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XPath check for attribute and show the value

I need to check a bunch of .xml files for a specific permission, which is an attribute. If there is such a permission attribute i have to find out which value the attribute has.
this is my code which produces NullPointerExceptions:
public static void checkXmlPermissions(String path)
{
FileInputStream file;
try
{
file = new FileInputStream(new File(path));
DocumentBuilderFactory builderFactory = DocumentBuilderFactory.newInstance();
//XPath initialisieren;
DocumentBuilder builder = builderFactory.newDocumentBuilder();
Document xmlDocument = builder.parse(file);
XPath xPath = XPathFactory.newInstance().newXPath();
String expression ="//*[#permission]"; //Expression;
NodeList nodeList = (NodeList) xPath.compile(expression).evaluate(xmlDocument, XPathConstants.NODESET);
for (int i = 0; i < nodeList.getLength(); i++)
{
String match = nodeList.item(i).getFirstChild().getNodeValue();
System.out.println(match);
}
}
catch(Exception ex)
{
System.out.println(ex);
}
}
I guess my mistake is the NodeList but I can´t find a solution by my own.

change:
String match = nodeList.item(i).getFirstChild().getNodeValue();
to:
String match = nodeList.item(i).getTextContent();

parsing xml string in java

I have been trying to use code I found from other posts, to parse an xml string but when trying to get to the node elements I keep getting null values. Can someone see the error or something I am missing?
try {
DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
DocumentBuilder db = dbf.newDocumentBuilder();
InputSource is = new InputSource();
is.setCharacterStream(new StringReader(post));
Document doc;
try {
doc = db.parse(is);
doc.getDocumentElement().normalize();
Element root = doc.getDocumentElement();
NodeList nodes = root.getElementsByTagName("entry");
} catch (SAXException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
the root element is null and nodelist null as well. The XML post starts with:
<entry xmlns="http://www.w3.org/2005/Atom"
xmlns:activity="http://activitystrea.ms/spec/
1.0/" xmlns:service="http://activitystrea.ms/
service-provider" xmlns:thr="http://purl.org/syndication/thread/1.0"
xmlns:gnip="http://www.post.com/schemas/2010"
xmlns:geo="http://www.georss.org/georss" xmlns:poco="http://portablecontacts.net/spec/1.0">
<id>...
EDIT:
NodeList nodes.. //The value of the local nodes is not used
[entry: null]
// it skips the following lines...
for(int i=0; i < nodes.getLength() - 1; i++){
System.out.println(nodes.item(i).toString());
}

replace NodeList nodes = root.getElementsByTagName("entry");
by NodeList nodes = root.getChildNodes();

XML parsing using XPath in Java

Hi!
I've spent some time to parse an XML document with XPath. It seeams to be a simple task but I got in troubles since the begining.
My code is :
public class QueryXML3 {
public static void main(String[] args) {
DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
factory.setNamespaceAware(true);
DocumentBuilder builder;
Document doc = null;
try {
builder = factory.newDocumentBuilder();
//doc = builder.parse("SampleExample.xml");
InputStream is = QueryXML3.class.getClassLoader().getResourceAsStream("SampleXml.xml");
doc = builder.parse(is);
XPathFactory xpathFactory = XPathFactory.newInstance();
// Create XPath object
XPath xpath = xpathFactory.newXPath();
Node parNode = getParameterNode(doc, xpath);
System.out.println("parameter node:" + parNode);
NodeList res = getParameterNodeList(doc, xpath );
System.out.println("List of nodes" + res);
} catch (ParserConfigurationException | SAXException | IOException e) {
e.printStackTrace();
}
}
public static Node getParameterNode(Document doc, XPath xpath) {
Node res = null;
try {
res = (Node) xpath.evaluate("/definitions/process", doc, XPathConstants.NODE);
} catch (XPathExpressionException e) {
e.printStackTrace();
}
return res;
}
public static NodeList getParameterNodeList(Document doc, XPath xpath) {
NodeList nodeList = null;
try {
nodeList = (NodeList) xpath.evaluate("/definitions/process", doc, XPathConstants.NODESET);
for (int i = 0; i > nodeList.getLength(); i++) {
System.out.print(nodeList.item(i).getNodeName() + " ");
}
} catch (XPathExpressionException e) {
e.printStackTrace();
}
return nodeList;
}
}
As a result i get this:
parameter node:[process: null]
List of nodes com.sun.org.apache.xml.internal.dtm.ref.DTMNodeList#2f17aadf
I just want to output all the nodes of my xml file and theire attributes...

