Develop Reference

How to make XML Parser aware of all Character Entity References?

I get arbitrary XML from a server and parse it using this Java code:
String xmlStr; // arbitrary XML input
DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
try {
DocumentBuilder builder = factory.newDocumentBuilder();
InputSource is = new InputSource(new StringReader(xmlStr));
return builder.parse(is);
}
catch (SAXException | IOException | ParserConfigurationException e) {
LOGGER.error("Failed to parse XML.", e);
}
Every once in a while, the XML input contains some unknown entity reference like and fails with an error, such as org.xml.sax.SAXParseException: The entity "nbsp" was referenced, but not declared.
I could solve this problem by preprocessing the original xmlStr and translating all problematic entity references before parsing. Here's a dummy implementation that works:
protected static String translateEntityReferences(String xml) {
String newXml = xml;
Map<String, String> entityRefs = new HashMap<>();
entityRefs.put(" ", " ");
entityRefs.put("«", "«");
entityRefs.put("»", "»");
// ... and 250 more...
for(Entry<String, String> er : entityRefs.entrySet()) {
newXml = newXml.replace(er.getKey(), er.getValue());
}
return newXml;
}
However, this is really unsatisfactory, because there are are a huge number of entity references which I don't want to all hard-code into my Java class.
Is there any easy way of teaching this entire list of character entity references to the DocumentBuilder?

If you can change your code to work with StAX instead of DOM, the trivial solution is to use the XMLInputFactory property IS_REPLACING_ENTITY_REFERENCES set to false.
public static void main(String[] args) throws Exception
{
String doc = "<doc> </doc>";
ByteArrayInputStream is = new ByteArrayInputStream(doc.getBytes());
XMLInputFactory xif = XMLInputFactory.newFactory();
xif.setProperty(javax.xml.stream.XMLInputFactory.IS_REPLACING_ENTITY_REFERENCES, false);
XMLStreamReader xr = xif.createXMLStreamReader(is);
while(xr.hasNext())
{
int t = xr.getEventType();
switch(t) {
case XMLEvent.ENTITY_REFERENCE:
System.out.println("Entity: "+ xr.getLocalName());
break;
case XMLEvent.START_DOCUMENT:
System.out.println("Start Document");
break;
case XMLEvent.START_ELEMENT:
System.out.println("Start Element: " + xr.getLocalName());
break;
case XMLEvent.END_DOCUMENT:
System.out.println("End Document");
break;
case XMLEvent.END_ELEMENT:
System.out.println("End Element: " + xr.getLocalName());
break;
default:
System.out.println("Other: ");
break;
}
xr.next();
}
}
Output:
Start Document
Start Element: doc
Entity: nbsp null
End Element: doc
But that may require too much rewrite in your code if you really need the full DOM tree in memory.
I spent an hour tracing through the DOM implementation and couldn't find any way to make the DOM parser read from an XMLStreamReader.
Also there is evidence in the code that the internal DOM parser implementation has an option similar to IS_REPLACING_ENTITY_REFERENCES but I couldn't find any way to set it from the outside.

How to get unique value of repeated nodes using dom parser

I have a XML having repeated nodes and I have top parse it using DOM parser. After a lot R&D I could find anything on internet which can help me. My xml looks like
<nos1>
<Name>aqwer</Name>
<class>sas</class>
<class>xcd</class>
<class>asd</class>
<Name>cfg</Name>
<Name>cfg</Name>
<nos1>
Any suggestion How can I parse this xml for repeated values.

You can use w3c dom document to parse your XML as follows:
DocumentBuilderFactory df = DocumentBuilderFactory.newInstance();
try
{
DocumentBuilder db = df.newDocumentBuilder();
InputStream is = new ByteArrayInputStream(response.getContent().getBytes("UTF-8"));
org.w3c.dom.Document doc = db.parse(is);
NodeList links = doc.getElementsByTagName("class");
for(int i=0; i< links.getLength(); i++)
{
Node link = links.item(i);
System.out.println(link.getTextContent());
}
}
catch(Exception ex)
{
}
Hope this helps you.

