如何使用多个输入中的任何循环指定输入目录？

Question

public class MultiXslt
{
    public static void main(String[] args) throws TransformerException,ParserConfigurationException, SAXException, IOException 
    {
        //source xslt
        StreamSource stylesource = new StreamSource("C:/Users/santhanamk/Desktop/newxslt/Xslt inputs/Idml0.xsl");

        DocumentBuilderFactory docbFactory = DocumentBuilderFactory
                .newInstance();
        DocumentBuilder dBuilder = docbFactory.newDocumentBuilder();

        //source XML
        Document sourceDoc = dBuilder.parse("C:/Users/santhanamk/Desktop/newxslt/input.xml");

        DOMSource source = new DOMSource(sourceDoc);

        TransformerFactory transformerFactory = TransformerFactory
                .newInstance();
        Transformer transformer = transformerFactory
                .newTransformer(stylesource);

        Document document = dBuilder.newDocument();
        DOMResult result = new DOMResult(document);

        transformer.transform(source, result);

        Node resultDoc = ((Document) result.getNode()).getDocumentElement();

        System.out.println(resultDoc.getChildNodes().getLength());

        // print the result
        StringWriter writer = new StringWriter();
        Transformer transformer2 = transformerFactory.newTransformer();
        transformer2.transform(new DOMSource(resultDoc), new StreamResult(writer));
        String s = writer.toString();

    }
}

实际上，我有一个xml文件和多个xsl文件（C：/ Users / santhanamk / Desktop / newxslt / Xslt输入/
xsl列表）。当我将xml和xsl0作为输入时，我需要以字符串形式获取输出。因此，在获得输出之后，我需要为xsl1提供与输入字符串相同的输出，以获取另一个输出（字符串）。然后，我需要将输出作为xsl2的输入字符串，以获取另一个输出。当给定的源目录（C：/
Users / santhanamk / Desktop / newxslt / Xslt输入/
xsl列表）没有任何新的xsl文件来加载输出字符串时，它将最终输出作为xml！

Answer 1

我想使用JAXP转换API，如果您想链接转换，那么http://docs.oracle.com/javaee/1.4/tutorial/doc/JAXPXSLT8.html就有一个示例，使用XMLFilter通过创建的SAXTransformerFactory.newXMLFilter。

这是一个示例Java代码，该代码显示了如何使用样式表文件名数组和设置过滤器链：

import java.io.FileInputStream;
import java.io.FileNotFoundException;
import javax.xml.parsers.ParserConfigurationException;
import javax.xml.parsers.SAXParserFactory;
import javax.xml.transform.Result;
import javax.xml.transform.Transformer;
import javax.xml.transform.TransformerConfigurationException;
import javax.xml.transform.TransformerException;
import javax.xml.transform.TransformerFactory;
import javax.xml.transform.sax.SAXSource;
import javax.xml.transform.sax.SAXTransformerFactory;
import javax.xml.transform.stream.StreamResult;
import javax.xml.transform.stream.StreamSource;
import org.xml.sax.InputSource;
import org.xml.sax.SAXException;
import org.xml.sax.XMLFilter;
import org.xml.sax.XMLReader;


public class JAXPTransChain1 {


    public static void main(String[] args) throws FileNotFoundException, SAXException, ParserConfigurationException, TransformerException {
        String[] stylesheets = new String[] {"sheet1.xsl", "sheet2.xsl", "sheet3.xsl"};
        String inputDoc = "input1.xml";
        chainSheets(stylesheets, inputDoc, new StreamResult(System.out));
    }

    private static void chainSheets(String[] stylesheets, String inputDoc, Result result) throws FileNotFoundException, ParserConfigurationException, SAXException, TransformerConfigurationException, TransformerException {
        InputSource input = new InputSource(new FileInputStream(inputDoc));

        SAXParserFactory spf = SAXParserFactory.newInstance();
        spf.setNamespaceAware(true);
        XMLReader reader = spf.newSAXParser().getXMLReader();

        SAXTransformerFactory stf = (SAXTransformerFactory)TransformerFactory.newInstance();

        XMLReader parent = reader;

        for (int i = 0; i < stylesheets.length; i++)
        {
            String sheetUri  = stylesheets[i];
            XMLFilter sheetFilter = stf.newXMLFilter(new StreamSource(new FileInputStream(sheetUri)));
            sheetFilter.setParent(parent);
            parent = sheetFilter;
        }

        Transformer proc = stf.newTransformer();

        SAXSource transSource = new SAXSource(parent, input);

        proc.transform(transSource, result);

    }

}

如果输入是像

<root>
    <foo>bar</foo>   
</root>

和样式表做例如

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">

    <xsl:template match="@* | node()">
        <xsl:copy>
            <xsl:apply-templates select="@* | node()"/>
        </xsl:copy>
    </xsl:template>

    <xsl:template match="/node()[last()]">
        <xsl:copy>
            <xsl:apply-templates select="@* | node()"/>
        </xsl:copy>
        <xsl:comment>sheet 1</xsl:comment>
    </xsl:template>

</xsl:stylesheet>

和

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">

    <xsl:template match="@* | node()">
        <xsl:copy>
            <xsl:apply-templates select="@* | node()"/>
        </xsl:copy>
    </xsl:template>

    <xsl:template match="/node()[last()]">
        <xsl:copy>
            <xsl:apply-templates select="@* | node()"/>
        </xsl:copy>
        <xsl:comment>sheet 2</xsl:comment>
    </xsl:template>

</xsl:stylesheet>

和

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">

    <xsl:template match="@* | node()">
        <xsl:copy>
            <xsl:apply-templates select="@* | node()"/>
        </xsl:copy>
    </xsl:template>

    <xsl:template match="/node()[last()]">
        <xsl:copy>
            <xsl:apply-templates select="@* | node()"/>
        </xsl:copy>
        <xsl:comment>sheet 3</xsl:comment>
    </xsl:template>

</xsl:stylesheet>

那么最终的输出是

<?xml version="1.0" encoding="UTF-8"?><root>
    <foo>bar</foo>   
</root><!--sheet 1--><!--sheet 2--><!--sheet 3-->

因此样式表全部按照输入数组的顺序应用。一旦知道所需顺序，就很容易从目录中的文件列表中建立这样的数组。