Seems like a basic question but I can't find this anywhere. Basically I've got a list of XML links like so: (all in one string)
I already have the "string" var which contains all the XML. Just extracting the HTML strings.
<?xml version="1.0" encoding="UTF-8"?>
<fql_query_response xmlns="http://api.facebook.com/1.0/" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" list="true">
<photo>
<src_small>http://photos-a.ak.fbcdn.net/hphotos-ak-ash4/486603_10151153207000351_1200565882_t.jpg</src_small>
</photo>
<photo>
<src_small>http://photos-c.ak.fbcdn.net/hphotos-ak-ash3/578919_10150988289678715_1110488833_t.jpg</src_small>
</photo>
I want to convert these into a arrayList, so something like URLArray[0] would be the first address as a string.
Can anyone tell me how to do this thanks?
DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
DocumentBuilder builder = factory.newDocumentBuilder();
InputSource is = new InputSource( new StringReader( xmlString) );
Document doc = builder.parse( is );
XPathFactory factory = XPathFactory.newInstance();
XPath xpath = factory.newXPath();
xpath.setNamespaceContext(new PersonalNamespaceContext());
XPathExpression expr = xpath.compile("//src_small/text()");
Object result = expr.evaluate(doc, XPathConstants.NODESET);
NodeList nodes = (NodeList) result;
List<String> urls = new ArrayList<String>();
for (int i = 0; i < nodes.getLength(); i++) {
urls.add (nodes.item(i).getNodeValue());
System.out.println(nodes.item(i).getNodeValue());
}
You are right, there should be some other resources out there that can help you. Maybe your searches just do not use the right keywords.
You basically have 2 choices:
Use an XML processing library. SAX, DOM, XPATH, & xmlreader are some keywords you can use to find some.
Just ignore the fact that your string is xml and perform normal string operations on it. splits, iterate through it, regular expressions, ect...
Yes for that you have to perform XML Parsing.
then store that in ArrayList.
ex:
ArrayList<String> aList = new ArrayList<String>();
aList.add("your string");
Related
The XML file is as :
Xml File
Code I have written:
List queryXmlUsingXpathAndReturnList(String xml, String xpathExpression) {
DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance()
DocumentBuilder dBuilder = dbFactory.newDocumentBuilder()
Document doc = dBuilder.parse(new ByteArrayInputStream(xml.getBytes(StandardCharsets.UTF_8)))
doc.getDocumentElement().normalize()
XPath xPath = XPathFactory.newInstance().newXPath()
NodeList nodeList = (NodeList) xPath.compile(xpathExpression).evaluate(doc, XPathConstants.NODESET)
List returnElements = new ArrayList<>()
nodeList.each { n ->
returnElements.add(n.getTextContent())
}
When I am passing the xpath as:
/Envelope/Body/CommandResponseData/OperationResult/Operation/ParameterList/ListParameter/StringElement
It returns all the values.
But I want to return only the ListParameter values whose name="PackageTypeList".
For that I am using the xpath as:
/Envelope/Body/CommandResponseData/OperationResult/Operation/ParameterList/ListParameter[#name='PackageTypeList']/StringElement
But it returns list as null.
I guess you miss "CommandResult" between "CommandResponseData" and "OperationResult" in your XPath-Expression.
I have an XML with the following structure.
<message>
<header>
</header>
<body>
</body>
<end>
</end>
</message>
Each header,body and end nodes contain fields that i need to extract into separate hash maps. What is the best way to go about this without using external libraries? The end result is to display a two-column view of the entire message. (field name, value)
You can use DocumentBuilderFactory and DocumentBuilder that comes along with java Api.
For Example, refer link.
It depends on the structure of your data and your hashmap: what is the key, what if the value.
Nevertheless, DOM and XPATH do the job:
String xml= // your xml
DocumentBuilderFactory builderFactory =DocumentBuilderFactory.newInstance();
DocumentBuilder builder = builderFactory.newDocumentBuilder();
Document document = builder.parse(new InputSource(new StringReader(xml)));
String expression="//header"; // Same for body, ...
XPathExpression expr = xpath.compile(expression) ;
NodeList nodes = (NodeList) expr.evaluate(document, XPathConstants.NODESET);
for (int k = 0; k < nodes.getLength(); k++) {
// Do what you want with that
hope it helps
I have such XML structure, when I use NodeList nList = doc.getElementsByTagName("stock"); it return me 3 stocks, 2 main stock tags and one which is under substocks. I want to get only two stock which is on upper level and ignore all which is under substock tags.
Is it possible in Java to make something like LINQ query in C#, say return me elements only where name is equals to "Sony".
Thanks!
