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How do I change the XML root element name in JAXB?

By Wayan in XML Last modified: June 5, 2023 0 Comment

In the following code snippet you will learn how to change the default root element name of the XML generated by the JAXB API. By default, the name of the class is use as the root element name. To change the root element name we can use the name property of the @XmlRootElement annotation. In the Customer model below we change the root element name into cust.

package org.kodejava.xml;

import org.kodejava.xml.support.Address;

import javax.xml.bind.annotation.XmlElement;
import javax.xml.bind.annotation.XmlRootElement;
import javax.xml.bind.annotation.XmlType;

@XmlRootElement(name = "cust")
@XmlType(propOrder = {"id", "name", "address"})
public class Customer {
    private Integer id;
    private String name;
    private Address address;

    @XmlElement
    public Integer getId() {
        return id;
    }

    public void setId(Integer id) {
        this.id = id;
    }

    @XmlElement
    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    @XmlElement
    public Address getAddress() {
        return address;
    }

    public void setAddress(Address address) {
        this.address = address;
    }
}

package org.kodejava.xml.support;

public class Address {
    private String street;
    private String city;
    private String province;
    private String zipcode;
    private String country;

    // Getters & Setters
}

package org.kodejava.xml;

import org.kodejava.xml.support.Address;

import javax.xml.bind.JAXBContext;
import javax.xml.bind.JAXBException;
import javax.xml.bind.Marshaller;

public class JAXBCustomRootElement {
    public static void main(String[] args) {
        Customer customer = new Customer();
        customer.setId(1);
        customer.setName("Johnny Mnemonic");

        Address address = new Address();
        address.setStreet("Sunset Road");
        address.setCity("Denpasar");
        address.setProvince("Bali");
        address.setZipcode("80225");
        address.setCountry("Indonesia");
        customer.setAddress(address);

        try {
            JAXBContext context = JAXBContext.newInstance(Customer.class);
            Marshaller marshaller = context.createMarshaller();
            marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, Boolean.TRUE);
            marshaller.marshal(customer, System.out);
        } catch (JAXBException e) {
            e.printStackTrace();
        }
    }
}

When we convert this POJO to XML using JAXB API we will get the following result:

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<cust>
    <id>1</id>
    <name>Johnny Mnemonic</name>
    <address>
        <city>Denpasar</city>
        <country>Indonesia</country>
        <province>Bali</province>
        <street>Sunset Road</street>
        <zipcode>80225</zipcode>
    </address>
</cust>

Maven Dependencies

<dependencies>
    <dependency>
        <groupId>javax.xml.bind</groupId>
        <artifactId>jaxb-api</artifactId>
        <version>2.3.1</version>
    </dependency>    
    <dependency>
        <groupId>com.sun.xml.bind</groupId>
        <artifactId>jaxb-ri</artifactId>
        <version>2.3.5</version>
        <type>pom</type>
    </dependency>
</dependencies>

How to convert object to XML using JAXB?

By Wayan in XML Last modified: June 5, 2023 1 Comment

JAXB, Java Architecture for XML Binding, it uses JAXB annotations to convert POJO to or from XML file. In this example you will learn how to convert an object / POJO into an XML document using JAXB. The process of converting an object into XML also known as marshalling. In this snippet our POJO is a class called Track as you can see below.

The first thing that we need to do is to annotate the Track POJO with a couple of JAXB annotations. For this example we add the @XmlRootElement, @XmlElement and @XmlAttribute annotations.

package org.kodejava.xml.support;

import javax.xml.bind.annotation.XmlAttribute;
import javax.xml.bind.annotation.XmlElement;
import javax.xml.bind.annotation.XmlRootElement;

@XmlRootElement
public class Track {
    private Integer id;
    private String title;

    public Track() {
    }

    public Integer getId() {
        return id;
    }

    @XmlAttribute
    public void setId(Integer id) {
        this.id = id;
    }

    @XmlElement
    public String getTitle() {
        return title;
    }

    public void setTitle(String title) {
        this.title = title;
    }

    @Override
    public String toString() {
        return "Track{" +
                "id=" + id +
                ", title='" + title + '\'' +
                '}';
    }
}

After we have the Track class annotated we can use the code snippet below to convert the POJO into XML. The step to take is to start by creating an instance of JAXBContext. From this context object we create the Marshaller object. To convert it to XML we called the marshall() method. The method have many overloads, in this case we just want to print the XML to the System.out.

package org.kodejava.xml;

import org.kodejava.xml.support.Track;

import javax.xml.bind.JAXBContext;
import javax.xml.bind.JAXBException;
import javax.xml.bind.Marshaller;

public class JAXBObjectToXml {
    public static void main(String[] args) {
        Track track = new Track();
        track.setId(1);
        track.setTitle("Hey Jude");

        try {
            JAXBContext context = JAXBContext.newInstance(Track.class);

            Marshaller marshaller = context.createMarshaller();
            marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
            marshaller.marshal(track, System.out);
        } catch (JAXBException e) {
            e.printStackTrace();
        }
    }
}

If you run this code you will get the following output:

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<track id="1">
    <title>Hey Jude</title>
</track>

Maven Dependencies

<dependencies>
    <dependency>
        <groupId>javax.xml.bind</groupId>
        <artifactId>jaxb-api</artifactId>
        <version>2.3.1</version>
    </dependency>
    <dependency>
        <groupId>com.sun.xml.bind</groupId>
        <artifactId>jaxb-ri</artifactId>
        <version>2.3.5</version>
        <type>pom</type>
    </dependency>
</dependencies>

How to create an XML file of a POJO using JAXB?

