在RestFul-Webservice(泽西岛)上下文中,我需要将对象图封送/序列化为XML和JSON。为简单起见,我尝试用 2-3 个类来解释这个问题:
人.java
@XmlRootElement
@XmlAccessorType(XmlAccessType.FIELD)
public class Person {
private String name;
// @XmlIDREF
@XmlElement(name = "house")
@XmlElementWrapper(name = "houses")
private Collection<House> houses;
public Person() {}
public Person(String name, Collection<House> houses) {
this.name = name;
this.houses = houses;
}
}
房子.java
@XmlAccessorType(XmlAccessType.FIELD)
public class House {
// @XmlID
public String name;
public String location;
public House() {}
public House(String name, String location) {
this.name = name;
this.location = location;
}
}
现在,当我序列化一个人时,XML 将如下所示:
<people>
<person>
<name>Edward</name>
<houses>
<house>
<name>MyAppartment</name>
<location>London</location>
</house>
<house>
<name>MySecondAppartment</name>
<location>London</location>
</house>
</houses>
</person>
<person>
<name>Thomas</name>
<houses>
<house>
<name>MyAppartment</name>
<location>London</location>
</house>
<house>
<name>MySecondAppartment</name>
<location>London</location>
</house>
</houses>
</person>
</people>
这里的问题是,相同的房屋被多次列出。现在,我添加了未注释的XmlIDREF
和XmlID
注释,这将导致类似于以下内容的XML:
<people>
<person>
<name>Edward</name>
<houses>
<house>MyAppartment</house>
<house>MySecondAppartment</house>
</houses>
</person>
<person>
<name>Thomas</name>
<houses>
<house>MyAppartment</house>
<house>MySecondAppartment</house>
</houses>
</person>
</people>
虽然第一个XML太冗长,但这个XML缺乏信息。如何创建(和取消封送)类似于以下内容的内容:
<people>
<person>
<name>Edward</name>
<houses>
<house>MyAppartment</house>
<house>MySecondAppartment</house>
</houses>
</person>
<person>
<name>Thomas</name>
<houses>
<house>MyAppartment</house>
<house>MySecondAppartment</house>
</houses>
</person>
<houses>
<house>
<name>MyAppartment</name>
<location>London</location>
</house>
<house>
<name>MySecondAppartment</name>
<location>London</location>
</house>
</houses>
</people>
解决方案应该是通用的,因为我不想为对象图中的每个新元素编写额外的类。
为了完整起见,这里是 restful 网络服务:
@Path("rest/persons")
public class TestService {
@GET
@Produces({ MediaType.TEXT_XML, MediaType.APPLICATION_JSON })
public Collection<Person> test() throws Exception {
Collection<Person> persons = new ArrayList<Person>();
Collection<House> houses = new HashSet<House>();
houses.add(new House("MyAppartment", "London"));
houses.add(new House("MySecondAppartment", "London"));
persons.add(new Person("Thomas", houses));
persons.add(new Person("Edward", houses));
return persons;
}
}
提前谢谢。
如果您尝试序列化为与您给出的最后一个 XML 示例匹配的格式,那么我相信您的对象图结构不正确,无法实现这一目标。
如果要提供Person
对象的集合及其关联的房屋,并提供House
对象的集合,则需要返回包含这两个集合的序列化 XML 消息。 看起来好像您的@XmlIDREF
和@XmlID
注释位于正确的位置,以便按预期进行 Person-House 关联(根据您的描述),但您只返回Person
对象的集合,而不是返回两个集合。
您的 Web 服务应该看起来更像这样(省略序列化,因为您很清楚如何序列化它):
@Path("rest/persons")
public class TestService {
@GET
@Produces({ MediaType.TEXT_XML, MediaType.APPLICATION_JSON })
public Map<String, Object> test() throws Exception {
Map<String, Object> peopleAndHouses = new HashMap<String, Object>();
Collection<Person> persons = new ArrayList<Person>();
Collection<House> houses = new HashSet<House>();
houses.add(new House("MyAppartment", "London"));
houses.add(new House("MySecondAppartment", "London"));
persons.add(new Person("Thomas", houses));
persons.add(new Person("Edward", houses));
peopleAndHouses.put("houses", houses);
peopleAndHouses.put("people", persons);
return peopleAndHouses;
}
}
还有其他方法可以实现这一点(例如,创建一个包装器对象,该对象具有人员和房屋的集合属性等),但希望您能理解这个想法。