So schreiben Sie Daten auf ein neues Excel-Blatt

Ich habe eine Lösung von OpenXML.

     string path = Context.Server.MapPath("~/ExcelData/test.xslx");
     ExportDataSet(ds, path);


  private void ExportDataSet(DataSet ds, string destination)
{
    using (var workbook = SpreadsheetDocument.Create(destination, DocumentFormat.OpenXml.SpreadsheetDocumentType.Workbook))
    {
        var workbookPart = workbook.AddWorkbookPart();

        workbook.WorkbookPart.Workbook = new DocumentFormat.OpenXml.Spreadsheet.Workbook();

        workbook.WorkbookPart.Workbook.Sheets = new DocumentFormat.OpenXml.Spreadsheet.Sheets();

        foreach (System.Data.DataTable table in ds.Tables)
        {

            var sheetPart = workbook.WorkbookPart.AddNewPart<WorksheetPart>();
            var sheetData = new DocumentFormat.OpenXml.Spreadsheet.SheetData();
            sheetPart.Worksheet = new DocumentFormat.OpenXml.Spreadsheet.Worksheet(sheetData);

            DocumentFormat.OpenXml.Spreadsheet.Sheets sheets = workbook.WorkbookPart.Workbook.GetFirstChild<DocumentFormat.OpenXml.Spreadsheet.Sheets>();
            string relationshipId = workbook.WorkbookPart.GetIdOfPart(sheetPart);

            uint sheetId = 1;
            if (sheets.Elements<DocumentFormat.OpenXml.Spreadsheet.Sheet>().Count() > 0)
            {
                sheetId =
                    sheets.Elements<DocumentFormat.OpenXml.Spreadsheet.Sheet>().Select(s => s.SheetId.Value).Max() + 1;
            }

            DocumentFormat.OpenXml.Spreadsheet.Sheet sheet = new DocumentFormat.OpenXml.Spreadsheet.Sheet() { Id = relationshipId, SheetId = sheetId, Name = table.TableName };
            sheets.Append(sheet);

            DocumentFormat.OpenXml.Spreadsheet.Row headerRow = new DocumentFormat.OpenXml.Spreadsheet.Row();

            List<String> columns = new List<string>();
            foreach (System.Data.DataColumn column in table.Columns)
            {
                columns.Add(column.ColumnName);

                DocumentFormat.OpenXml.Spreadsheet.Cell cell = new DocumentFormat.OpenXml.Spreadsheet.Cell();
                cell.DataType = DocumentFormat.OpenXml.Spreadsheet.CellValues.String;
                cell.CellValue = new DocumentFormat.OpenXml.Spreadsheet.CellValue(column.ColumnName);
                headerRow.AppendChild(cell);
            }


            sheetData.AppendChild(headerRow);

            foreach (System.Data.DataRow dsrow in table.Rows)
            {
                DocumentFormat.OpenXml.Spreadsheet.Row newRow = new DocumentFormat.OpenXml.Spreadsheet.Row();
                foreach (String col in columns)
                {
                    DocumentFormat.OpenXml.Spreadsheet.Cell cell = new DocumentFormat.OpenXml.Spreadsheet.Cell();
                    cell.DataType = DocumentFormat.OpenXml.Spreadsheet.CellValues.String;
                    cell.CellValue = new DocumentFormat.OpenXml.Spreadsheet.CellValue(dsrow[col].ToString()); //
                    newRow.AppendChild(cell);
                }

                sheetData.AppendChild(newRow);
            }

        }
    }
}

Referenzen :

using DocumentFormat.OpenXml.Packaging;
using DocumentFormat.OpenXml.Spreadsheet;
using DocumentFormat.OpenXml;