<city>
<stock>
<name>Sony</name>
</stock>
<stock>
<name>Panasonic</name>
<substocks>
<stock>
<name>Panasonic Shop 2</name>
</stock>
</substocks>
</stock>
</city>
I recommend you to use XPath with javax.xml.xpath package:
final InputStream is = new FileInputStream('your.xml');
final DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
final DocumentBuilder builder = factory.newDocumentBuilder();
final Document doc = builder.parse(is);
final XPathFactory xPathfactory = XPathFactory.newInstance();
final XPath xpath = xPathfactory.newXPath();
final XPathExpression expr = xpath.compile("/city/stock/name[text()='Sony']");
and then:
final NodeList nl = (NodeList) expr.evaluate(doc, XPathConstants.NODESET);
Take a look on XPath and its java implementation JXPath. Other possible approach is parsing XML using JAXB and operating objects list using LambdaJ.
There is also dom4j library which has powerful navigation with XPath:
import org.dom4j.Document;
import org.dom4j.io.SAXReader;
SAXReader reader = new SAXReader();
Document document = reader.read("test.xml");
List list = document.selectNodes("/city/stock/name[text()='Sony']");
for (Iterator iter = list.iterator(); iter.hasNext(); ) {
// TODO: place you logic here
}
More examples are here
Try jcabi-xml (see this blog post) with a one-liner:
Collection<XML> found = new XMLDocument("your document here").nodes(
"/city/stock/name[text()='Sony']"
);
I have an XML document, and an XPath expression for that doc. I have to update the doc by using XPath at runtime.
How can I do this using Java?
The below is my xml:
<?xml version="1.0" encoding="ISO-8859-1"?>
<PersonList>
<Person>
<Name>Sonu Kapoor</Name>
<Age>24</Age>
<Gender>M</Gender>
<PostalCode>54879</PostalCode>
</Person>
<Person>
<Name>Jasmin</Name>
<Age>28</Age>
<Gender>F</Gender>
<PostalCode>78745</PostalCode>
</Person>
<Person>
<Name>Josef</Name>
<Age>232</Age>
<Gender>F</Gender>
<PostalCode>53454</PostalCode>
</Person>
</PersonList>
I have to change the values of name and age under //PersonList/Person[2]/Name.
Use setNodeValue. First, get a NodeList, for example:
myNodeList = (NodeList) xpath.compile("//MyXPath/text()")
.evaluate(myXmlDoc, XPathConstants.NODESET);
Then set the value of e.g. the first node:
myNodeList.item(0).setNodeValue("Hi mom!");
More examples e.g. here.
As mentioned in two other answers here, as well as in your previous question: technically, XPath is not a way to "update" an XML document, but only to locate nodes within an XML document. But I presume the above is what you want.
EDIT: Responding to your comment... Are you asking how to write your DOM to an XML file after you've finished editing the DOM? If so, here are two examples of how to do it:
http://www.java2s.com/Code/Java/XML/WriteDOMout.htm
http://download.oracle.com/javaee/1.4/tutorial/doc/JAXPXSLT4.html
You can delete the file and create a new one.
Document doc = DocumentBuilderFactory.newInstance().newDocumentBuilder().parse(
new InputSource("data.xml"));
XPath xpath = XPathFactory.newInstance().newXPath();
NodeList nodes = (NodeList) xpath.evaluate("//employee/name[text()='old']", doc,
XPathConstants.NODESET);
for (int idx = 0; idx < nodes.getLength(); idx++) {
nodes.item(idx).setTextContent("new value");
}
Transformer xformer = TransformerFactory.newInstance().newTransformer();
xformer.transform(new DOMSource(doc), new StreamResult(new File("data_new.xml")));
XPath is used to select parts of an XML document.It has no provision for updating. But since it returns DOM objects (Elements, if memory serves, or maybe Nodes) you can then use DOM methods for altering the document.
XPath can be used to select nodes in a document, not for modification
You apply the xpath expression to your document and get an element (in your case). Once you have this Element, you can use the Element methods to change values (name and age in your case)
Starting from a NodeList it should work like that:
NodeList nodes = getNodeListFromXPathExpression(); // you know how
if (nodes.length == 0)
return; // empty nodelist, xpath didn't select anything
Node first = node.getItem(0); // take the first from the list, your element
// this is a shortcut for your example:
// first is the actual selected element (a node)
// .getFirst() returns the first child node, the "text node" (="Jasmine", ="28")
// .setNodeValue() replace the actual value of that text node with a new string
first.getFirstChild().setNodeValue("New Name or new age");
Consider using XQuery Update instead of XPath. This allows you to write
replace value of node //PersonList/Person[2]/Name with "Anonymous"
This is much easier than using the Java DOM API.
I've created a small project for using XPATH to create/update XML:
https://github.com/shenghai/xmodifier
the code to change your xml is like:
Document document = readDocument("personList.xml");
XModifier modifier = new XModifier(document);
modifier.addModify("//PersonList/Person[2]/Name", "newName");
modifier.modify();
This is a super cool function where you can able to modify any tag value for any XML document using its xpath. You need to pass three arguments xml,xpathExpression and newValue and it returns the XML file as String with modified value.
