Making JAXB generate an XML processing instruction

Question

I am generating XML dynamically using JAXB.

Now, I want to convert it to HTML using XSL. How can i include

<?xml-stylesheet type="text/xsl" href="">

in the dynamically generated XML?

Do you mean you want to include a stylesheet reference in the JAXB output? — skaffman, Jan 28 '10 at 08:45

score 11 · Answer 1 · edited Nov 12 '19 at 13:30

11

All the solutions here are pretty ugly and verbose. Simply set the line inside the Mashaller object specifying the additional header.

Marshaller jaxbMarshaller = ...
jaxbMarshaller.setProperty("com.sun.xml.bind.xmlHeaders", 
    "<?xml-stylesheet type='text/xsl' href='nameoffile.xsl' ?>");

This example will output an XML object to a file using a stylesheet and format the elements nicely for humans to read. The object myXmlObject is of class MyXmlClass, and will be written to file, formatted by a stylesheet given by xslUrl:

JAXBContext context = JAXBContext.newInstance(MyXmlClass.class);
Marshaller marshaller = context.createMarshaller();
//Need to use a Writer to marshal with the XSL
FileWriter fw = new FileWriter(file);
//Do this or else the XML is all one line and not human friendly...
marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, Boolean.TRUE);
marshaller.setProperty("com.sun.xml.bind.xmlHeaders",
        "<?xml-stylesheet type='text/xsl' href=\"" +
        xslUrl +
        "\" ?>");
marshaller.marshal(myXmlObject, fw);

Update

In recent version of JAXB we need to use property key as com.sun.xml.internal.bind.xmlHeaders like below.

Marshaller jaxbMarshaller = ...
jaxbMarshaller.setProperty("com.sun.xml.internal.bind.xmlHeaders", 
    "<?xml-stylesheet type='text/xsl' href='nameoffile.xsl' ?>");

edited Nov 12 '19 at 13:30

Mahendran Sakkarai

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answered Sep 26 '13 at 16:09

mat_boy

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1

This is the elegant way to do it, but it won't work if you are doing `marshal()` to a `File` (for example). It requires the output to be a Writer or Stream. Once you are aware of this it should be simple matter to wrap your `File` in a `FileWriter`. – Kevin Sadler Nov 04 '13 at 15:11
@KevinSadler Your edit has been rejected by the community. I added the edit, but it has been uglyfied. Please, prettify it such that the edit will be linked to you! Thanks! – mat_boy May 13 '14 at 07:46
OK I have reformatted it and altered the comment slightly, I hope it will be accepted as I think it adds a couple of useful things. – Kevin Sadler May 13 '14 at 14:33
Just a slight note: the supported properties depend on the used implemenation of JAXB – VeikkoW Sep 08 '14 at 14:44
For the record: using the default Jaxb implementation in Java 8, which uses the marshaller javax.xml.bind.helpers.AbstractMarshallerImpl under the hood, this very elegant method unfortunately does not work anymore. – mtj Sep 21 '16 at 06:21
1

only working using `com.sun.xml.internal.bind.xmlHeaders` – MushyPeas Oct 12 '19 at 16:00

score 4 · Answer 2 · answered Jan 31 '10 at 15:33

You could use a StringWriter to first write the stylesheet information into it, and then marshal the object into it:

StringWriter writer = new StringWriter();
//add processing instructions "by hand" with escaped quotation marks
//or single marks
writer.println("<?xml version='1.0'?>");
writer.println("<?xml-stylesheet type=\"text/xsl\" href=\"\">");

//create and configure marshaller to leave out processing instructions
Marshaller marshaller = context.createMarshaller();
marshaller.setProperty(Marshaller.JAXB_FRAGMENT, Boolean.TRUE);

//marshal to the StringWriter
marshaller.marshal(someObject,writer);
//get the string representation 
String str = writer.toString();

Of course you can also directly print to every other output stream you want, e.g. files or Sytstem.out.

score 2 · Answer 3 · answered Oct 30 '12 at 17:58

See how it's done in rexsl-core, part of ReXSL XSL/JAXB/JAX-RS framework: XslResolver:

final String header = String.format(
  "\n<?xml-stylesheet type='text/xsl' href='%s'?>",
  StringEscapeUtils.escapeXml("my-stylesheet.xsl")
);
marshaller.setProperty("com.sun.xml.bind.xmlHeaders", header);

score 1 · Answer 4 · edited Nov 05 '12 at 13:41

1

JAXBContext jaxbContext;
try {
       jaxbContext = JAXBContext.newInstance(new Class[] {SomeObject.class});

       StringWriter writer = new StringWriter();
       writer.write("<?xml version='1.0'?>");
       writer.write("\n");
       writer.write("<?xml-stylesheet type=\"text/xsl\" href=\"\">");
       writer.write("\n");

       Marshaller jaxbMarshaller = jaxbContext.createMarshaller();
       jaxbMarshaller.marshal(someobject, writer);
}

edited Nov 05 '12 at 13:41

Artemix

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answered Nov 05 '12 at 13:17

Ashok

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Please try to answer in words and not only in code. And since you answer so late, how does your answer differ from the other ones? – Olle Sjögren Nov 05 '12 at 13:38

Making JAXB generate an XML processing instruction

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