Ok, so I got some url link like https://stackoverflow.com/ and I'm trying to parse it in document but getting error. Why? Because this is not xml file, so the question is how can I get data as xml if i got only url?
My code:
public class URLReader {
public static void main(String[] args) throws Exception {
// or if you prefer DOM:
DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
DocumentBuilder db = dbf.newDocumentBuilder();
Document doc = db.parse(new URL("https://stackoverflow.com/").openStream());
int nodes = doc.getChildNodes().getLength();
System.out.println(nodes + " nodes found");
}
}
To parse HTML you may use JSOUP: https://jsoup.org/
This library provides also some features to transform HTML to XHTML, which some sort of XML:
Document document = Jsoup.parse(html);
document.outputSettings().syntax(Document.OutputSettings.Syntax.xml);
document.outputSettings().escapeMode(org.jsoup.nodes.Entities.EscapeMode.xhtml);
String xhtml=document.html();
I'm reading correctly an xml file, but I'm not able to write it.
Here is the file: a configuration file for key-value settings.
<?xml version="1.0" encoding="UTF-8"?>
<configuration>
<setting key="Password" value="d92e1dedba95d2cf00d4c567e57e3342"/>
<setting key="ExceptionFileLog" value="exception.txt"/>
<setting key="ActionFileLog" value="actions.txt" />
<setting key="ShowInfoMessage" value="false" />
</configuration>
I correctly open and read file using javax.xml.parsers.DocumentBuilder:
private Document _doc = null;
public XmlConfig(String filePath) throws Exception
{
File xml = new File(filePath);
DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
_doc = dBuilder.parse(xml);
_doc.getDocumentElement().normalize();
}
So far so good, but I'm not able to write and persist changes to the file:
public boolean updateValue(String key, String value)
{
NodeList settlist = _doc.getElementsByTagName(SETTNAME);
for(int i = 0; i < settlist.getLength(); i++)
{
Element sett = (Element) settlist.item(i);
if(sett.getNodeType() == Node.ELEMENT_NODE)
{
if(null != sett.getAttribute("key") && sett.getAttribute("key").equals(key))
{
sett.setAttribute("value", value);
return true;
}
}
}
return false;
}
So, if I print xml file from _doc (Document object) the changes are correctly written, but xml file is not updated!
I suppose that I'm opening,reading and writing xml file in memory and I need a way to persist changes on disk.
I have no idea, any suggestion will be appreciated.
save the changed xml file using the following code
Transformer transformer = TransformerFactory.newInstance().newTransformer();
Result output = new StreamResult(xml); // xml is a object of File i.e. File xml = new File(filePath);
Source input = new DOMSource(_doc);
transformer.transform(input, output);
it will store the updated values in the xml file.
reference from how-to-save-parsed-and-changed-dom-document-in-xml-file
I am new to XML. I want to read the following XML on the basis of request name. Please help me on how to read the below XML in Java -
<?xml version="1.0"?>
<config>
<Request name="ValidateEmailRequest">
<requestqueue>emailrequest</requestqueue>
<responsequeue>emailresponse</responsequeue>
</Request>
<Request name="CleanEmail">
<requestqueue>Cleanrequest</requestqueue>
<responsequeue>Cleanresponse</responsequeue>
</Request>
</config>
If your XML is a String, Then you can do the following:
String xml = ""; //Populated XML String....
DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
DocumentBuilder builder = factory.newDocumentBuilder();
Document document = builder.parse(new InputSource(new StringReader(xml)));
Element rootElement = document.getDocumentElement();
If your XML is in a file, then Document document will be instantiated like this:
Document document = builder.parse(new File("file.xml"));
The document.getDocumentElement() returns you the node that is the document element of the document (in your case <config>).
Once you have a rootElement, you can access the element's attribute (by calling rootElement.getAttribute() method), etc. For more methods on java's org.w3c.dom.Element
More info on java DocumentBuilder & DocumentBuilderFactory. Bear in mind, the example provided creates a XML DOM tree so if you have a huge XML data, the tree can be huge.
