Excel.Range objrange;

objexcel = new Excel.Application();
if (Directory("C:\\csharp\\error report1.xls") = "")
{
    wbexcel.NewSheet();
}

else
{
    wbexcel.Open("C:\\csharp\\error report1.xls");
    objsht = ("sheet1");
}
objsht.Activate();

So what's the problem? Please describe it since it's easier to see directly than start creating a blank project myself and then copy and paste your code. — Mats Fredriksson
– Mats Fredriksson, Commented Jan 21, 2009 at 11:35
for starts the directory part is throwing an error 'system.io.directory is a type but used like a variable" — tksy
– tksy, Commented Jan 21, 2009 at 11:48
This would have stopped that error: if(!Directory.Exists(@"C:\csharp\error report.xls")) — dkroy
– dkroy, Commented Mar 29, 2013 at 13:20
if (Directory("C:\\csharp\\error report1.xls") = "") should be if (!File.Exists("C:\\csharp\\error report1.xls")) — Stephen MacDougall
– Stephen MacDougall, Commented Oct 11, 2013 at 19:09

abatishchev · Accepted Answer · 2021-03-01 01:06:11Z

104

You need to have installed Microsoft Visual Studio Tools for Office (VSTO).

VSTO can be selected in the Visual Studio installer under Workloads > Web & Cloud > Office/SharePoint Development.

After that create a generic .NET project and add a reference to Microsoft.Office.Interop.Excel via 'Add Reference... > Assemblies' dialog.

Application excel = new Application();
Workbook wb = excel.Workbooks.Open(path);

Missing.Value is a special reflection struct for unnecessary parameters replacement

In newer versions, the assembly reference required is called Microsoft Excel 16.0 Object Library. If you do not have the latest version installed you might have Microsoft Excel 15.0 Object Library, or an older version, but it is the same process to include.

edited Mar 1, 2021 at 1:06

answered Jan 21, 2009 at 15:07

abatishchev

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13 Comments

abatishchev Over a year ago

@Saravanan: See MSDN: Object ReadOnly. I guess you need to pass true.

Saravanan Over a year ago

:Please See my Question... stackoverflow.com/questions/8906670/…

HEEN Over a year ago

when I add that I get error as

CS0104: 'DataTable' is an ambiguous reference between 'System.Data.DataTable' and 'Microsoft.Office.Interop.Excel.DataTable'

abatishchev Over a year ago

@coder: remove using System.Data; or use "using DataTable = Microsoft.Office.Interop.Excel.DataTable;

HEEN Over a year ago

@abatishchev: when I remove using System.Data; i get namespace error for dataset which I am using for another set of code..

|

Community · Accepted Answer · 2014-06-04 18:39:32Z

65

FileInfo fi = new FileInfo("C:\\test\\report.xlsx");
if(fi.Exists)
{
    System.Diagnostics.Process.Start(@"C:\test\report.xlsx");
}
else
{
    //file doesn't exist
}

edited Jun 4, 2014 at 18:39

CommunityBot

11 silver badge

answered Oct 18, 2011 at 21:49

Mennan

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5 Comments

Roland Kwee Over a year ago

Works for me. Certainly simpler than the accepted 2009 answer because no additional tools are needed, and it is essential a one-liner.

Eda Over a year ago

Great, just what I was looking for, thanks! I simplified it by just passing the fi inside the start as System.Diagnostics.Process.Start(fi.ToString());

shahsani Over a year ago

This is by far, the most easy way to do it. Thank you

nharrer Over a year ago

This is the best solution. However in .NET core it has to be done a bit differently: stackoverflow.com/a/60970719/1018443

Jenna Leaf Over a year ago

What if your app users do not have Microsoft Office installed on their computer? If they don't, your code needs to have Microsoft.Office.Interop.Excel and Microsoft.VisualStudio.Tools.Applications.Runtime loaded to open up Excel for data view.

LarsTech · Accepted Answer · 2014-01-28 18:04:13Z

11

private void btnChoose2_Click(object sender, EventArgs e)
{
  OpenFileDialog openfileDialog1 = new OpenFileDialog();
  if (openfileDialog1.ShowDialog() == System.Windows.Forms.DialogResult.OK)
  {
    this.btnChoose2.Text = openfileDialog1.FileName;
    String filename = DialogResult.ToString();

    var excelApp = new Excel.Application();
    excelApp.Visible = true;
    excelApp.Workbooks.Open(btnChoose2.Text);
  }
}

edited Jan 28, 2014 at 18:04

LarsTech

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answered Jan 28, 2014 at 17:45

Flane

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1 Comment

shaosh Over a year ago

The excel does not show without excelApp.Visible = true. Thanks!

user1848210 · Accepted Answer · 2016-08-20 03:04:09Z

5

Imports

 using Excel= Microsoft.Office.Interop.Excel;
 using Microsoft.VisualStudio.Tools.Applications.Runtime;

Here is the code to open an excel sheet using C#.

