转换excel到XML
如何使用excel工作簿中的预定义模式创buildxml？
有一个窍门。 您可以创buildXSL （样式表）来将您的XML转换为简单的HTML标记。 使用XslCompiledTransform.Transform()转换并使用XLS （Excel Spreadsheet）扩展名进行保存。 它将由MS Excel打开，所有的Excelfunction将可用。 据我所知这是最好的解决scheme，如果你真的只需要快速转储数据，所以可以在Excel中打开之后。
这是一个可以帮助你的function，我没有太多的理解你的问题。 简单地把你的文件名和xls与数据集中的数据生成。
public static void exportToExcel(DataSet source, string fileName) { System.IO.StreamWriter excelDoc; excelDoc = new System.IO.StreamWriter(fileName); const string startExcelXML = "<xml version>\r\n<Workbook " + "xmlns=\"urn:schemas-microsoft-com:office:spreadsheet\"\r\n" + " xmlns:o=\"urn:schemas-microsoft-com:office:office\"\r\n " + "xmlns:x=\"urn:schemas- microsoft-com:office:" + "excel\"\r\n xmlns:ss=\"urn:schemas-microsoft-com:" + "office:spreadsheet\">\r\n <Styles>\r\n " + "<Style ss:ID=\"Default\" ss:Name=\"Normal\">\r\n " + "<Alignment ss:Vertical=\"Bottom\"/>\r\n <Borders/>" + "\r\n <Font/>\r\n <Interior/>\r\n <NumberFormat/>" + "\r\n <Protection/>\r\n </Style>\r\n " + "<Style ss:ID=\"BoldColumn\">\r\n <Font " + "x:Family=\"Swiss\" ss:Bold=\"1\"/>\r\n </Style>\r\n " + "<Style ss:ID=\"StringLiteral\">\r\n <NumberFormat" + " ss:Format=\"@\"/>\r\n </Style>\r\n <Style " + "ss:ID=\"Decimal\">\r\n <NumberFormat " + "ss:Format=\"0.0000\"/>\r\n </Style>\r\n " + "<Style ss:ID=\"Integer\">\r\n <NumberFormat " + "ss:Format=\"0\"/>\r\n </Style>\r\n <Style " + "ss:ID=\"DateLiteral\">\r\n <NumberFormat " + "ss:Format=\"mm/dd/yyyy;@\"/>\r\n </Style>\r\n " + "</Styles>\r\n "; const string endExcelXML = "</Workbook>"; int rowCount = 0; int sheetCount = 1; /* <xml version> <Workbook xmlns="urn:schemas-microsoft-com:office:spreadsheet" xmlns:o="urn:schemas-microsoft-com:office:office" xmlns:x="urn:schemas-microsoft-com:office:excel" xmlns:ss="urn:schemas-microsoft-com:office:spreadsheet"> <Styles> <Style ss:ID="Default" ss:Name="Normal"> <Alignment ss:Vertical="Bottom"/> <Borders/> <Font/> <Interior/> <NumberFormat/> <Protection/> </Style> <Style ss:ID="BoldColumn"> <Font x:Family="Swiss" ss:Bold="1"/> </Style> <Style ss:ID="StringLiteral"> <NumberFormat ss:Format="@"/> </Style> <Style ss:ID="Decimal"> <NumberFormat ss:Format="0.0000"/> </Style> <Style ss:ID="Integer"> <NumberFormat ss:Format="0"/> </Style> <Style ss:ID="DateLiteral"> <NumberFormat ss:Format="mm/dd/yyyy;@"/> </Style> </Styles> <Worksheet ss:Name="Sheet1"> </Worksheet> </Workbook> */ excelDoc.Write(startExcelXML); excelDoc.Write("<Worksheet ss:Name=\"Sheet" + sheetCount + "\">"); excelDoc.Write("<Table>"); excelDoc.Write("<Row>"); for (int x = 0; x < source.Tables[0].Columns.Count; x++) { excelDoc.Write("<Cell ss:StyleID=\"BoldColumn\"><Data ss:Type=\"String\">"); excelDoc.Write(source.Tables[0].Columns[x].ColumnName); excelDoc.Write("</Data></Cell>"); } excelDoc.Write("</Row>"); foreach (DataRow x in source.Tables[0].Rows) { rowCount++; //if the number of rows is > 64000 create a new page