We Keep Coding

Retrieve XML Element names with Java from unknown message format

I am parsing XML from lots of JMS messaging topics, so the structure of each message varies a lot and I'd like to make one general tool to parse them all.
To start, all I want to do is get the element names:
<gui-action>
<action>some action</action>
<params>
<param1>blue</param1>
<param2>tall</param2>
<params>
</gui-action>
I just want to retrieve the strings "gui-action", "action", "params", "param1", and "param2." Duplicates are just fine.
I've tried using org.w3c.dom.Node, Element, NodeLists and I'm not having much luck. I keep getting the element values, not the names.
private Element root;
private Document doc;
private NodeList nl;
//messageStr is passed in elsewhere in the code
//but is a string of the full XML message.
doc = xmlParse( messageStr );
root = doc.getDocumentElement();
nl = root.getChildNodes();
int size = nl.getLength();
for (int i=0; i<size; i++) {
log.info( nl.item(i).getNodeName() );
}
public Document xmlParse( String xml ){
DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
DocumentBuilder db;
InputSource is;
try {
//Using factory get an instance of document builder
db = dbf.newDocumentBuilder();
is = new InputSource(new StringReader( xml ) );
doc = db.parse( is );
} catch(ParserConfigurationException pce) {
pce.printStackTrace();
} catch(SAXException se) {
se.printStackTrace();
} catch(IOException ioe) {
ioe.printStackTrace();
}
return doc;
//parse using builder to get DOM representation of the XML file
}
My logged "parsed" XML looks like this:
#text
action
#text
params
#text

Figured it out. I was iterating over only the child nodes, and not including the parent. So now I just filter out the #texts, and include the parent. Derp.
log.info(root.getNodeName() );
for (int i=0; i<size; i++) {
nodeName = nl.item(i).getNodeName();
if( nodeName != "#text" ) {
log.info( nodeName );
}
}
Now if anyone knows a way to get a NodeList of the entire document, that would be awesome.

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);

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

Cannot run Xpath queries from JAVA in XML files having <DOCTYPE> tag

I have made the following method which runs hard-coded xPath queries in a hard-coded XML file. The method works perfect with one exception. Some xml files contains the following tag
<!DOCTYPE WorkFlowDefinition SYSTEM "wfdef4.dtd">
When i try to run a query in that file i get the following exception:
java.io.FileNotFoundException:
C:\ProgramFiles\code\other\xPath\wfdef4.dtd(The system cannot find the file specified).
The question is : What can i do to instruct my program not to take under consideration this DTD file?
I have also noted that the path C:\ProgramFiles\code\other\xPath\wfdef4.dtd is the one i run my application from and not the one that the actual xml file is located.
Thank you in advace
Here is my method:
public String evaluate(String expression,File file){
XPathFactory factory = XPathFactory.newInstance();
xPath = XPathFactory.newInstance().newXPath();
StringBuffer strBuffer = new StringBuffer();
try{
InputSource inputSource = new InputSource(new FileInputStream(file));
//evaluates the expression
NodeList nodeList = (NodeList)xPath.evaluate(expression,
inputSource,XPathConstants.NODESET);
//does other stuff, irrelevant with my question.
for (int i = 0 ; i <nodeList.getLength(); i++){
strBuffer.append(nodeList.item(i).getTextContent());
}
}catch (Exception e) {
e.printStackTrace();
}
return strBuffer.toString();
}

And the answer is :
xPath = XPathFactory.newInstance().newXPath();
DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
//add this line to ignore dth DTD
dbf.setFeature("http://apache.org/xml/features/nonvalidating/load-external-dtd", false);