You are really asking how to serialize an Element to a string - use either a Transformer or DOMImplementationLS.
The NodeList type has no toString() contract and the implementation does not override the default Object.toString(). You need to iterate over the nodes and serialize each Element as above.

You could easily parse an XML file in java using a 3rd party package such as JSoup or JDom.
As an example, here is some simple output of an XML files elements using JSoup:
<note>
<to>Tove</to>
<from>Jani</from>
<heading>Reminder</heading>
<body>Don't forget me this weekend!</body>
</note>
Java code printing all elements and the selected <from>-element:
String xml = "<note>\n"
+ "<to>Tove</to>\n"
+ "<from>Jani</from>\n"
+ "<heading>Reminder</heading>\n"
+ "<body>Don't forget me this weekend!</body>\n"
+ "</note>";
Document doc = Jsoup.parse(xml, "", Parser.xmlParser());
for (Element e : doc.children()) {
System.out.println(e);
}
Element fromElement = doc.select("from").first();
System.out.println("\nThis is the <from>-element content:\n" + fromElement);

Getting null values when trying to parse an XML using XPATH in Java

When running this code in Eclipse I get only null values:
String recordsAsXML = "<vGVC:Row xmlns:vGVC=\"urn:com.versai:Bacon:GetViewContentsResponse.v1.0\"><vGVC:Column><vGVC:Value>aaa0</vGVC:Value><vGVC:ObjectName>2ff31656-b10c-11e0-99f6-e01321378d2a</vGVC:ObjectName></vGVC:Column><vGVC:ModifiedAfterQuery>false</vGVC:ModifiedAfterQuery></vGVC:Row>";
DocumentBuilderFactory domFactory = DocumentBuilderFactory.newInstance();
domFactory.setNamespaceAware(true);
DocumentBuilder builder = domFactory.newDocumentBuilder();
InputSource is = new InputSource();
is.setCharacterStream(new StringReader(recordsAsXML));
Document doc = builder.parse(is);
XPathFactory factory = XPathFactory.newInstance();
XPath xpath = factory.newXPath();
XPathExpression expr = xpath.compile("//vGVC:Value/text()");
Object result = expr.evaluate(doc, XPathConstants.NODESET);
NodeList nodes = (NodeList) result;
for (int i = 0; i < nodes.getLength(); i++) {
System.out.println(nodes.item(i).getNodeValue());
}
I'm trying to extract the "aaa0" value.
Could you please help me understand what I'm doing wrong?
Thanks in advance, Michal.

You should use namespace context to let XPath know about your namespace, for example like this:
xpath.setNamespaceContext(new NamespaceContext() {
#Override public Iterator<?> getPrefixes(String namespaceURI) { return null; }
#Override public String getPrefix(String namespaceURI) { return "vGVC"; }
#Override public String getNamespaceURI(String prefix) { return "urn:com.versai:Bacon:GetViewContentsResponse.v1.0"; }
});
This should be done before calling evaluate, and if you have more than one namespace to handle, you can add them all in one implementation and use if statements to return the right prefix/URI.
More details in the javadocs here: http://download.oracle.com/javase/6/docs/api/javax/xml/namespace/NamespaceContext.html

Create XML document using nodeList

I need to create a XML Document object using the NodeList. Can someone pls help me to do this. This is my Java code:
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.xpath.*;
import org.w3c.dom.*;
public class ReadFile {
public static void main(String[] args) {
String exp = "/configs/markets";
String path = "testConfig.xml";
try {
Document xmlDocument = DocumentBuilderFactory.newInstance().newDocumentBuilder().parse(path);
XPath xPath = XPathFactory.newInstance().newXPath();
XPathExpression xPathExpression = xPath.compile(exp);
NodeList nodes = (NodeList)
xPathExpression.evaluate(xmlDocument,
XPathConstants.NODESET);
} catch (Exception ex) {
ex.printStackTrace();
}
}
}
I want to have an XML file like this:
<configs>
<markets>
<market>
<name>Real</name>
</market>
<market>
<name>play</name>
</market>
</markets>
</configs>
Thanks in advance.

