JAXB解组具有相同名称的多个节点

I have an XML element with nodes of the same name that I need to unmarshal. I need service in a List or Array. Here is my XML:

我有一个XML元素,其中包含我需要解组的同名节点。我需要List或Array中的服务。这是我的XML:

<provider name="foo">
    <service active="true" name="alias" timeout="N/A">value1</service>
    <service active="true" name="caption" timeout="N/A">value2</service>
    <service active="true" name="expect_manifest_file" timeout="15m">value3</service>
    <service active="true" name="expect_SD_and_HD_ADI" timeout="15m">value4</service>
</provider>

I only know how to get the last element in the XML. How do I make a list of all <service> nodes and can the list include the active and name attribute? Here is my object:

我只知道如何获取XML中的最后一个元素。如何列出所有节点,列表中是否包含active和name属性?这是我的对象:

@XmlRootElement (name="provider")
public class Customer {
    String service;

    public String getService() {
        return service;
    }
    @XmlElement(name="service")
    public void setService(String service) {
        this.service = service;
    }
}

1 个解决方案

#1

I'd make another approach according to your needs, maybe this is what you want, let me know.

我会根据你的需要采取另一种方法,也许这就是你想要的,让我知道。

First I will create Service class which will handle the information of your Service node:

首先,我将创建Service类,它将处理您的Service节点的信息:

@XmlRootElement(name = "service")
@XmlAccessorType (XmlAccessType.FIELD)

public class Service {

private boolean active;
private String name;

public boolean isActive() {
    return active;
}

public void setActive(boolean active) {
    this.active = active;
}

public String getName() {
    return name;
}

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

}

Then I will create the class "Customer" which is needed to hold the services you are including in your XML:

然后我将创建“Customer”类,这是保存XML中包含的服务所需的:

@XmlRootElement(name = "provider")
@XmlAccessorType (XmlAccessType.FIELD)

public class Customer {

@XmlElementWrapper
@XmlElement(name = "service")
private List<Service> services = null;

public List<Service> getServices() {
    return services;
}

public void setServices(List<Service> services) {
    this.services = services;
}

}

I prepared some methods in a main class to make sure this is working:

我在主类中准备了一些方法以确保它正常工作:

public class JaxbHandler {

public void marshal(String path, Customer customer) throws JAXBException {
    JAXBContext jaxbContext = JAXBContext.newInstance(Customer.class);
    Marshaller jaxbMarshaller = jaxbContext.createMarshaller();

    jaxbMarshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);

    jaxbMarshaller.marshal(customer, System.out);
    jaxbMarshaller.marshal(customer, new File(path));
}

public Customer unmarshal(String path) throws JAXBException {
    Customer cust = null;
    JAXBContext jaxbContext = JAXBContext.newInstance(Customer.class);
    Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();

    cust = (Customer) jaxbUnmarshaller.unmarshal(new File(path));

    return cust;
}

public static void main(String[] args) throws JAXBException {
    String path = "c:/dev/sample.xml";

    JaxbHandler sampleHandler = new JaxbHandler();

    List<Service> myServices = new ArrayList<Service>();

    Service s1 = new Service();
    s1.setActive(true);
    s1.setName("Service A");

    Service s2 = new Service();
    s2.setActive(true);
    s2.setName("Service B");

    myServices.add(s1);
    myServices.add(s2);

    Customer cust = new Customer();
    cust.setServices(myServices);

    sampleHandler.marshal(path, cust);

    Customer backAgain = sampleHandler.unmarshal(path);



}

}

Please let me know if this works for you. Happy coding :)

如果这对你有用,请告诉我。快乐编码:)

#1