Java XML Reading

Question

I've been wondering how to read XML files, but before you answer, read the whole post.

For example I have:

<?xml version="1.0" encoding="UTF-8"?>

<messages>

<incoming id="0" class="HelloIlikeyou" />

</messages>

What I want, is get all values from the tag . I want to place it in a dictionary, which key is incoming/outgoing, and then it will contain a list of Pair as value, with as key the id value and as value the class value.

So I got this:

HashMap<String, List<Pair<Integer, String>>> headers = new HashMap<>();

Then it will store this:

HashMap.get("incoming").add(new Pair<>("0", "HelloIlikeyou"));

But I don't know how to do it, I already got a part but it aint working:

File xml = new File(file);
        DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
        DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
        Document doc = dBuilder.parse(xml);
        doc.getDocumentElement().normalize();

        NodeList nodes = doc.getElementsByTagName("messages");

        for (int i = 0; i < nodes.getLength(); i++) {

            Node node = nodes.item(i);

                System.out.println("Type: " + node.getNodeValue() + " packet ID " + node.getUserData("id"));    
            }

What do you mean "it aint working" (sic)? Getting exceptions? Not returning any data? Computer catches on fire? — James Cronen
– James Cronen, Commented Jun 27, 2013 at 14:57
You are still operating on the messages node, you have to iterate over node.getChildNodes() — Manuel Manhart
– Manuel Manhart, Commented Jun 27, 2013 at 15:07
could anyone answer this? [enter link description here][1] [1]: stackoverflow.com/questions/24757825/… — Sharmishtha Kulkarni
– Sharmishtha Kulkarni, Commented Jul 15, 2014 at 12:09

Alks · Accepted Answer · 2013-06-27 14:59:15Z

3

You can use JAXB, i think that is the best way. take a look of this: Jaxb tutorial

answered Jun 27, 2013 at 14:59

Alks

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2 Comments

Nikola Yovchev Over a year ago

JAXB, imo, provides the worst api among all xml serialization libs. XStream is far superior in api fluency and convenience. The reason behind it might be JAXB is a ref implementation and can't use exotic stuff like XStream.

Mgetz Over a year ago

Please provide a compelling reason for using this particular tool rather than just suggesting it.

Manuel Manhart · Accepted Answer · 2013-06-27 15:11:48Z

2

This is what you want:

    public static void main(final String[] args)
    throws ParserConfigurationException, SAXException, IOException {
File xml = new File(file);
DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
Document doc = dBuilder.parse(xml);
doc.getDocumentElement().normalize();

NodeList nodes = doc.getElementsByTagName("messages");

for (int i = 0; i < nodes.getLength(); i++) {
    Node node = nodes.item(i);
    for (int j = 0; j < node.getChildNodes().getLength(); j++) {

    Node child = node.getChildNodes().item(j);

    if (!child.getNodeName().equals("#text")) {
        NamedNodeMap attributes = child.getAttributes();

        System.out.println("Type: " + child.getNodeName()
            + " packet ID " + attributes.getNamedItem("id")
            + " - class: " + attributes.getNamedItem("class"));
    }
    }
}
}

This gives me the following output:

Type: incoming packet ID id="0" - class: class="HelloIlikeyou"

answered Jun 27, 2013 at 15:11

Manuel Manhart

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1 Comment

user2528595 Over a year ago

Thanks, but I had to add: if (attributes == null) { continue; } To prevent null errors (my project works on a way it will stop after a bug and because that was a null bug I got an error). Thanks anyway for this fix.

Joop Eggen · Accepted Answer · 2013-06-27 15:00:59Z

0

Node node = nodes.item(i);
if (node instanceOf Element) {
    Element elem = (Element)node;
    String id = elem.getAttribute("id");
    ...

So you were almost there. The W3C classes are a bit old-stylish.

answered Jun 27, 2013 at 15:00

Joop Eggen

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Comments

facundofarias · Accepted Answer · 2016-08-29 17:00:36Z

0

Use one of the many available libraries that will do that for you, for example XStream:

http://x-stream.github.io/

edited Aug 29, 2016 at 17:00

facundofarias

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answered Jun 27, 2013 at 14:56

Nikola Yovchev

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