There are of course a lot of good solutions based on what you need. If it is just configuration, you should have a look at Jakarta commons-configuration and commons-digester.
You could always use the standard JDK method of getting a document :
import java.io.File;
import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import org.w3c.dom.Document;
[...]
File file = new File("some/path");
DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
DocumentBuilder db = dbf.newDocumentBuilder();
Document document = db.parse(file);
Answer from Guillaume on Stack OverflowThere are of course a lot of good solutions based on what you need. If it is just configuration, you should have a look at Jakarta commons-configuration and commons-digester.
You could always use the standard JDK method of getting a document :
import java.io.File;
import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import org.w3c.dom.Document;
[...]
File file = new File("some/path");
DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
DocumentBuilder db = dbf.newDocumentBuilder();
Document document = db.parse(file);
XML Code:
<?xml version="1.0"?>
<company>
<staff id="1001">
<firstname>yong</firstname>
<lastname>mook kim</lastname>
<nickname>mkyong</nickname>
<salary>100000</salary>
</staff>
<staff id="2001">
<firstname>low</firstname>
<lastname>yin fong</lastname>
<nickname>fong fong</nickname>
<salary>200000</salary>
</staff>
</company>
Java Code:
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.parsers.DocumentBuilder;
import org.w3c.dom.Document;
import org.w3c.dom.NodeList;
import org.w3c.dom.Node;
import org.w3c.dom.Element;
import java.io.File;
public class ReadXMLFile {
public static void main(String argv[]) {
try {
File fXmlFile = new File("/Users/mkyong/staff.xml");
DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
Document doc = dBuilder.parse(fXmlFile);
doc.getDocumentElement().normalize();
System.out.println("Root element :" + doc.getDocumentElement().getNodeName());
NodeList nList = doc.getElementsByTagName("staff");
System.out.println("----------------------------");
for (int temp = 0; temp < nList.getLength(); temp++) {
Node nNode = nList.item(temp);
System.out.println("\nCurrent Element :" + nNode.getNodeName());
if (nNode.getNodeType() == Node.ELEMENT_NODE) {
Element eElement = (Element) nNode;
System.out.println("Staff id : "
+ eElement.getAttribute("id"));
System.out.println("First Name : "
+ eElement.getElementsByTagName("firstname")
.item(0).getTextContent());
System.out.println("Last Name : "
+ eElement.getElementsByTagName("lastname")
.item(0).getTextContent());
System.out.println("Nick Name : "
+ eElement.getElementsByTagName("nickname")
.item(0).getTextContent());
System.out.println("Salary : "
+ eElement.getElementsByTagName("salary")
.item(0).getTextContent());
}
}
} catch (Exception e) {
e.printStackTrace();
}
}
}
Output:
----------------
Root element :company
----------------------------
Current Element :staff
Staff id : 1001
First Name : yong
Last Name : mook kim
Nick Name : mkyong
Salary : 100000
Current Element :staff
Staff id : 2001
First Name : low
Last Name : yin fong
Nick Name : fong fong
Salary : 200000
I recommended you reading this: Normalization in DOM parsing with java - how does it work?
Example source.
How to read XML file using java - Stack Overflow
Java - Reading XML file - Stack Overflow
How do i parse XML files in java ?
What's the simplest, most concise way to load an XML file from a remote server?
Videos
One of the possible implementations:
File file = new File("userdata.xml");
DocumentBuilderFactory documentBuilderFactory = DocumentBuilderFactory
.newInstance();
DocumentBuilder documentBuilder = documentBuilderFactory.newDocumentBuilder();
Document document = documentBuilder.parse(file);
String usr = document.getElementsByTagName("user").item(0).getTextContent();
String pwd = document.getElementsByTagName("password").item(0).getTextContent();
when used with the XML content:
<credentials>
<user>testusr</user>
<password>testpwd</password>
</credentials>
results in "testusr" and "testpwd" getting assigned to the usr and pwd references above.
Reading xml the easy way:
http://www.mkyong.com/java/jaxb-hello-world-example/
package com.mkyong.core;
import javax.xml.bind.annotation.XmlAttribute;
import javax.xml.bind.annotation.XmlElement;
import javax.xml.bind.annotation.XmlRootElement;
@XmlRootElement
public class Customer {
String name;
int age;
int id;
public String getName() {
return name;
}
@XmlElement
public void setName(String name) {
this.name = name;
}
public int getAge() {
return age;
}
@XmlElement
public void setAge(int age) {
this.age = age;
}
public int getId() {
return id;
}
@XmlAttribute
public void setId(int id) {
this.id = id;
}
}
.
package com.mkyong.core;
import java.io.File;
import javax.xml.bind.JAXBContext;
import javax.xml.bind.JAXBException;
import javax.xml.bind.Marshaller;
public class JAXBExample {
public static void main(String[] args) {
Customer customer = new Customer();
customer.setId(100);
customer.setName("mkyong");
customer.setAge(29);
try {
File file = new File("C:\\file.xml");
JAXBContext jaxbContext = JAXBContext.newInstance(Customer.class);
Marshaller jaxbMarshaller = jaxbContext.createMarshaller();
// output pretty printed
jaxbMarshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
jaxbMarshaller.marshal(customer, file);
jaxbMarshaller.marshal(customer, System.out);
} catch (JAXBException e) {
e.printStackTrace();
}
}
}