By Wayan in XML Last modified: June 5, 2023 0 Comment

The code snippet below show you how to convert POJO into XML file using JAXB. To do this we can pass the output file where we want the XML to be saved to the marshaller object.

package org.kodejava.xml;

import org.kodejava.xml.support.Track;

import javax.xml.bind.JAXBContext;
import javax.xml.bind.JAXBException;
import javax.xml.bind.Marshaller;
import java.io.File;

public class JAXBObjectToXmlFile {
    public static void main(String[] args) {
        Track track = new Track();
        track.setId(2);
        track.setTitle("She Loves You");

        try {
            JAXBContext context = JAXBContext.newInstance(Track.class);

            Marshaller marshaller = context.createMarshaller();
            marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);

            File output = new File("Track.xml");
            marshaller.marshal(track, output);
        } catch (JAXBException e) {
            e.printStackTrace();
        }
    }
}

This snippet will create a file called Track.xml with the following content:

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<track id="2">
    <title>She Loves You</title>
</track>

package org.kodejava.xml.support;

import javax.xml.bind.annotation.XmlAttribute;
import javax.xml.bind.annotation.XmlElement;
import javax.xml.bind.annotation.XmlRootElement;

@XmlRootElement
public class Track {
    private Integer id;
    private String title;

    public Track() {
    }

    public Integer getId() {
        return id;
    }

    @XmlAttribute
    public void setId(Integer id) {
        this.id = id;
    }

    @XmlElement
    public String getTitle() {
        return title;
    }

    public void setTitle(String title) {
        this.title = title;
    }

    @Override
    public String toString() {
        return "Track{" +
                "id=" + id +
                ", title='" + title + '\'' +
                '}';
    }
}

Maven Dependencies

<dependencies>
    <dependency>
        <groupId>javax.xml.bind</groupId>
        <artifactId>jaxb-api</artifactId>
        <version>2.3.1</version>
    </dependency>
    <dependency>
        <groupId>com.sun.xml.bind</groupId>
        <artifactId>jaxb-ri</artifactId>
        <version>2.3.5</version>
        <type>pom</type>
    </dependency>
</dependencies>

How do I get attributes of element during SAX parsing?

By Wayan in XML Last modified: June 2, 2023 0 Comment

This example show you how to get the attributes of elements in an XML file using the SAX parser.

package org.kodejava.xml;

import org.xml.sax.Attributes;
import org.xml.sax.SAXException;
import org.xml.sax.helpers.DefaultHandler;

import javax.xml.parsers.SAXParser;
import javax.xml.parsers.SAXParserFactory;
import java.io.InputStream;

public class SAXElementAttribute {
    public static void main(String[] args) {
        SAXElementAttribute demo = new SAXElementAttribute();
        demo.run();
    }

    private void run() {
        try {
            // Create SAXParserFactory instance and a SAXParser
            SAXParserFactory factory = SAXParserFactory.newInstance();
            SAXParser parser = factory.newSAXParser();

            // Get an InputStream to the elements.xml file and parse
            // its contents using the SAXHandler.
            InputStream is =
                    getClass().getResourceAsStream("/elements.xml");
            DefaultHandler handler = new SAXHandler();
            parser.parse(is, handler);
        } catch (Exception e) {
            e.printStackTrace();
        }
    }

    static class SAXHandler extends DefaultHandler {
        @Override
        public void startElement(String uri, String localName,
                                 String qName, Attributes attributes)
                throws SAXException {

            int attributeLength = attributes.getLength();
            if ("person".equals(qName)) {
                for (int i = 0; i < attributeLength; i++) {
                    // Get attribute names and values
                    String attrName = attributes.getQName(i);
                    String attrVal = attributes.getValue(i);
                    System.out.print(attrName + " = " + attrVal + "; ");
                }
                System.out.println();
            }
        }
    }
}

The elements.xml file is as follows:

<?xml version="1.0" encoding="UTF-8"?>
<root>
    <persons>
        <person name="Foo" age="25" />
        <person name="Bar" age="22" />
    </persons>
</root>

How do I handle error when parsing an XML file using SAX?

By Wayan in XML Last modified: June 2, 2023 0 Comment

The ErrorHandler interface implemented by the org.xml.sax.helpers.DefaultHandler class provides some methods for error handling mechanism in SAX parsing. The methods are warning(), error(), and fatalError().

package org.kodejava.xml;

import org.xml.sax.SAXException;
import org.xml.sax.SAXParseException;
import org.xml.sax.helpers.DefaultHandler;

import javax.xml.parsers.SAXParser;
import javax.xml.parsers.SAXParserFactory;
import java.io.InputStream;

public class SAXErrorHandlerDemo {
    public static void main(String[] args) throws Exception {
        SAXErrorHandlerDemo demo = new SAXErrorHandlerDemo();
        demo.run();
    }

    public void run() throws Exception {
        // Creates the SAXParserFactory and SAXParser instance.
        SAXParserFactory factory = SAXParserFactory.newInstance();
        SAXParser parser = factory.newSAXParser();

        // Parse the error.xml file using MySAXHandler as the 
        // DefaultHandler implementation.
        InputStream is = getClass().getResourceAsStream("/error.xml");
        DefaultHandler handler = new MySAXHandler();
        parser.parse(is, handler);
    }

    // Override the error handling methods defined by the ErrorHandler
    // interface. This method will handler exceptions thrown by the
    // parsing process.
    static class MySAXHandler extends DefaultHandler {
        @Override
        public void warning(SAXParseException e) throws SAXException {
            System.out.println("warning   : " + e.getMessage());
        }

        @Override
        public void error(SAXParseException e) throws SAXException {
            System.out.println("error     : " + e.getMessage());
        }

        @Override
        public void fatalError(SAXParseException e) throws SAXException {
            System.out.println("fatalError: " + e.getMessage());
        }
    }
}