Related question.
Update Here's an example to get "value" of element <requestqueue>
protected String getString(String tagName, Element element) {
NodeList list = element.getElementsByTagName(tagName);
if (list != null && list.getLength() > 0) {
NodeList subList = list.item(0).getChildNodes();
if (subList != null && subList.getLength() > 0) {
return subList.item(0).getNodeValue();
}
}
return null;
}
You can effectively call it as,
String requestQueueName = getString("requestqueue", element);
In case you just need one (first) value to retrieve from xml:
public static String getTagValue(String xml, String tagName){
return xml.split("<"+tagName+">")[1].split("</"+tagName+">")[0];
}
In case you want to parse whole xml document use JSoup:
Document doc = Jsoup.parse(xml, "", Parser.xmlParser());
for (Element e : doc.select("Request")) {
System.out.println(e);
}
If you are just looking to get a single value from the XML you may want to use Java's XPath library. For an example see my answer to a previous question:
How to use XPath on xml docs having default namespace
It would look something like:
import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.xpath.XPath;
import javax.xml.xpath.XPathConstants;
import javax.xml.xpath.XPathFactory;
import org.w3c.dom.Document;
import org.w3c.dom.NodeList;
public class Demo {
public static void main(String[] args) {
DocumentBuilderFactory domFactory = DocumentBuilderFactory.newInstance();
try {
DocumentBuilder builder = domFactory.newDocumentBuilder();
Document dDoc = builder.parse("E:/test.xml");
XPath xPath = XPathFactory.newInstance().newXPath();
Node node = (Node) xPath.evaluate("/Request/#name", dDoc, XPathConstants.NODE);
System.out.println(node.getNodeValue());
} catch (Exception e) {
e.printStackTrace();
}
}
}
There are a number of different ways to do this. You might want to check out XStream or JAXB. There are tutorials and the examples.
If the XML is well formed then you can convert it to Document. By using the XPath you can get the XML Elements.
String xml = "<stackusers><name>Yash</name><age>30</age></stackusers>";
Form XML-String Create Document and find the elements using its XML-Path.
Document doc = getDocument(xml, true);
public static Document getDocument(String xmlData, boolean isXMLData) throws Exception {
DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
dbFactory.setNamespaceAware(true);
dbFactory.setIgnoringComments(true);
DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
Document doc;
if (isXMLData) {
InputSource ips = new org.xml.sax.InputSource(new StringReader(xmlData));
doc = dBuilder.parse(ips);
} else {
doc = dBuilder.parse( new File(xmlData) );
}
return doc;
}
Use org.apache.xpath.XPathAPI to get Node or NodeList.
System.out.println("XPathAPI:"+getNodeValue(doc, "/stackusers/age/text()"));
NodeList nodeList = getNodeList(doc, "/stackusers");
System.out.println("XPathAPI NodeList:"+ getXmlContentAsString(nodeList));
System.out.println("XPathAPI NodeList:"+ getXmlContentAsString(nodeList.item(0)));
public static String getNodeValue(Document doc, String xpathExpression) throws Exception {
Node node = org.apache.xpath.XPathAPI.selectSingleNode(doc, xpathExpression);
String nodeValue = node.getNodeValue();
return nodeValue;
}
public static NodeList getNodeList(Document doc, String xpathExpression) throws Exception {
NodeList result = org.apache.xpath.XPathAPI.selectNodeList(doc, xpathExpression);
return result;
}
Using javax.xml.xpath.XPathFactory
System.out.println("javax.xml.xpath.XPathFactory:"+getXPathFactoryValue(doc, "/stackusers/age"));
static XPath xpath = javax.xml.xpath.XPathFactory.newInstance().newXPath();
public static String getXPathFactoryValue(Document doc, String xpathExpression) throws XPathExpressionException, TransformerException, IOException {
Node node = (Node) xpath.evaluate(xpathExpression, doc, XPathConstants.NODE);
String nodeStr = getXmlContentAsString(node);
return nodeStr;
}
Using Document Element.