A question on how to solve the string problem in Java

I've created a simple xml file here:
http://roberthan.host56.com/productsNew.xml
which is quite simple, the root node is [products] while all other element nodes are [product]. Under each [product] node, there are two child nodes, [code] and [name], so it basically looks like:
[product]
[code]ddd[/code]
[name]ssss[/name]
[/product]
I've also written up the following Java code to parse this XML file and take out the text content of the [product] node, and add it to a JComboBox.
docBuilder = docFactory.newDocumentBuilder();
doc = docBuilder.parse("http://roberthan.host56.com/productsNew.xml");
NodeList productNodes = doc.getElementsByTagName("product");
productlist.clear();
for (i = 0; i < productNodes.getLength(); i++)
{
Node childNode = productNodes.item(i);
if (childNode.hasChildNodes()) {
NodeList nl = childNode.getChildNodes();
Node nameNode = nl.item(2);
productlist.add(nameNode.getTextContent());
}
}
final JComboBox productComboB = new JComboBox();
Iterator iterator = productlist.iterator();
while(iterator.hasNext())
{
productComboB.addItem(iterator.next().toString());
}
The code is quite straightforward, I firstly parse the xml and get all the product nodes and put them into a nodelist, and the productList is an arrayList. I loop through the all the [product] nodes, for each of them, if it has child nodes, then I take the second child node (which is the [name] node) and put the text content of it in the array list, and finally, I loop through the arrayList and add each item to the combo box.
The problem I got is, if I select the [code] child node, which means "Node nameNode = nl.item(1)", it will work perfectly; however, if I change that item(1) to item(2) to extract all the [name] nodes, the combo box will have a drop down list, but all the items are blank, like I have inserted 10 empty strings.
Also, if I try to add a "Hello World" string into the combo box after the above code, the "Hello World" item will appear after the 10 empty items.
I have spent the whole afternoon debugging this but still no breakthrough, the XML is actually quite simple and the Java is straightforward too. Could anyone share some thoughts with me on this please. Thanks a lot!

It is because the node list contains text nodes also.
If you add the following snippet to your code you will find that
for(int j = 0;j<nl.getLength();j++){
System.out.println(nl.item(j).getNodeName());
}
It will give the following output for each iteration of the product
#text
code
#text
name
#text
This means you have to get the 3rd element to get the name node.
Node nameNode = nl.item(3);
But I'll suggest you to use XPath to solve this problem.
NodeList nodelist = XPathAPI.selectNodeList(doc, "//products/product/name");
for (int i = 0; i < nodelist.getLength(); i++) {
productlist.add(nodelist.item(i).getTextContent());
}

XPath using this expression will easily solve your problem:
String XPATH_EXPRESSION1 = "//name/text()";
e.g.,
public static final String PRODUCTS_NEW = "http://roberthan.host56.com/productsNew.xml";
public static final String XPATH_EXPRESSION1 = "//name/text()";
public XmlFun() {
URL productsUrl;
try {
productsUrl = new URL(PRODUCTS_NEW);
List<String> nameList = xPathExtract(productsUrl.openStream());
} catch (MalformedURLException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
} catch (ParserConfigurationException e) {
e.printStackTrace();
} catch (SAXException e) {
e.printStackTrace();
} catch (XPathExpressionException e) {
e.printStackTrace();
}
}
private List<String> xPathExtract(InputStream inStream) throws ParserConfigurationException, SAXException, IOException, XPathExpressionException {
DocumentBuilderFactory domFactory = DocumentBuilderFactory.newInstance();
DocumentBuilder builder = domFactory.newDocumentBuilder();
Document domDoc = builder.parse(inStream);
XPathFactory xFactory = XPathFactory.newInstance();
XPath xpath = xFactory.newXPath();
XPathExpression xExpr = xpath.compile(XPATH_EXPRESSION1);
NodeList nodes = (NodeList)xExpr.evaluate(domDoc, XPathConstants.NODESET);
List<String> resultList = new ArrayList<String>();
for (int i = 0; i < nodes.getLength(); i++) {
String node = nodes.item(i).getNodeValue();
resultList.add(node);
}
return resultList;
}