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how to get null while parsing xml with geotools parser

I tried to get null value from my xml file such below
<top:kadastroParselFeature gml:id="KADASTRO_PARSEL.1">
<top:olcek xsi:nil="true" />
</top:kadastroParselFeature>
but my parser throws such an exception :
java.lang.RuntimeException: Parsing failed for kadastroParselFeature: java.lang.NullPointerException
at org.geotools.xml.impl.ParseExecutor.visit(ParseExecutor.java:164)
at org.geotools.xml.impl.BindingWalker$BindingExecutionChain.execute(BindingWalker.java:220)
at org.geotools.xml.impl.BindingWalker.walk(BindingWalker.java:186)
at org.geotools.xml.impl.ElementHandlerImpl.endElement(ElementHandlerImpl.java:236)
at org.geotools.xml.impl.ParserHandler.endElement(ParserHandler.java:719)
My parser configuration like this :
org.geotools.xml.Configuration configuration = new org.geotools.gml2.GMLConfiguration();
org.geotools.xml.Parser parser = new org.geotools.xml.Parser( configuration );
parser.setFailOnValidationError(false);
HashMap<Object, Object> parsedMap = (HashMap<Object, Object>) parser.parse( isx);
if I remove xsi:nill="true" from element it returns ""(empty string) but i need null .
there is a usage of my parser at http://docs.geotools.org/stable/userguide/library/xml/geometry.html

I found a solution for my work the problem at my gml writer version if i changed the version from gml2 to gml3 the problem not seen. and ı changed my parser config. like this
org.geotools.xml.Configuration configuration = new org.geotools.gml3.ApplicationSchemaConfiguration("sampleURL", "sample.xsd");
org.geotools.xml.Parser parser = new org.geotools.xml.Parser( configuration );

Efficient way of parsing XML in Java

I have to parse an XML file with following structure:
<root>
<object_1>
<pro1> abc </pro1>
<pro2> pqr </pro2>
<pro3> xyz </pro3>
<children>
<object_a>
<pro1> abc </pro1>
<pro2> pqr </pro2>
<pro3> xyz </pro3>
<children>
.
.
.
</children>
</object_a>
</children>
</object_1>
<object_2>
.
.
.
</object_n>
</root>
Aim is to parse this multilevel nesting. A few classes are defined in Java.
Class Object_1
Class Object_2
.
.
.
Class Object_N
with their respective properties.
The following code is working for me, but then this is not the best way of doing things.
File file = new File(fileName);
DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
Document doc = dBuilder.parse(file);
doc.getDocumentElement().normalize();
if(doc ==null) return;
Node node = doc.getFirstChild();
NodeList lst = node.getChildNodes();
Node children = null ;
int len = lst.getLength();
for(int index=0;index<len;index++)
{
Node child = lst.item(index);
String name = child.getNodeName();
if(name=="Name")
name = child.getNodeValue();
else if(name=="Comment")
comment = child.getNodeValue());
else if(name=="children")
children = child;
}
if(children==null) return;
lst = children.getChildNodes();
len = lst.getLength();
Class<?> obj=null;
AbsModel model = null;
for(int index=0;index<len;index++)
{
Node childNode = lst.item(index);
String modelName = childNode.getNodeName();
try {
obj = Class.forName(modelName);
} catch (ClassNotFoundException e) {
e.printStackTrace();
}
if(obj!=null)
model = (AbsModel) obj.newInstance();
else
model = new GenericModel();
model.restoreDefaultPropFromXML(childNode);
addChild(model);
}
}
Is there a better way of parsing this XML.

Consider using JAXB, which is part of Java since version 6. You should be able to parse (“unmarshall”) your XML file into your own classes with almost no code, just adding a few annotations expliciting the mapping between your object structure and your XML structure.

StAX and or JAXB is almost always the way to go.
If the XML is really dynamic (like attributes specify the property name) ie <prop name="property" value="" /> then you will need to use StAX only or live with what JAXB will map it to (a POJO with name and value properties) and post process.
Personally I find combining StAX and JAXB the best. I parse to the elements I want and then use JAXB to turn the element into a POJO.
See Also:
My own utility library that will turn an XML Stream into an iterator of objects.
Parsing very large XML files and marshalling to Java Objects
http://tedone.typepad.com/blog/2011/06/unmarshalling-benchmark-in-java-jaxb-vs-stax-vs-woodstox.html

While JAXB may be the best choice I'd also like to mention jOOX which provides a JQuery-like API and makes working with XML documents really pleasant.

