如何在C＃中使用OpenXml创建/打开Excel文件

Question

I have a console application in which we are creating xlsx files using OPENXML, we are able to create xlsx file & save it into specific folder in application.

But Now we want to show that file as a Save/Open dialog pop up. Then we can able to specify a particular path to save/ to open the existing files.

I am new to this OpenXml, Can anyone please help me on this to proceed further? How can I acheive this? Do we have any built-in DLL for this?

Thanks.

Answer 1

se the Save file dialog. It will prompts the user to select a location for saving a file. After that you can use saveFileDialog.FileName.ToString() property to get the full path. See the sample code below:

//Save a file in a particular format as specified in the saveAsType parameter
     private void OpenSaveFileDialog(int saveAsType)
     {
        SaveFileDialog saveFileDialog = new SaveFileDialog();
        saveFileDialog.InitialDirectory = Convert.ToString(Environment.SpecialFolder.MyDocuments);
        saveFileDialog.Filter = "CSV|*.csv|Excel|*.xlsx";
        saveFileDialog.FilterIndex = saveAsType;
        saveFileDialog.Title = "Save Data";
        saveFileDialog.FileName = "My File";
        saveFileDialog.ShowDialog();

        if (saveFileDialog.FileName != "")
        {
            //File Path =   m_fileName         
             m_fileName = saveFileDialog.FileName.ToString();
             //FilterIndex property is one-based.
             switch (saveFileDialog.FilterIndex)
             {
                case 1:
                    m_fileType = 1;
                    break;
                case 2:
                    m_fileType = 2;
                    break;
              }
        }
      }

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