 在c#中将dataset输出成xml内容的xls文件，解决其它输出不能在excel中显示“001”的前导0的问题
记得以前直接把gridview导出称excel，虽然很简单，但是却不能控制具体细节，尤其是字符串的“00001”到了excel中必定变为整数1,一直没找到解决办法，直到在codeproject上找到这个函数，它直接把数据写称excel可以读取的xml文档，仅仅把扩展名改为xls，但是效果却很好。 public static void DataSetExportToExcel(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();
}
|
[返回] 中文XML论坛 - 专业的XML技术讨论区 →
XML.ORG.CN讨论区 - XML技术 → 『 WORD to XML, HTML to XML 』 → 在c#中将dataset输出成xml内容的xls文件，解决其它输出不能在excel中显示“001”的前导0的问题
(订阅本版)
2007/9/14 17:02:00
新浪微博
