It is possible with SAX, DOM, XMLPull parse a XML file with the following header:
<JsonResult xmlns:i="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://schemas.datacontract.org/2004/07/System.Web.Mvc">
<ContentEncoding xmlns:d2p1="http://schemas.datacontract.org/2004/07/System.Text" i:nil="true"/>
<ContentType i:nil="true"/>
<Data xmlns:d2p1="http://schemas.datacontract.org/2004/07/ProximoColectivo.Entities" i:type="d2p1:FeatureCollection">
o should always start with <?xml version = 1.0>
To do so.
Yes you can parse an xml file which does not start with <?xml version = 1.0>
. For eg
public static void main(String args[]) throws ParserConfigurationException, SAXException, IOException {
String str = "<abc/>";
DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
Document doc = dBuilder.parse(new ByteArrayInputStream(str.getBytes()));
System.out.println("Root element :" + doc.getDocumentElement().getNodeName());
}
will work. But your String xml file will not work. For xml parsers to work there should be a root tag and then all tags inside it. So
String str = "<abc/><pqr/>";
will not work but
String str = "<abc><pqr/></abc>";
will work. What you can do as a workaround for this is
Enumeration<InputStream> combinedStreams = Collections.enumeration(
Arrays.asList(new InputStream[] {
new ByteArrayInputStream("<abc>".getBytes()),
yourXMLFileInputStreamStream,
new ByteArrayInputStream("</abc>".getBytes()),
}));
SequenceInputStream xmlStream = new SequenceInputStream(combinedStreams);
Document doc = dBuilder.parse(xmlStream);
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.