    Microsoft.Office.Interop.Excel.Application excel = new Microsoft.Office.Interop.Excel.Application();
    Microsoft.Office.Interop.Excel.Workbook wbv = excel.Workbooks.Open("C:\\YourExcelSheet.xlsx");
    Microsoft.Office.Interop.Excel.Worksheet wx = excel.ActiveSheet as Microsoft.Office.Interop.Excel.Worksheet;

    wbv.Close(true, Type.Missing, Type.Missing);
    excel.Quit();

Here is a video mate on how to open an excel worksheet using C# https://www.youtube.com/watch?v=O5Dnv0tfGv4

edited Aug 20, 2016 at 3:04

answered Aug 20, 2016 at 0:39

user1848210

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Comments

rudenaggar · Accepted Answer · 2013-04-17 06:23:41Z

4

you should open like this

        Excel.Application xlApp ;
        Excel.Workbook xlWorkBook ;
        Excel.Worksheet xlWorkSheet ;
        object misValue = System.Reflection.Missing.Value;

        xlApp = new Excel.ApplicationClass();
        xlWorkBook = xlApp.Workbooks.Open("csharp.net-informations.xls", 0, true, 5, "", "", true, Microsoft.Office.Interop.Excel.XlPlatform.xlWindows, "\t", false, false, 0, true, 1, 0);
        xlWorkSheet = (Excel.Worksheet)xlWorkBook.Worksheets.get_Item(1);

source : http://csharp.net-informations.com/excel/csharp-open-excel.htm

ruden

answered Apr 17, 2013 at 6:23

rudenaggar

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Comments

G S · Accepted Answer · 2009-01-21 11:39:27Z

2

For opening a file, try this:

objexcel.Workbooks.Open(@"C:\YourPath\YourExcelFile.xls",
    missing, missing, missing, missing, missing, missing, missing,
    missing, missing, missing, missing, missing,missing, missing);

You must supply those stupid looking 'missing' arguments. If you were writing the same code in VB.Net you wouldn't have needed them, but you can't avoid them in C#.

answered Jan 21, 2009 at 11:39

G S

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2 Comments

Callum Rogers Over a year ago

Unless you're using VS2010 :) Oh, and you want Missing.Value

Chjquest Over a year ago

private object _value = Missing.Value; replace the missing with _value. And it works for me.

RememberME · Accepted Answer · 2011-11-12 04:41:46Z

2

Microsoft.Office.Interop.Excel.Application excapp;

excapp = new Microsoft.Office.Interop.Excel.Application();

object misval=System.Reflection.Missing.Value;

Workbook wrkbuk = new Workbook();

Worksheet wrksht = new Worksheet();

wrkbuk = excapp.Workbooks._Open(@"C:\Users\...\..._template_v1.0.xlsx", misval, misval, 
misval, misval, misval, misval, misval, misval, misval, misval, misval, misval);

wrksht = (Microsoft.Office.Interop.Excel.Worksheet)wrkbuk.Worksheets.get_Item(2);

edited Nov 12, 2011 at 4:41

RememberME

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answered Jul 15, 2011 at 10:50

Jairaj

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Comments

Mats Fredriksson · Accepted Answer · 2009-01-21 14:57:21Z

It's easier to help you if you say what's wrong as well, or what fails when you run it.

But from a quick glance you've confused a few things.

The following doesn't work because of a couple of issues.

if (Directory("C:\\csharp\\error report1.xls") = "")

What you are trying to do is creating a new Directory object that should point to a file and then check if there was any errors.

What you are actually doing is trying to call a function named Directory() and then assign a string to the result. This won't work since 1/ you don't have a function named Directory(string str) and you cannot assign to the result from a function (you can only assign a value to a variable).

What you should do (for this line at least) is the following

FileInfo fi = new FileInfo("C:\\csharp\\error report1.xls");
if(!fi.Exists)
{
    // Create the xl file here
}
else
{
    // Open file here
}

As to why the Excel code doesn't work, you have to check the documentation for the Excel library which google should be able to provide for you.

Mkkt Bkkt · Accepted Answer · 2009-01-21 15:29:42Z

1

Is this a commercial application or some hobbyist / open source software?

I'm asking this because in my experience, all free .NET Excel handling alternatives have serious problems, for different reasons. For hobbyist things, I usually end up porting jExcelApi from Java to C# and using it.

But if this is a commercial application, you would be better off by purchasing a third party library, like Aspose.Cells. Believe me, it totally worths it as it saves a lot of time and time ain't free.

answered Jan 21, 2009 at 15:29

Mkkt Bkkt

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Comments

Janus Troelsen · Accepted Answer · 2017-06-12 12:15:02Z

1

For editing Excel files from within a C# application, I recently started using NPOI. I'm very satisfied with it.

edited Jun 12, 2017 at 12:15

Janus Troelsen

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answered Apr 4, 2012 at 15:06

bvgheluwe

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Comments

Dharmesh Porwal · Accepted Answer · 2015-01-12 07:53:34Z

Code :

 private void button1_Click(object sender, EventArgs e)
     {

        textBox1.Enabled=false;

            OpenFileDialog ofd = new OpenFileDialog();
            ofd.Filter = "Excell File |*.xlsx;*,xlsx";
            if (ofd.ShowDialog() == DialogResult.OK)
            {
                string extn = Path.GetExtension(ofd.FileName);
                if (extn.Equals(".xls") || extn.Equals(".xlsx"))
                {
                    filename = ofd.FileName;

                    if (filename != "")
                    {
                        try
                        {
                            string excelfilename = Path.GetFileName(filename);


                        }
                        catch (Exception ew)
                        {
                            MessageBox.Show("Errror:" + ew.ToString());
                        }
                    }
                }
            }

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