Java Transformer错误：无法编译样式表

Question

我想用Java中的XSLT转换XML。为此，我使用了javax.xml.transform包装。但是，我得到了例外javax.xml.transform.TransformerConfigurationException: Could not compile stylesheet。这是我正在使用的代码：

public static String transform(String XML, String XSLTRule) throws TransformerException {

    Source xmlInput = new StreamSource(XML);
    Source xslInput = new StreamSource(XSLTRule);

    TransformerFactory tFactory = TransformerFactory.newInstance();
    Transformer transformer = tFactory.newTransformer(xslInput); // this is the line that throws the exception

    Result result = new StreamResult();
    transformer.transform(xmlInput, result);

    return result.toString();
}

请注意，我标记了引发异常的行。

当我输入方法时，的值为XSLTRule：

<xsl:stylesheet version='1.0'
 xmlns:xsl='http://www.w3.org/1999/XSL/Transform'
 xmlns:msxsl='urn:schemas-microsoft-com:xslt'
 exclude-result-prefixes='msxsl'
 xmlns:ns='http://www.ibm.com/wsla'>
    <xsl:strip-space elements='*'/>
    <xsl:output method='xml' indent='yes'/>
    <xsl:template match='@* | node()'>
        <xsl:copy>
            <xsl:apply-templates select='@* | node()'/>
        </xsl:copy>
    </xsl:template>
    <xsl:template match="/ns:SLA
                            /ns:ServiceDefinition
                               /ns:WSDLSOAPOperation
                                  /ns:SLAParameter[@name='Performance']"/>
</xsl:stylesheet>

Answer 1

该构造

public StreamSource(String systemId)

从URL构造StreamSource。我认为您正在传递XSLT的内容。试试这个：

File xslFile = new File("path/to/your/xslt");
TransformerFactory factory = TransformerFactory.newInstance();
Templates xsl = factory.newTemplates(new StreamSource(xslFile));

你还必须设置OutputStream你的StreamResult意愿写：

ByteArrayOutputStream baos = new ByteArrayOutputStream();
Result result = new StreamResult(baos);
transformer.transform(xmlInput, result);
return baos.toString();