使用xslt时嵌套的xml数据没有作为输出进入excel

Question

我正在尝试使用 xslt 模板将 xml 数据转换为 excel。但我只得到标题，而不是数据。请检查以下 xml 数据和 xslt 模板以及我的 C# 代码供您参考。

example.xml

<?xml version="1.0"?>
<projects>
  <project>
    <name>Shock</name>
    <language>Ruby</language>
    <owner>Brian May</owner>
    <state>New</state>
    <Date>31/10/2008 0:00:00</Date>
  </project>
  <project>
    <name>Shock1</name>
    <language>Ruby1</language>
    <owner>Brian May1</owner>
    <state>New1</state>
    <Date>30/10/2008 0:00:00</Date>
  </project>
</projects>

test.xsl

<?xml version="1.0" encoding="utf-8"?>
<?mso-application progid="Excel.Sheet"?>
<xsl:stylesheet version="1.0"
xmlns:html="http://www.w3.org/TR/REC-html40"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
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">
  <xsl:template match="/">
    <Workbook>
      <Styles>
        <Style ss:ID="Default" ss:Name="Normal">
          <Alignment ss:Vertical="Bottom" />
          <Borders />
          <Font />
          <Interior />
          <NumberFormat />
          <Protection />
        </Style>
        <Style ss:ID="s21">
          <Font ss:Size="22" ss:Bold="1" />
        </Style>
        <Style ss:ID="s22">
          <Font ss:Size="14" ss:Bold="1" />
        </Style>
        <Style ss:ID="s23">
          <Font ss:Size="12" ss:Bold="1" />
        </Style>
        <Style ss:ID="s24">
          <Font ss:Size="10" ss:Bold="1" />
        </Style>
      </Styles>
      <Worksheet ss:Name="Page1">
        <Table>
          <xsl:call-template name="XMLToXSL" />
        </Table>
      </Worksheet>
    </Workbook>
  </xsl:template>
  <xsl:template name="XMLToXSL">
    <Row ss:Index="1" >
      <Cell>
        <Data ss:Type="String">name</Data>
      </Cell>
      <Cell>
        <Data ss:Type="String">language</Data>
      </Cell>
      <Cell>
        <Data ss:Type="String">owner</Data>
      </Cell>
      <Cell>
        <Data ss:Type="String">state</Data>
      </Cell>
      <Cell>
        <Data ss:Type="String">Date</Data>
      </Cell>
    </Row>
    <xsl:for-each select="/projects">
      <xsl:for-each select="/project">
        <Row ss:Index="1">
          <Cell ss:Index="2">
            <Data ss:Type="String">
              <xsl:value-of select="name" />
            </Data>
          </Cell>
          <Cell ss:Index="2">
            <Data ss:Type="String">
              <xsl:value-of select="language" />
            </Data>
          </Cell>
          <Cell ss:Index="2">
            <Data ss:Type="String">
              <xsl:value-of select="owner" />
            </Data>
          </Cell>
          <Cell ss:Index="2">
            <Data ss:Type="String">
              <xsl:value-of select="state" />
            </Data>
          </Cell>
          <Cell ss:Index="2">
            <Data ss:Type="String">
              <xsl:value-of select="Date" />
            </Data>
          </Cell>
        </Row>
      </xsl:for-each>
    </xsl:for-each>
  </xsl:template>
  <xsl:template match="DataList">
  </xsl:template>
</xsl:stylesheet>

C#代码：

XmlDocument xdoc = new XmlDocument();
xdoc.Load(@"C:\Xmls\example.xml");
XslCompiledTransform xct = new XslCompiledTransform();
xct.Load(@"C:\Templates\Test.xsl");
XmlTextWriter writer = new XmlTextWriter(@"C:\Outputs\output.xls", null);
writer.WriteProcessingInstruction("xml", "version='1.0'");
xct.Transform(xdoc, null, writer);
writer.Close();

Excel 输出.xls：

|  name  |  language  |  owner  |  state  |  Date   |
-----------------------------------------------------

但我的预期输出应该是，

|  name  |  language  |  owner    |  state  |        Date        |
------------------------------------------------------------------
| Shock  | Ruby       | Brian May | New     | 31/10/2008 0:00:00 |
------------------------------------------------------------------
| Shock1 | Ruby1      | Brian May1| New1    | 30/10/2008 0:00:00 |
------------------------------------------------------------------

目前只在excel中打印标题， 如何按行将数据导入excel。

Answer 1

将xsl:for-each select="/project" 替换为xsl:for-each select="project"。

我也会简化 C# 的使用

XslCompiledTransform xct = new XslCompiledTransform();
xct.Load(@"C:\Templates\Test.xsl");
xct.Transform(@"C:\Xmls\example.xml", @"C:\Outputs\output.xls");

虽然我认为您的代码不会导致问题，但 XSLT 的 XPath 错误。