to continue output if (rowCount == 64000) { rowCount = 0; sheetCount++; excelDoc.Write("</Table>"); excelDoc.Write(" </Worksheet>"); excelDoc.Write("<Worksheet ss:Name=\"Sheet" + sheetCount + "\">"); excelDoc.Write("<Table>"); } excelDoc.Write("<Row>"); //ID=" + rowCount + " for (int y = 0; y < source.Tables[0].Columns.Count; y++) { System.Type rowType; rowType = x[y].GetType(); switch (rowType.ToString()) { case "System.String": string XMLstring = x[y].ToString(); XMLstring = XMLstring.Trim(); XMLstring = XMLstring.Replace("&", "&"); XMLstring = XMLstring.Replace(">", ">"); XMLstring = XMLstring.Replace("<", "<"); excelDoc.Write("<Cell ss:StyleID=\"StringLiteral\">" + "<Data ss:Type=\"String\">"); excelDoc.Write(XMLstring); excelDoc.Write("</Data></Cell>"); break; case "System.DateTime": //Excel has a specific Date Format of YYYY-MM-DD followed by //the letter 'T' then hh:mm:sss.lll Example 2005-01-31T24:01:21.000 //The Following Code puts the date stored in XMLDate //to the format above DateTime XMLDate = (DateTime)x[y]; string XMLDatetoString = ""; //Excel Converted Date XMLDatetoString = XMLDate.Year.ToString() + "-" + (XMLDate.Month < 10 ? "0" + XMLDate.Month.ToString() : XMLDate.Month.ToString()) + "-" + (XMLDate.Day < 10 ? "0" + XMLDate.Day.ToString() : XMLDate.Day.ToString()) + "T" + (XMLDate.Hour < 10 ? "0" + XMLDate.Hour.ToString() : XMLDate.Hour.ToString()) + ":" + (XMLDate.Minute < 10 ? "0" + XMLDate.Minute.ToString() : XMLDate.Minute.ToString()) + ":" + (XMLDate.Second < 10 ? "0" + XMLDate.Second.ToString() : XMLDate.Second.ToString()) + ".000"; excelDoc.Write("<Cell ss:StyleID=\"DateLiteral\">" + "<Data ss:Type=\"DateTime\">"); excelDoc.Write(XMLDatetoString); excelDoc.Write("</Data></Cell>"); break; case "System.Boolean": excelDoc.Write("<Cell ss:StyleID=\"StringLiteral\">" + "<Data ss:Type=\"String\">"); excelDoc.Write(x[y].ToString()); excelDoc.Write("</Data></Cell>"); break; case "System.Int16": case "System.Int32": case "System.Int64": case "System.Byte": excelDoc.Write("<Cell ss:StyleID=\"Integer\">" + "<Data ss:Type=\"Number\">"); excelDoc.Write(x[y].ToString()); excelDoc.Write("</Data></Cell>"); break; case "System.Decimal": case "System.Double": excelDoc.Write("<Cell ss:StyleID=\"Decimal\">" + "<Data ss:Type=\"Number\">"); excelDoc.Write(x[y].ToString()); excelDoc.Write("</Data></Cell>"); break; case "System.DBNull": excelDoc.Write("<Cell ss:StyleID=\"StringLiteral\">" + "<Data ss:Type=\"String\">"); excelDoc.Write(""); excelDoc.Write("</Data></Cell>"); break; default: throw (new Exception(rowType.ToString() + " not handled.")); } } excelDoc.Write("</Row>"); } excelDoc.Write("</Table>"); excelDoc.Write(" </Worksheet>"); excelDoc.Write(endExcelXML); excelDoc.Close(); }
如果您在Excel中有一些数据，并想将它们转换为XML，那么您应该有一个合适的XML模式定义。 我假设你有一个合适的模式，那么你只需要遵循几个简单的步骤在Excel中将其转换为XML。 让我知道，如果你正在寻找如何在Excel中实现这个步骤…希望你有最新版本的Microsoft Office ..