(de)serialize an object(Hashmap) from an XML of this form with XStream

XML:
JAVA Hashmap:
map = {key1=text1,key2=text2}
this doesn't work. why?
String xml = "<nodes><node id=\"key1\"><![CDATA[text1]]></node><node id="\key2\"><![CDATA[text2]]></node></nodes>";
XStream xs = new XStream();
xs.alias("nodes", Map.class);
xs.alias("node", String.class);
xs.useAttributeFor("id",String.class);
Map<String,String> map= (Map<String,String>) xs.fromXML(xml);
System.out.println(map);

If you can define your XML structure you should check the Map Converter and adjust your XML.
If not, you should write your own custom converter. You can see this thread to check an implementation similar to your needs.

ready made parser in java

i have some user defined tag. for example data here , jssj .I have a file(not xml) which contains some data embeded in tags.I need a parser for this which will identify my tags and will extract the data in proper format.
Eg
<newpage> thix text </newpage>
<tagD>
<tagA> kk</tagA>
</tagD>
tags can also have some attributes as simlar to html tags. Eg
<mytag height="f" width ="d" > bla bla bla </mytag>
<mytag attribute="val"> bla bla bla</mytag>

You could look at a parser generator like antlr.
Unless your tag syntax can be represented with a (simple) regular grammar (in which case you could try to scan the file with regexes), you will need a proper parser. It is actually not very hard to do at all - just the first time tastes like biting bullets...

You can use JAXB, already included in Java. It's quite simple.
First you need to create a binding to your XML code. The binding provides a map between Java objects and the XML code.
An example would be:
#XmlRootElement(name = "YourRootElement", namespace ="http://someurl.org")
#XmlAccessorType(XmlAccessType.FIELD)
#XmlType(name = "", propOrder = {
"intValue",
"stringArray",
"stringValue"}
)
public class YourBindingClass {
protected int intValue;
#XmlElement(nillable = false)
protected List<String> stringArray;
#XmlElement(name = "stringValue", required = true)
protected String stringValue;
public int getIntValue() {
return intValue;
}
public void setIntValue(int value) {
this.intValue = value;
}
public List<String> getStringArray() {
if (stringArray == null) {
stringArray = new ArrayList<String>();
}
return this.stringArray;
}
public String getStringValue() {
return stringValue;
}
public void setStringValue(String value) {
this.stringValue = value;
}
}
Then, to encode your Java objects into XML, you can use:
YourBindingClass yourBindingClass = ...;
JAXBContext jaxbContext = JAXBContext.newInstance(YourBindingClass.class);
Marshaller marshaller = jaxbContext.createMarshaller();
marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
marshaller.setProperty(Marshaller.JAXB_FRAGMENT, false);
/** If you need to specify a schema */
SchemaFactory sf = SchemaFactory.newInstance(XMLConstants.W3C_XML_SCHEMA_NS_URI);
Schema schema = sf.newSchema(new URL("http:\\www.someurl.org"));
marshaller.setSchema(schema);
marshaller.setProperty(Marshaller.JAXB_SCHEMA_LOCATION, true);
ByteArrayOutputStream stream = new ByteArrayOutputStream();
marshaller.marshal(yourBindingClass, stream);
System.out.println(stream);
To parse your XML back to objects:
InputStream resourceAsStream = ... // Your XML, File, etc.
JAXBContext jaxbContext = JAXBContext.newInstance(YourBindingClass.class);
Unmarshaller unmarshaller = jaxbContext.createUnmarshaller();
Object r = unmarshaller.unmarshal(resourceAsStream);
if (r instanceof YourBindingClass) ...
Example starting from a Java object:
YourBindingClass s = new YourBindingClass();
s.setIntValue(1);
s.setStringValue("a");
s.getStringArray().add("b1");
s.getStringArray().add("b2");
// marshal ...
Result:
<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<ns2:YourRootElement xmlns:ns2="http://someurl.org">
<intValue>1</intValue>
<stringArray>b1</stringArray>
<stringArray>b2</stringArray>
<stringValue>a</stringValue>
</ns2:YourRootElement>
If you don't know the input format, that means you probably don't have a XML schema. If you don't have a schema you don't have some it's benefits such as:
It is easier to describe allowable document content
It is easier to validate the correctness of data
It is easier to define data facets (restrictions on data)
It is easier to define data patterns (data formats)
It is easier to convert data between different data types
Anyway, the previous code also works with XML code that contains 'unknown' tags. However your XML code still have to present the required fields and follow the declared patterns.
So the following XML code is also valid. The only restriction is: the tag 'stringValue' should be there. Note that 'stringArrayQ' was not previously declared.
<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<ns2:YourRootElement xmlns:ns2="http://someurl.org">
<stringValue>a</stringValue>
<stringArrayQ>b1</stringArrayQ>
</ns2:YourRootElement>

Are these XML tags? If so, look into one of the many Java XML libraries already available. If they're some kind of custom tagging format, then you're just going to have to write it yourself.

For xml tags - use DOM parser or SAX parser.

You example is XML with this modification:
<root>
<newpage> thix text </newpage>
<tagD>
<tagA> kk</tagA>
</tagD>
</root>
You can use any XML parser you want to parse it.
Edit:
Attributes are a normal part of XML.
<root>
<newpage> thix text </newpage>
<tagD>
<tagA> kk</tagA>
</tagD>
<mytag height="f" width ="d" > bla bla bla </mytag>
<mytag attribute="val"> bla bla bla</mytag>
</root>
Every XML parser can deal with them.
Edit:
If you were able to use Python, you could do something like this:
import lxml.etree
doc = lxml.etree.parse("foo.xml")
print doc.xpath("//mytag[1]/#width")
# => ['d']
That's what i call simple.

