Question

I have a XML and XSD file corresponds to it. I have just started to learn Spring Framework and I use Spring 3. I should write a code that takes that XML file and assigns it to an object at Java. I searched about it but how can I do it with using Spring (maybe some useful tricks or anything else?)

Answer 1

I recently used Spring OXM & JAXB for that. The class is org.springframework.oxm.jaxb.Jaxb2Marshaller. You can, ofcourse, use any other implementation of org.springframework.oxm.Unmarshaller.

But first You’ll need to generate the objects based on Your XSD. For that I used maven-jaxb2-plugin.

<plugin>
    <groupId>org.jvnet.jaxb2.maven2</groupId>
    <artifactId>maven-jaxb2-plugin</artifactId>
    <executions>
        <execution>
            <id>generate-oxm</id>
            <goals>
                <goal>generate</goal>
            </goals>
            <configuration>
                <schemaDirectory>src/main/resources/META-INF/xsd</schemaDirectory>
                <generatePackage>com.stackoverflow.xjc</generatePackage>
            </configuration>
        </execution>
    </executions>
</plugin>

And then configure the marshaller:

@Configuration
public class ApplicationConfiguration {

    @Autowired
    private ResourcePatternResolver resourceResolver;

    @Bean
    public Jaxb2Marshaller oxmMarshaller() throws IOException {
        Jaxb2Marshaller marshaller = new Jaxb2Marshaller();
        marshaller.setContextPath("com.stackoverflow.xjc");
        marshaller.setSchemas(resourceResolver.getResources("classpath:/META-INF/xsd/*.xsd"));
        return marshaller;
    }
}

Than just:

File xmlFile = new File("my.xml");
Source source = new StreamSource(new FileInputStream(xmlFile));
JAXBElement<MyXmlRootElemClass> result = oxmMarshaller.unmarshal(source);
MyXmlRootElemClass theObject = result.getValue();