返回JSONArray而不是JSONObject,Jersey JAX-RS

Question

我正在使用Jersey来使我的一些服务RESTful.

我的REST服务调用返回给我

{"param1":"value1", "param2":"value2",...."paramN":"valueN"}

但是,我希望它能够回归

["param1":"value1", "param2":"value2",...."paramN":"valueN"]

我需要在下面的代码中做出哪些更改？

@GET
@Produces(MediaType.APPLICATION_JSON)
public List<com.abc.def.rest.model.SimplePojo> getSomeList() {
    /* 
            Do something
    */
    return listOfPojos;
}

我的web.xml文件的一部分看起来像这样

    <servlet>
        <servlet-name>Abc Jersey REST Service</servlet-name>
        <servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
        <init-param>
            <param-name>com.sun.jersey.config.property.packages</param-name>
            <param-value>com.abc.def.rest</param-value>
        </init-param>
        <load-on-startup>0</load-on-startup>
    </servlet>

谢谢!

Answer 1

您可以使用Person POJO定义您的服务方法,如下所示:

@GET
@Produces("application/json")
@Path("/list")
public String getList(){
    List<Person> persons = new ArrayList<>();
    persons.add(new Person("1", "2"));
    persons.add(new Person("3", "4"));
    persons.add(new Person("5", "6"));
    // takes advantage to toString() implementation to format as [a, b, c]
    return persons.toString();
}

POJO课程:

@XmlRootElement
public class Person {
    @XmlElement(name="fn")
    String fn;

    @XmlElement(name="ln")
    String ln;

    public Person(){        
    }

    public Person(String fn, String ln) {
        this.fn = fn;
        this.ln = ln;
    }    

    @Override
    public String toString(){
        try {
            // takes advantage of toString() implementation to format {"a":"b"}
            return new JSONObject().put("fn", fn).put("ln", ln).toString();
        } catch (JSONException e) {
            return null;
        }
    }
}

结果如下:

[{"fn":"1","ln":"2"}, {"fn":"3","ln":"4"}, {"fn":"5","ln":"6"}]