System.out.println("DocumentElementText:"+getDocumentElementText(doc, "age"));
public static String getDocumentElementText(Document doc, String elementName) {
return doc.getElementsByTagName(elementName).item(0).getTextContent();
}
Get value in between two strings.
String nodeVlaue = org.apache.commons.lang.StringUtils.substringBetween(xml, "<age>", "</age>");
System.out.println("StringUtils.substringBetween():"+nodeVlaue);
Full Example:
public static void main(String[] args) throws Exception {
String xml = "<stackusers><name>Yash</name><age>30</age></stackusers>";
Document doc = getDocument(xml, true);
String nodeVlaue = org.apache.commons.lang.StringUtils.substringBetween(xml, "<age>", "</age>");
System.out.println("StringUtils.substringBetween():"+nodeVlaue);
System.out.println("DocumentElementText:"+getDocumentElementText(doc, "age"));
System.out.println("javax.xml.xpath.XPathFactory:"+getXPathFactoryValue(doc, "/stackusers/age"));
System.out.println("XPathAPI:"+getNodeValue(doc, "/stackusers/age/text()"));
NodeList nodeList = getNodeList(doc, "/stackusers");
System.out.println("XPathAPI NodeList:"+ getXmlContentAsString(nodeList));
System.out.println("XPathAPI NodeList:"+ getXmlContentAsString(nodeList.item(0)));
}
public static String getXmlContentAsString(Node node) throws TransformerException, IOException {
StringBuilder stringBuilder = new StringBuilder();
NodeList childNodes = node.getChildNodes();
int length = childNodes.getLength();
for (int i = 0; i < length; i++) {
stringBuilder.append( toString(childNodes.item(i), true) );
}
return stringBuilder.toString();
}
OutPut:
StringUtils.substringBetween():30
DocumentElementText:30
javax.xml.xpath.XPathFactory:30
XPathAPI:30
XPathAPI NodeList:<stackusers>
<name>Yash</name>
<age>30</age>
</stackusers>
XPathAPI NodeList:<name>Yash</name><age>30</age>
following links might help
http://labe.felk.cvut.cz/~xfaigl/mep/xml/java-xml.htm
http://developerlife.com/tutorials/?p=25
http://www.java-samples.com/showtutorial.php?tutorialid=152
There are two general ways of doing that. You will either create a Domain Object Model of that XML file, take a look at this
and the second choice is using event driven parsing, which is an alternative to DOM xml representation. Imho you can find the best overall comparison of these two basic techniques here. Of course there are much more to know about processing xml, for instance if you are given XML schema definition (XSD), you could use JAXB.
There are various APIs available to read/write XML files through Java.
I would refer using StaX
Also This can be useful - Java XML APIs
You can make a class which extends org.xml.sax.helpers.DefaultHandler and call
start_<tag_name>(Attributes attrs);
and
end_<tag_name>();
For it is:
start_request_queue(attrs);
etc.
And then extends that class and implement xml configuration file parsers you want. Example:
...
public void startElement(String uri, String name, String qname,
org.xml.sax.Attributes attrs)
throws org.xml.sax.SAXException {
Class[] args = new Class[2];
args[0] = uri.getClass();
args[1] = org.xml.sax.Attributes.class;
try {
String mname = name.replace("-", "");
java.lang.reflect.Method m =
getClass().getDeclaredMethod("start" + mname, args);
m.invoke(this, new Object[] { uri, (org.xml.sax.Attributes)attrs });
}
catch (IllegalAccessException e) {
throw new RuntimeException(e);
}
catch (NoSuchMethodException e) {
throw new RuntimeException(e); }
catch (java.lang.reflect.InvocationTargetException e) {
org.xml.sax.SAXException se =
new org.xml.sax.SAXException(e.getTargetException());
se.setStackTrace(e.getTargetException().getStackTrace());
}
and in a particular configuration parser:
public void start_Request(String uri, org.xml.sax.Attributes attrs) {
// make sure to read attributes correctly
System.err.println("Request, name="+ attrs.getValue(0);
}
Since you are using this for configuration, your best bet is apache commons-configuration. For simple files it's way easier to use than "raw" XML parsers.
