How to create an XML file in Java from scratch?

Question

I want to create an XML file from scratch and I am having an issue with the function parse, I have the below code. The documentation states this:

"abstract Document parse(InputSource is) Parse the content of the given input source as an XML document and return a new DOM Document object."

//Line with the issue on the parse function
Document document = docBuilder.parse(new InputSource(new StringReader(str)));

An the error that I get is this: " The method parse(InputStream) in the type DocumentBuilder is not applicable for the arguments (InputSource)"

What could be wrong? Thanks.

private static Document toXmlDocument(String str) throws ParserConfigurationException, SAXException, IOException{

  DocumentBuilderFactory docBuilderFactory = DocumentBuilderFactory.newInstance();
  DocumentBuilder docBuilder = docBuilderFactory.newDocumentBuilder();
  Document document = docBuilder.parse(new InputSource(new StringReader(str)));

  return document;
}

 public static void main(String[] args) {

       try{

       String xmlStr = "<?xml version=\"1.0\" encoding=\"UTF-8\"?>\n"+
       "<xbrli:xbrl\\n"+
       "xmlns:xbrli=\"http://www.xbrl.org/2003/instance\" "+
       "xmlns:link=\"http://www.xbrl.org/2003/linkbase\" "+
       "xmlns:xlink=\"http://www.w3.org/1999/xlink\"><\n"+
       "</xbrli:xbrl>";

       Document doc = toXmlDocument(xmlStr);

       }
       catch(Exception e ){
           e.printStackTrace();
       }
      }

Answer 1

DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
DocumentBuilder db = dbf.newDocumentBuilder();
Document d = db.parse(new ByteArrayInputStream(string.getBytes()));

This will work for you. You don't need to wrap the InputStream in an InputSource, as that's an option as well.