If you want to pass XML as file, you need to change the function accordingly. But the logic will be same.
public String updateXML(String xml, String xpathExpression, String newValue)
{
try {
//Creating document builder
DocumentBuilderFactory builderFactory = DocumentBuilderFactory.newInstance();
DocumentBuilder builder = builderFactory.newDocumentBuilder();
Document document = builder.parse(new org.xml.sax.InputSource(new StringReader(xml)));
//Evaluating xpath expression using Element
XPath xpath = XPathFactory.newInstance().newXPath();
Element element = (Element)xpath.evaluate(xpathExpression, document, XPathConstants.NODE);
//Setting value in the text
element.setTextContent(value);
//Transformation of document to xml
StringWriter stringWriter = new StringWriter();
Transformer xformer = TransformerFactory.newInstance().newTransformer();
xformer.transform(new DOMSource(document), new StreamResult(stringWriter));
xml = stringWriter.toString();
}
catch (Exception e)
{
e.printStackTrace();
}
return xml;
}
Here is the code to change the content with vtd-xml... vtd-xml is unique in that it is the only API that offers incremental update capability.
import com.ximpleware.*;
import java.io.*;
public class changeName {
public static void main(String s[]) throws VTDException,java.io.UnsupportedEncodingException,java.io.IOException{
VTDGen vg = new VTDGen();
if (!vg.parseFile("input.xml", false))
return;
VTDNav vn = vg.getNav();
AutoPilot ap = new AutoPilot(vn);
XMLModifier xm = new XMLModifier(vn);
ap.selectXPath("//PersonList/Person[2]");
int i=0;
while((i=ap.evalXPath())!=-1){
if (vn.toElement(VTDNav.FIRST_CHILD,"Name")){
int k=vn.getText();
if (i!=-1)
xm.updateToken(k, "Jonathan");
vn.toElement(VTDNav.PARENT);
}
if (vn.toElement(VTDNav.FIRST_CHILD,"Age")){
int k=vn.getText();
if (i!=-1)
xm.updateToken(k, "42");
vn.toElement(VTDNav.PARENT);
}
}
xm.output("new.xml");
}
}
I have small Strings with XML, like:
String myxml = "<resp><status>good</status><msg>hi</msg></resp>";
which I want to query to get their content.
What would be the simplest way to do this?
XPath using Java 1.5 and above, without external dependencies:
String xml = "<resp><status>good</status><msg>hi</msg></resp>";
XPathFactory xpathFactory = XPathFactory.newInstance();
XPath xpath = xpathFactory.newXPath();
InputSource source = new InputSource(new StringReader(xml));
String status = xpath.evaluate("/resp/status", source);
System.out.println("satus=" + status);
Using dom4j, similar to McDowell's solution:
String myxml = "<resp><status>good</status><msg>hi</msg></resp>";
Document document = new SAXReader().read(new StringReader(myxml));
String status = document.valueOf("/resp/msg");
System.out.println("status = " + status);
XML handling is a bit simpler using dom4j. And several other comparable XML libraries exist. Alternatives to dom4j are discussed here.
Here is example of how to do that with XOM:
String myxml = "<resp><status>good</status><msg>hi</msg></resp>";
Document document = new Builder().build(myxml, "test.xml");
Nodes nodes = document.query("/resp/status");
System.out.println(nodes.get(0).getValue());
I like XOM more than dom4j for its simplicity and correctness. XOM won't let you create invalid XML even if you want to ;-) (e.g. with illegal characters in character data)
You could try JXPath
After your done with simple ways to query XML in java. Look at XOM.
#The comments of this answer:
You can create a method to make it look simpler
String xml = "<resp><status>good</status><msg>hi</msg></resp>";
System.out.printf("satus= %s\n", getValue("/resp/status", xml ) );
The implementation:
public String getValue( String path, String xml ) {
return XPathFactory
.newInstance()
.newXPath()
.evaluate( path , new InputSource(
new StringReader(xml)));
}
convert this string into a DOM object and visit the nodes:
Document dom= DocumentBuilderFactory().newDocumentBuilder().parse(new InputSource(new StringReader(myxml)));
Element root= dom.getDocumentElement();
for(Node n=root.getFirstChild();n!=null;n=n.getNextSibling())
{
System.err.prinlnt("Current node is:"+n);
}
Here is a code snippet of querying your XML with VTD-XML
import com.ximpleware.*;
public class simpleQuery {
public static void main(String[] s) throws Exception{
String myXML="<resp><status>good</status><msg>hi</msg></resp>";
VTDGen vg = new VTDGen();
vg.setDoc(myXML.getBytes());
vg.parse(false);
VTDNav vn = vg.getNav();
AutoPilot ap = new AutoPilot(vn);
ap.selectXPath("/resp/status");
int i = ap.evalXPath();
if (i!=-1)
System.out.println(" result ==>"+vn.toString(i));
}
}
You can use Jerry to query XML similar to jQuery.
jerry(myxml).$("status")