You should read all elements and after reading eliminate the duplicates via a Set. Here is an example using XMLBeam, but any other library will do.
public class TestMultipleElements {
#XBDocURL("resource://test.xml")
public interface Projection {
#XBRead("/nos1/Name")
List<String> getNames();
#XBRead("/nos1/class")
List<String> getClasses();
}
#Test
public void uniqueElements() throws IOException {
Projection projection = new XBProjector().io().fromURLAnnotation(Projection.class);
for (String name : new HashSet<String>(projection.getNames())) {
System.out.println("Found Name:" + name);
}
for (String clazz : new HashSet<String>(projection.getClasses())) {
System.out.println("Found Name:" + clazz);
}
}
}
This prints out:
Found Name:aqwer
Found Name:cfg
Found Name:xcd
Found Name:sas
Found Name:asd

How to convert xml node into string without changing order of attributes? [duplicate]

This question already has answers here:
Order of XML attributes after DOM processing
(12 answers)
Closed 8 years ago.
I am trying to convert a XML Node to String using the following code :
private String nodeToString(final Node node) {
final StringWriter stringWriter = new StringWriter();
try {
final Transformer transformer = TransformerFactory.newInstance().newTransformer();
transformer.setOutputProperty(OutputKeys.OMIT_XML_DECLARATION, "yes");
transformer.setOutputProperty(OutputKeys.INDENT, "no");
transformer.transform(new DOMSource(node), new StreamResult(stringWriter));
} catch (final TransformerException e) {
JOptionPane.showMessageDialog(this, e.getMessage(), "Error", JOptionPane.ERROR_MESSAGE);
}
return stringWriter.toString();
}
My problem is that it formats attributes of XML node in alphabetical orders. Is there any property I could apply to ignore formatting of Node attributes ?

The DOM API does not preserve attribute order:
NamedNodeMaps are not maintained in any particular order
If you have a Node then you have already lost any attribute ordering. Consider this XML:
<?xml version="1.0" encoding="UTF-8"?>
<!-- attrs.xml -->
<attrs
a="a"
z="z"
b="b"
m="m" />
There are no guarantees about the ordering of the output of this application:
import javax.xml.parsers.DocumentBuilderFactory;
import org.w3c.dom.*;
public class Attrs {
public static void main(String[] args) throws Exception {
NamedNodeMap attrs = DocumentBuilderFactory.newInstance()
.newDocumentBuilder()
.parse("attrs.xml")
.getElementsByTagName("attrs")
.item(0)
.getAttributes();
for (int i = 0; i < attrs.getLength(); i++) {
Attr attribute = (Attr) attrs.item(i);
System.out.println(attribute.getName() + "=" + attribute.getValue());
}
}
}
If they are alphabetical then that is only an implementation side-effect, not a requirement. If attribute order is significant to you then you are using the wrong tools.

I figure it out how to do this, I have read xml file, and read only specific node from that xml file as a string. And applied operations on string to match my conditions. By doing this obviously I cannot leverage the Parser API, but that fulfilled my requirements. Following is my code snippet:
/**
* #param in InputStream of xml file
*/
private String getNodeString(InputStream in) throws IOException {
String nodeString = "";
InputStreamReader is = new InputStreamReader(in);
StringBuilder sb = new StringBuilder();
BufferedReader br = new BufferedReader(is);
String read = br.readLine();
String fileData;
while (read != null) {
//System.out.println(read);
sb.append(read);
read = br.readLine();
}
fileData = sb.toString().trim();
// Start index of node
int start = fileData.indexOf("<" + mSignedNode);
// End index of node, next node name
int end = fileData.indexOf("</Configuration>");
nodeString = fileData.substring(start, end);
return nodeString.trim();
}
The method is quite dirty, but you can pass parameters to find start index and end index.
Hope this would help someone, rather just closing their question ;)

Parse some elements from a xml

i want to know if is possible to me to parse some atributes from a xml file, to be a object in java
I don´t wanna to create all fields that are in xml.
So, how can i do this?
For exemple below there is a xml file, and i want only the data inside the tag .
<emit>
<CNPJ>1109</CNPJ>
<xNome>OESTE</xNome>
<xFant>ABATEDOURO</xFant>
<enderEmit>
<xLgr>RODOVIA</xLgr>
<nro>S/N</nro>
<xCpl>402</xCpl>
<xBairro>GOMES</xBairro>
<cMun>314</cMun>
<xMun>MINAS</xMun>
<UF>MG</UF>
<CEP>35661470</CEP>
<cPais>58</cPais>
<xPais>Brasil</xPais>
<fone>03</fone>
</enderEmit>
<IE>20659</IE>
<CRT>3</CRT>