NamespaceContext and using namespaces with XPath

Resolving an xpath that includes namespaces in Java appears to require the use of a NamespaceContext object, mapping prefixes to namespace urls and vice versa. However, I can find no mechanism for getting a NamespaceContext other than implementing it myself. This seems counter-intuitive.
The question: Is there any easy way to acquire a NamespaceContext from a document, or to create one, or failing that, to forgo prefixes altogether and specify the xpath with fully qualified names?

It is possible to get a NamespaceContext instance without writing your own class. Its class-use page shows you can get one using the javax.xml.stream package.
String ctxtTemplate = "<data xmlns=\"http://base\" xmlns:foo=\"http://foo\" />";
NamespaceContext nsContext = null;
XMLInputFactory factory = XMLInputFactory.newInstance();
XMLEventReader evtReader = factory
.createXMLEventReader(new StringReader(ctxtTemplate));
while (evtReader.hasNext()) {
XMLEvent event = evtReader.nextEvent();
if (event.isStartElement()) {
nsContext = ((StartElement) event)
.getNamespaceContext();
break;
}
}
System.out.println(nsContext.getNamespaceURI(""));
System.out.println(nsContext.getNamespaceURI("foo"));
System.out.println(nsContext
.getNamespaceURI(XMLConstants.XMLNS_ATTRIBUTE));
System.out.println(nsContext
.getNamespaceURI(XMLConstants.XML_NS_PREFIX));
Forgoing prefixes altogether is likely to lead to ambiguous expressions - if you want to drop namespace prefixes, you'd need to change the document format. Creating a context from a document doesn't necessarily make sense. The prefixes have to match the ones used in the XPath expression, not the ones in any document, as in this code:
String xml = "<data xmlns=\"http://base\" xmlns:foo=\"http://foo\" >"
+ "<foo:value>"
+ "hello"
+ "</foo:value>"
+ "</data>";
String expression = "/stack:data/overflow:value";
class BaseFooContext implements NamespaceContext {
#Override
public String getNamespaceURI(String prefix) {
if ("stack".equals(prefix))
return "http://base";
if ("overflow".equals(prefix))
return "http://foo";
throw new IllegalArgumentException(prefix);
}
#Override
public String getPrefix(String namespaceURI) {
throw new UnsupportedOperationException();
}
#Override
public Iterator<String> getPrefixes(
String namespaceURI) {
throw new UnsupportedOperationException();
}
}
XPathFactory factory = XPathFactory.newInstance();
XPath xpath = factory.newXPath();
xpath.setNamespaceContext(new BaseFooContext());
String value = xpath.evaluate(expression,
new InputSource(new StringReader(xml)));
System.out.println(value);
Neither the implementation returned by the StAX API nor the one above implement the full class/method contracts as defined in the doc. You can get a full, map-based implementation here.

I've just been working through using xpath and NamespaceContexts myself. I came across a good treatment of the issue on developerworks.

I found a convenient implementation in "Apache WebServices Common Utilities" called NamespaceContextImpl.
You can use the following maven dependency to obtain this class:
<dependency>
<groupId>org.apache.ws.commons</groupId>
<artifactId>ws-commons-util</artifactId>
<version>1.0.1</version>
</dependency>
I've use it in the following manner (I know its built for sax, but after reading the code, its o.k):
NamespaceContextImpl nsContext = new NamespaceContextImpl();
nsContext.startPrefixMapping("foo", "my.name.space.com");
You don't need to called endPrefixMapping.

If you are using the Spring framework you can reuse their NamespaceContext implementation
org.springframework.util.xml.SimpleNamespaceContext
This is a similar answer like the one from Asaf Mesika. So it doesn't give you automatic a NamespaceContext based on your document. You have to construct it yourself. Still it helps you because it at least gives you an implementation to starts with.
When we faced a similar problem, Both the spring SimpleNamespaceContext and the "Apache WebServices Common Utilities" worked. We wanted to avoid to the addition jar dependency on Apache WebServices Common Utilities and used the Spring one, because our application is Spring based.

If you are using Jersey 2 and only have a default XML namespace (xmlns="..."), you can use SimpleNamespaceResolver:
<?xml version="1.0" encoding="UTF-8"?>
<Outer xmlns="http://host/namespace">
<Inner />
</Outer>
DocumentBuilderFactory docBuilderFactory = DocumentBuilderFactory.newInstance();
docBuilderFactory.setNamespaceAware(true);
DocumentBuilder docBuilder = docBuilderFactory.newDocumentBuilder();
Document document = docBuilder.parse(new File("document.xml"));
String query = "/t:Outer/t:Inner";
XPath xpath = XPathFactory.newInstance().newXPath();
String xmlns = document.getDocumentElement().getAttribute("xmlns");
xpath.setNamespaceContext(new SimpleNamespaceResolver("t", xmlns));
NodeList nodeList = (NodeList) xpath.evaluate(query, document, XPathConstants.NODESET);
//nodeList will contain the <Inner> element
You can also specify xmlns manually if you want.