在RestFul-Webservice(Jersey)上下文中,我需要将Object图形编组/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:
Person.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;
}
}
House.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Person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,多次列出相同的房屋。现在我添加了未注释的XmlIDREF
和XmlID
注释,这将导致XML与此类似:
<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,但这个缺乏信息。我如何创建(和解组)类似于:
的东西<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,这是宁静的网络服务:
@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;
}
}
提前致谢。
答案 0 :(得分:2)
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形结构不正确无法实现。
如果要提供Person
个对象及其相关房屋的集合,并提供House
个对象的集合,则需要返回包含两个集合的序列化XML消息。看起来好像您在正确的地方有@XmlIDREF
和@XmlID
注释,可以根据您的描述制作Person-House关联,但您只返回{{1对象而不是返回两个集合。
您的网络服务应该看起来更像这样(省略序列化,因为您似乎清楚如何序列化它):
Person
还有其他方法可以实现这一目标(例如,创建一个包含人和房屋集合属性的包装器对象等),但希望你能得到这个想法。