I want to create xml file using xsl file that will use in memory data only (no use of a source file). I have the code below but I'm not sure what to use instand of the source file

part of the xsl for example :

<ExternalReference>
                    <xsl:value-of select="erstefnImport:retriveXslExternalReference()" />
                </ExternalReference>

the code part :

if (xsltFile != null) {
            Transformer transformer = getTransformer(xsltFile);
            Source source = null;
            try {
                source = new StreamSource(new FileInputStream(null));// ??
            } catch (FileNotFoundException e1) {
                // TODO Auto-generated catch block
                e1.printStackTrace();
            }
            ByteArrayOutputStream byteOutputStream = new ByteArrayOutputStream();
            StreamResult outputTarget = new StreamResult(byteOutputStream);

            if (transformer != null) {
                try {
                    transformer.transform(source, outputTarget);
                    rst = new String(byteOutputStream.toByteArray());
                } catch (Exception e) {
                    Log.error(this, "Error while transforming xml.", e);

                }
            }

        }

I tried to use Document as a source but the output is only :

<?xml version="1.0" encoding="iso-8859-1"?>

the code :

if (xsltFile != null) {
            Transformer transformer = getTransformer(xsltFile);
            Source source = null;
            DocumentBuilderFactory docFactory = DocumentBuilderFactory.newInstance();
            Document docResult = null;
            try {
                docResult = docFactory.newDocumentBuilder().newDocument();
            } catch (ParserConfigurationException e2) {
                e2.printStackTrace();
            }
            DOMSource domSource = new DOMSource(docResult);
            ByteArrayOutputStream byteOutputStream = new ByteArrayOutputStream();
            StreamResult outputTarget = new StreamResult(byteOutputStream);

            if (transformer != null) {
                try {
                    transformer.transform(domSource, outputTarget);
                    rst = new String(byteOutputStream.toByteArray());
                } catch (Exception e) {
                    Log.error(this, "Error while transforming xml.", e);

                }
            }

        }

CodePudding user response：

XSLT 1.0 and that way the APIs of XSLT 1.0 processors are pretty much built around the idea of processing a source tree as the input to an XSLT stylesheet.

This is different in XSLT 2 and 3, in XSLT 2 you can start processing with a named template, in XSLT 3 as well, there you can also start by calling a public function.

As you are using Java, you could easily consider switching to Saxon 10, available in the open-source HE edition from Sourceforge and Maven, to use XSLT 3 and then make sure you use its own API https://www.saxonica.com/html/documentation10/using-xsl/embedding/s9api-transformation.html which has methods like callTemplate (https://www.saxonica.com/html/documentation10/using-xsl/embedding/s9api-transformation.html) or callFunction (https://www.saxonica.com/html/documentation10/javadoc/net/sf/saxon/s9api/Xslt30Transformer.html#callFunction-net.sf.saxon.s9api.QName-net.sf.saxon.s9api.XdmValue:A-).

In XSLT 3 there is a predefined name for the initial template that would be

<xsl:template name="xsl:initial-template">
  <root>
    <foo>bar</foo>
  </root>
</xsl:template>

in XSLT there is no predefined named so you can chose one like main e.g.

<xsl:template name="main">
  <root>
    <foo>bar</foo>
  </root>
</xsl:template>

and then use the referenced callTemplate method of the API to use the main name (new QName("", "main"), I think).

CodePudding user response：

I fount that using (almost)empty xml source file will work . Just need to use a root tag (map in this example) in that source xml , and use this tag name in the xsl tempele :

<xsl:template match="map">