XML node list as JavaFX observable list?

Question

got a problem...

I have some XML file with root element, 3 element with same name and different attribute and also each element has got some elements.

I want to have element attributes called "id" in FX observable list. No idea what to do.

Learn about X-Path which is a technique to select distinct nodes out of an xml. — Heri, Oct 24 '16 at 20:59

score 0 · Accepted Answer · edited May 23 '17 at 12:19

Since you didn´t give me much information, I have to explain it with some dummy-code.
So let´s say you have this XML:

<root id="0">
    <sub1 id="1">
        <hell1>hell1</hell1>
    </sub1>
    <sub2>
        <hell2 id="2">hell2</hell2>
    </sub2>
    <sub3>sub3</sub3>
</root>

Please note, that the "id"-attributes are in different element levels, as you said.

From here I would also advise you to use xpath like Heri. The simplest xpath to get all elements with "id"-attributes from XML is:

//*[@id]

If you don´t know how to work with XML and Xpath in Java you can look it up here:
How to read XML using XPath in Java
Which is the best library for XML parsing in java

If i wanted to get all "id"-attribute values from my XML it would look like this:

// My Test-XML
final String xml = "<?xml version=\"1.0\" encoding=\"UTF-8\"?>\n" +
        "<root id=\"0\">\n" +
        "\t<sub1 id=\"1\">\n" +
        "\t\t<hell1>hell1</hell1>\n" +
        "\t</sub1>\n" +
        "\t<sub2>\n" +
        "\t\t<hell2 id=\"2\">hell2</hell2>\n" +
        "\t</sub2>\n" +
        "\t<sub3>sub3</sub3>\n" +
        "</root>";
DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
DocumentBuilder builder = factory.newDocumentBuilder();
// Here you have to put your XML-Source (String, InputStream, File)
        Document doc = builder.parse(new InputSource(new ByteArrayInputStream(xml.getBytes(StandardCharsets.UTF_8))));
XPathFactory xPathFactory = XPathFactory.newInstance();
XPath xpath = xPathFactory.newXPath();
// The xpath from above
XPathExpression expr = xpath.compile("//*[@id]");
// Here you get the node-list of all elements with an attribute named "id"
NodeList nodeList = (NodeList) expr.evaluate(doc, XPathConstants.NODESET);
for (int i = 0; i < nodeList.getLength(); i++) {
    // run through all nodes
    Node node = nodeList.item(i);
    // fetch the values from the "id"-attribute
    final String idValue = node.getAttributes().getNamedItem("id").getNodeValue();
    // Do something awesome!!! 
    System.out.println(idValue);
}

I assume that you know how to put the values into your observable list then. :)
If you have any more questions, feel free to ask.

PS: If you show me your XML, I can help you in a more efficient way.

XML node list as JavaFX observable list?

1 Answers1