You should do it like this:
you create a new org.w3c.dom.Document newXmlDoc where you store the nodes in your NodeList,
you create a new root element, and append it to newXmlDoc
then, for each node n in your NodeList, you import n in newXmlDoc, and then you append n as a child of root
Here is the code:
public static void main(String[] args) {
String exp = "/configs/markets/market";
String path = "src/a/testConfig.xml";
try {
Document xmlDocument = DocumentBuilderFactory.newInstance()
.newDocumentBuilder().parse(path);
XPath xPath = XPathFactory.newInstance().newXPath();
XPathExpression xPathExpression = xPath.compile(exp);
NodeList nodes = (NodeList) xPathExpression.
evaluate(xmlDocument, XPathConstants.NODESET);
Document newXmlDocument = DocumentBuilderFactory.newInstance()
.newDocumentBuilder().newDocument();
Element root = newXmlDocument.createElement("root");
newXmlDocument.appendChild(root);
for (int i = 0; i < nodes.getLength(); i++) {
Node node = nodes.item(i);
Node copyNode = newXmlDocument.importNode(node, true);
root.appendChild(copyNode);
}
printTree(newXmlDocument);
} catch (Exception ex) {
ex.printStackTrace();
}
}
public static void printXmlDocument(Document document) {
DOMImplementationLS domImplementationLS =
(DOMImplementationLS) document.getImplementation();
LSSerializer lsSerializer =
domImplementationLS.createLSSerializer();
String string = lsSerializer.writeToString(document);
System.out.println(string);
}
The output is:
<?xml version="1.0" encoding="UTF-16"?>
<root><market>
<name>Real</name>
</market><market>
<name>play</name>
</market></root>
Some notes:
I've changed exp to /configs/markets/market, because I suspect you want to copy the market elements, rather than the single markets element
for the printXmlDocument, I've used the interesting code in this answer
I hope this helps.
If you don't want to create a new root element, then you may use your original XPath expression, which returns a NodeList consisting of a single node (keep in mind that your XML must have a single root element) that you can directly add to your new XML document.
See following code, where I commented lines from the code above:
public static void main(String[] args) {
//String exp = "/configs/markets/market/";
String exp = "/configs/markets";
String path = "src/a/testConfig.xml";
try {
Document xmlDocument = DocumentBuilderFactory.newInstance()
.newDocumentBuilder().parse(path);
XPath xPath = XPathFactory.newInstance().newXPath();
XPathExpression xPathExpression = xPath.compile(exp);
NodeList nodes = (NodeList) xPathExpression.
evaluate(xmlDocument,XPathConstants.NODESET);
Document newXmlDocument = DocumentBuilderFactory.newInstance()
.newDocumentBuilder().newDocument();
//Element root = newXmlDocument.createElement("root");
//newXmlDocument.appendChild(root);
for (int i = 0; i < nodes.getLength(); i++) {
Node node = nodes.item(i);
Node copyNode = newXmlDocument.importNode(node, true);
newXmlDocument.appendChild(copyNode);
//root.appendChild(copyNode);
}
printXmlDocument(newXmlDocument);
} catch (Exception ex) {
ex.printStackTrace();
}
}
This will give you the following output:
<?xml version="1.0" encoding="UTF-16"?>
<markets>
<market>
<name>Real</name>
</market>
<market>
<name>play</name>
</market>
</markets>

you can try the adoptNode() method of Document. Maybe you will need to iterate over your NodeList. You can access the individual Nodes with nodeList.item(i).If you want to wrap your search results in an Element, you can use createElement() from the Document and appendChild() on the newly created Element