how to parse xml to java object? [closed]

Answer 1

For performing Unmarshall using JAXB:

1) Convert given XML to XSD(by yourself or by online convertor),

2) Create a JAXB project in eclipse,

3) Create XSD file and paste that converted XSD content in it,

4) Right click on **XSD file--> Generate--> JAXB Classes-->follow the instructions(this will create all nessasary .java files in src, i.e., one package-info, object factory and pojo class),

5) Create another .java file in src to operate unmarshall operation, and run it.

Happy Coding !!

Answer 2

I find jackson fasterxml is one good choice to serializing/deserializing bean with XML.

Refer: How to use spring to marshal and unmarshal xml?

Answer 3

JAXB is an ideal solution. But you do not necessarily need xsd and xjc for that. More often than not you don't have an xsd but you know what your xml is. Simply analyze your xml, e.g.,

<customer id="100">
    <age>29</age>
    <name>mkyong</name>
</customer>

Create necessary model class(es):

@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;
    }

}

Try to unmarshal:

JAXBContext jaxbContext = JAXBContext.newInstance(Customer.class);
Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();
Customer customer = (Customer) jaxbUnmarshaller.unmarshal(new File("C:\\file.xml"));

Check results, fix bugs!

Answer 4

One thing that is really important to understand considering you have an XML file as :

<customer id="100">
    <Age>29</Age>
    <NAME>mkyong</NAME>
</customer>

I am sorry to inform you but :

@XmlElement
public void setAge(int age) {
    this.age = age;
}

will not help you, as it tries to look for "age" instead of "Age" element name from the XML.

I encourage you to manually specify the element name matching the one in the XML file :

@XmlElement(name="Age")
public void setAge(int age) {
    this.age = age;
}

And if you have for example :

@XmlRootElement
@XmlAccessorType (XmlAccessType.FIELD)
public class Customer {
...

It means it will use java beans by default, and at this time if you specify that you must not set another

@XmlElement(name="NAME")

annotation above a setter method for an element <NAME>..</NAME> it will fail saying that there cannot be two elements on one single variables.

I hope that it helps.

Answer 5

JAXB is a reliable choice as it does xml to java classes mapping smoothely. But there are other frameworks available, here is one such:

https://code.google.com/p/xmappr/