没有错误：但是通过Javax.xml.transform没有发生XSLT转换

Posted 2021-03-26

XSLT在线测试并且工作正常，但在Java上运行时，对XML没有任何影响，并且在服务器控制台上没有报告错误

请参考XML onl Link。当我在http://xsltransform.net上测试时，我通过XSLT转换获得了正确的结果。但是当我在Java上应用相同的XSLT时，没有任何转换发生，并且服务器控制台上也没有错误。

使用的库

import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.transform.Transformer;
import javax.xml.transform.TransformerFactory;
import javax.xml.transform.dom.DOMSource;
import javax.xml.transform.stream.StreamResult;
import javax.xml.transform.stream.StreamSource;

import org.w3c.dom.Document;
import org.w3c.dom.NodeList;
import org.w3c.dom.Node;
import org.w3c.dom.Element;

Java代码

         File fXmlFile = new File("C:\Java\Web Project\Merged_Excel\WebContent\source\XML\XML_Template.xml");

          DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
          dbFactory.setNamespaceAware(true);
          DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
          doc = (Document) dBuilder.parse(fXmlFile);

          doc.getDocumentElement().normalize();

          /*..... XML buliding.....*/

              String sorter="<?xml version="1.0" encoding="utf-8"?>"+
                            "<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:ss="urn:schemas-microsoft-com:office:spreadsheet">"+
                            "  <xsl:output method="xml" encoding="utf-8" indent="yes"/>"+
                            "  <xsl:template match="@* | node()">"+
                            "    <xsl:copy>"+
                            "      <xsl:apply-templates select="@* | node()"/>"+
                            "    </xsl:copy>"+
                            "  </xsl:template>"+
                            "  <xsl:template match="ss:Row">"+
                            "    <xsl:copy>"+
                            "      <xsl:apply-templates select="*">"+
                            "        <xsl:sort select="@ss:Index" order="ascending" data-type="number"/>"+
                            "      </xsl:apply-templates>"+
                            "    </xsl:copy>"+
                            "  </xsl:template>"+
                            "</xsl:stylesheet>";

              StreamSource stylesource = new StreamSource(new StringReader(sorter));
              System.out.println(stylesource);
              /*also tried to fetch External-XSLT but couldn't suceed
              Source stylesource = new StreamSource(new File("C:\Java\Web Project\Merged_Excel\WebContent\source\XML\XSLT.xslt"));*/
              StringWriter writer = new StringWriter();
              TransformerFactory factory = TransformerFactory.newInstance();

              Transformer trans = factory.newTransformer(stylesource);
              trans.transform(new DOMSource(doc), new StreamResult(writer));
              System.out.println(writer.toString());

答案

以下变化解决了问题（不知道根本原因）

1.对setnamespace感知错误

dbFactory.setNamespaceAware(false);

2.从xslt字符串中删除命名空间标识符

String sorter="<?xml version="1.0" encoding="utf-8"?>"+
                            "<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:ss="urn:schemas-microsoft-com:office:spreadsheet">"+
                            "  <xsl:output method="xml" encoding="utf-8" indent="yes"/>"+
                            "  <xsl:template match="@* | node()">"+
                            "    <xsl:copy>"+
                            "      <xsl:apply-templates select="@* | node()"/>"+
                            "    </xsl:copy>"+
                            "  </xsl:template>"+
                            "  <xsl:template match="Row">"+ //if namespace sensitive then use ss:Row
                            "    <xsl:copy>"+
                            "      <xsl:apply-templates select="*">"+
                            "        <xsl:sort select="@Index" order="ascending" data-type="number"/>"+ //if namespace sensitive then use ss:Index
                            "      </xsl:apply-templates>"+
                            "    </xsl:copy>"+
                            "  </xsl:template>"+
                            "</xsl:stylesheet>";