JAVA - Parse XML [duplicate]

Question

I have this XML:

<?xml version="1.0" encoding="UTF-8"?> <soapenv:Envelope> <soapenv:Body> <response> <note identifier="4719b11c-230e-4533-b1a5-76ac1dc69ef7" hash="956F3D94-A8B375ED-BA15739F-387B4F5E-75C6ED82" dat_prij="2016-09-27T23:06:20+02:00"/> <item uid="9f9cb5fb-75e6-4a2f-be55-faef9533d9ff-ff" active="true"/> </response> </soapenv:Body> </soapenv:Envelope>

I need parse item uid, in this case its value "9f9cb5fb...", but I dont know how to parse with Java standard nodes. Thank you for advice.

Edited: I tried this code, but not working:

DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance(); DocumentBuilder builder = factory.newDocumentBuilder(); InputSource is = new InputSource(new StringReader(response)); Document doc = builder.parse(is); Element root = doc.getDocumentElement(); NodeList nList = doc.getElementsByTagName("soapenv:Envelope"); NodeList body = nList.getElementsByTagName("soapenv:Body"); NodeList resp = body.getElementsByTagName("response"); Node node = nList.item("item"); System.out.println(""); //Just a separator Element eElement = (Element) node; System.out.println(eElement.getAttribute("uid"));

@CosmeLopes - I add additional information to question

David Horňák
– David Horňák

2016-09-27 22:27:40 +00:00
Commented Sep 27, 2016 at 22:27 — David Horňák
– David Horňák, Commented Sep 27, 2016 at 22:27

MGorgon · Accepted Answer · 2016-09-27 22:55:49Z

Using DOM:

 DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance(); DocumentBuilder builder = factory.newDocumentBuilder(); InputSource is = new InputSource(new StringReader(response)); Document doc = builder.parse(is); NodeList nList = doc.getElementsByTagName("item"); Node namedItem = nList.item(0).getAttributes().getNamedItem("uid"); System.out.println(namedItem.getNodeValue());

Using SAX:

 SAXParserFactory factory = SAXParserFactory.newInstance(); SAXParser saxParser = factory.newSAXParser(); DefaultHandler handler = new DefaultHandler() { @Override public void startElement(String uri, String localName, String qName, Attributes attributes) throws SAXException { if (qName.equals("item")) { System.out.println(attributes.getValue("uid")); } } }; saxParser.parse(new InputSource(new StringReader(response)), handler);

You may want to read some tutorials about XML, DOM and SAX: https://docs.oracle.com/javase/tutorial/jaxp/dom/readingXML.html https://docs.oracle.com/javase/tutorial/jaxp/sax/parsing.html

Or you can just use Regex and some string replacing:

 Pattern pattern = Pattern.compile("uid=\"[^\"]*\""); Matcher matcher = pattern.matcher(response); if(matcher.find()) { String attr = matcher.group(); attr = attr.replace("\"", ""); attr = attr.replace("uid=", ""); System.out.println(attr); }

Note:

your XML isn't valid SOAP message. It's missing some namespaces declaration - no soapenv prefix is declared, and your elements and attributes are undefined.

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