For Java XML parsing where you don't have the XSD and don't want to create a complete object graph to represent the XML, JDOM is a great tool. It allows you to easily walk the XML tree and pick the elements you are interested in.
Here's some sample code that uses JDOM to pick arbitrary values from the XML doc:
// reading can be done using any of the two 'DOM' or 'SAX' parser
// we have used saxBuilder object here
// please note that this saxBuilder is not internal sax from jdk
SAXBuilder saxBuilder = new SAXBuilder();
// obtain file object
File file = new File("/tmp/emit.xml");
try {
// converted file to document object
Document document = saxBuilder.build(file);
//You don't need this or the ns parameters in getChild()
//if your XML document has no namespace
Namespace ns = Namespace.getNamespace("http://www.example.com/namespace");
// get root node from xml. emit in your sample doc?
Element rootNode = document.getRootElement();
//getChild() assumes one and only one, enderEmit element. Use a lib and error
//checking as needed for your document
Element enderEmitElement = rootNode.getChild("enderEmit", ns);
//now we get two of the child from
Element xCplElement = enderEmitElement.getChild("xCpl", ns);
//should be 402 in your example
String xCplValue = xCplElement.getText();
System.out.println("xCpl: " + xCplValue);
Element cMunElement = enderEmitElement.getChild("cMun", ns);
//should be 314 in your example
String cMunValue = cMunElement.getText();
System.out.println("cMun: " + cMunValue);
} catch (JDOMException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}

You can use JAXB to unmarshal the xml into Java object, with which you can read selective elements easily. With JAXB, the given XML can be represented in Java as follows :
enderEmit element :
#XmlRootElement
public class EnderEmit{
private String xLgr;
//Other elements.Here you can define properties for only those elements that you want to load
}
emit element (This represents your XML file):
#XmlRootElement
public class Emit{
private String cnpj;
private String xnom;
private EnderEmit enderEmit;
..
//Add elements that you want to load
}
Now by using the below lines of code, you can read your xml to an object :
String filePath="filePath";
File file = new File(filePath);
JAXBContext jaxbContext = JAXBContext.newInstance(Emit.class);
jaxbUnmarshaller = jaxbContext.createUnmarshaller();
Emit emit = (Emit) jaxbUnmarshaller.unmarshal(file);
The line will give you an emit object for the given xml.

Try to use StringUtils.subStringBetween
try
{
String input = "";
br = new BufferedReader(new FileReader(FILEPATH));
String result = null;
while ((input = br.readLine()) != null) // here we read the file line by line
{
result = StringUtils.substringBetween(input, ">", "<"); // using StringUtils.subStringBetween to get the data what you want
if(result != null) // if the result should not be null because some of the line not having the tags
{
System.out.println(""+result);
}
}
}
catch (IOException e)
{
e.printStackTrace();
}
finally
{
try
{
if (br != null)
{
br.close();
}
}
catch (IOException ex)
{
ex.printStackTrace();
}
}

How to parse an XML that has an & in its tag name using XPath in Java?

I want to parse an XML whose tag contains an & for example: <xml><OC&C>12.4</OC&C></xml>. I tried to escape & to & but that didn't fix the issue for tag name (it fixes it for values only), currently my code is throwing an exception, see complete function below.
public static void main(String[] args) throws Exception
{
String xmlString = "<xml><OC&C>12.4</OC&C></xml>";
xmlString = xmlString.replaceAll("&", "&");
String path = "xml";
InputSource inputSource = new InputSource(new StringReader(xmlString));
try
{
Document xmlDocument = DocumentBuilderFactory.newInstance().newDocumentBuilder().parse(inputSource);
XPath xPath = XPathFactory.newInstance().newXPath();
XPathExpression xPathExpression = xPath.compile(path);
System.out.println("Compiled Successfully.");
}
catch (SAXException e)
{
System.out.println("Error while retrieving node Path:" + path + " from " + xmlString + ". Returning null");
}
}

Hmmm... I don't think that it is a legal XML name. I'd think about using a regex to replace OC&C to something legal first, and then parse it.

It's not "an XML". It's a non-XML. XML doesn't allow ampersands in names. Therefore, you can't parse it successfully using an XML parser.

xml could not be name of any XML element. So, your XML fragment could never be parsed anyway. Then you could try something like that.
<name><![CDATA[<OC&C>12.4</OC&C>]]></name>