Question

我试图解析一个非常简单的例子：

<?xml version="1.0" encoding="UTF-8"?>
<root xmlns:openSearch='http://a9.com/-/spec/opensearch/1.1/'>
    <openSearch:totalResults>100</openSearch:totalResults>
</root>

我使用的样式表如下：

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" 
                version="1.0" 
                xmlns:app='http://www.w3.org/2007/app' >
<xsl:output method="xml" indent="yes"/>
<xsl:preserve-space elements="*"/>
<xsl:template match="/">
    <results>
        <xsl:attribute name="total-results">
                <xsl:value-of 
                 select="atom:root/openSearch:totalResults"/>
        </xsl:attribute>
    </results>
</xsl:template>
</xsl:stylesheet>

这适用于libxslt，没问题。我现在尝试在java中执行相同的任务，我正在尝试使用javax.xml.transform包来执行此操作。它不是预期的结果，而是为total-results属性提供一个空值。但是当我将值改为：

时

                <xsl:value-of select="root/totalResults"/>

有效。更改xml和xslt不是一个选项。我应该在某个地方设置一个参数吗？代码非常简单：

InputSource xmlSource = new InputSource( new StringReader(xml) );

DocumentBuilder builder = factory.newDocumentBuilder();
document = builder.parse(xmlSource);

// Use a Transformer for output
TransformerFactory tFactory = TransformerFactory.newInstance();


StreamSource stylesource = new StreamSource(new StringReader(styleSheet));
Transformer transformer = tFactory.newTransformer(stylesource);

StringWriter writer = new StringWriter();

DOMSource source = new DOMSource(document);
StreamResult result = new StreamResult(writer);
transformer.transform(source, result);

stringResult = writer.toString();

Answer 1

在样式表中，您缺少“atom”和“openSearch”的名称空间声明。以下作品：

在样式表
删除“atom”命名空间，因为没有相关信息命名空间
将工厂设置为名称空间感知：factory.setNamespaceAware(true);

这是Scala中的完整代码（抱歉，我懒得从文件解析xml和样式表或在Java中进行字符串连接）：

  def testxsl = {
      val xml = """<?xml version="1.0" encoding="UTF-8"?>
<root xmlns:openSearch='http://a9.com/-/spec/opensearch/1.1/'>
    <openSearch:totalResults>100</openSearch:totalResults>
</root>
      """
      val styleSheet = """<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0" 
        xmlns:app='http://www.w3.org/2007/app' 
        xmlns:openSearch='http://a9.com/-/spec/opensearch/1.1/'>
<xsl:output method="xml" indent="yes"/>
<xsl:preserve-space elements="*"/>
<xsl:template match="/">
    <results>
        <xsl:attribute name="total-results">
                <xsl:value-of select="root/openSearch:totalResults"/>
        </xsl:attribute>
    </results>
</xsl:template>
</xsl:stylesheet>
        """
    val xmlSource = new InputSource( new StringReader(xml) );
    val factory = DocumentBuilderFactory.newInstance();
    factory.setNamespaceAware(true);
    val builder = factory.newDocumentBuilder();
    val document = builder.parse(xmlSource);

    // Use a Transformer for output
    val tFactory = TransformerFactory.newInstance();


    val stylesource = new StreamSource(new StringReader(styleSheet));
    val transformer = tFactory.newTransformer(stylesource);

    val writer = new StringWriter();

    val source = new DOMSource(document);
    val result = new StreamResult(writer);
    transformer.transform(source, result);
    writer.toString(); 
  }

javax.xml.transform.Transformer忽略前缀？

1 个答案: