Modify existing XML stylesheet processing instruction in Java

Question

I'm reading an existing XML file and outputting it (using DOM).

<?xml version="1.0" encoding="UTF-8"?>
<?xml-stylesheet type="text/xsl" href="test"?>
<Books>
    <Book name="MyBook" />
</Books>

But how do I modify the XML stylesheet? -> href here set "test".

Mads Hansen · Accepted Answer · 2011-02-19 14:26:03Z

4

Something like this should work (untested)

Element root = doc.getDocumentElement();
XPath xpath = XPathFactory.newInstance().newXPath();
String expression = "/processing-instruction('xml-stylesheet')";
ProcessingInstruction pi;
pi = (ProcessingInstruction)xpath.evaluate(expression, doc, XPathConstants.NODE);
pi.setData("type='text/xsl' href='foo.xsl'");

answered Feb 19, 2011 at 14:26

Mads Hansen

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Comments

Theresa Forster · Accepted Answer · 2011-02-19 10:07:43Z

0

Thats a bit tricky, but why not read the file first into a String and do a replace before sending it via a stream into the dom parser.

answered Feb 19, 2011 at 10:07

Theresa Forster

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3 Comments

xmlhandler Over a year ago

OK can be closed got a solution :)

xmlhandler Over a year ago

i parsed and found the xml-stylesheet and simply replaced it with my own attributes

StackOverflowed Over a year ago

xmlhandler: Questions aren't "closed", you can just Approve the answer that helped you (by clicking the checkbox beside the answer).

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