See the XML how-to
I want to unmarshall a element into a Class Property, which Type is Object, as to keep it generic.
I tried to build the class and marshall into xml, unmarshall it back,the result was fine.
But when i try to do this with an normal generated Xml-Document(although it has the same structure), the value of the result Class Object Property is null.
Here is my Teststructure:
#XmlRootElement
public class TestStructure {
private Object test;
public Object getTest() {
return test;
}
public void setTest(Object test) {
this.test = test;
}
}
I try to marshall this, and get this xml-document:
<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<testStructure>
<test xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xs="http://www.w3.org/2001/XMLSchema" xsi:type="xs:string">foo</test>
</testStructure>
However, if i try to arbitrarly build this structure Via Domsource to get the same Xml document:
DocumentBuilderFactory docFactory = DocumentBuilderFactory.newInstance();
DocumentBuilder docBuilder = docFactory.newDocumentBuilder();
// root element
Document doc = docBuilder.newDocument();
Element rootElement = doc.createElement("testStructure");
doc.appendChild(rootElement);
Element test = doc.createElement("test");
test.appendChild(doc.createTextNode("foo"));
test.setAttribute("xmlns:xsi", "http://www.w3.org/2001/XMLSchema-instance");
test.setAttribute("xsi:type", "xs:string");
test.setAttribute("xmlns:xs", "http://www.w3.org/2001/XMLSchema");
rootElement.appendChild(test);
and Unmarshal the document:
JAXBContext context2 = JAXBContext.newInstance(TestStructure.class);
Unmarshaller m2 = context2.createUnmarshaller();
TestStructure testobject2 =
( TestStructure ) m2.unmarshal(doc);
System.out.println(testobject2.getTest());
The attribute appears to be "null".
So, what went wrong?
woah, thanks to W A , i got the solution :)
You just have to modify the Element Attributes:
Element test = doc.createElement("test");
test.appendChild(doc.createTextNode("foo"));
test.setAttributeNS("http://www.w3.org/2001/XMLSchema-instance", "xsi:type", "xs:string");
test.setAttribute("xmlns:xs", "http://www.w3.org/2001/XMLSchema");
Using only code Java I can get the root name with these lines.
Element root = document.getDocumentElement();
and get the name with root.getNodeName()
But in a Android enviroment, how can I get for example, the name 'aluno' as root name?
<soap:Envelope xmlns:soap="http://schemas.xmlsoap.org/soap/envelope/"
xmlns:xsd="http://www.w3.org/2001/XMLSchema"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<soap:Body>
<ns1:autenticaAlunoResponse xmlns:ns1="http://xfire.codehaus.org/AlunoService">
<aluno xmlns="urn:bean.wsi.br">
<matricula xmlns="http://bean.wsi.br">61203475</matricula>
<turma xmlns="http://bean.wsi.br"><codigo>2547</codigo>
<nome>B</nome>
</turma>
</aluno>
</ns1:autenticaAlunoResponse>
</soap:Body>
</soap:Envelope>
Update (copied from a comment below):
I'm using Ksoap2 and trying to parse using SAX.
DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
DocumentBuilder db;
db = dbf.newDocumentBuilder();
InputSource is = new InputSource();
is.setCharacterStream(new StringReader(xml));
Document doc = db.parse(is);
In your example 'aluno' appears as a tag name. If you work with Jsoup you can find the element by tag and then use tagName method to retrieve its name:
Document doc;
Elements tagName;
String name;
try {
doc = Jsoup.connect(url).userAgent("Mozilla").get();
} catch (IOException ioe) {
ioe.printStackTrace();
}
doc.select("aluno");
name = tagName.tagName();