本文介绍了返回 JSONArray 而不是 JSONObject,Jersey JAX-RS的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我正在使用 Jersey 使我的一些服务成为 RESTful.
I am using Jersey to make some of my services RESTful.
我的 REST 服务调用返回给我
My REST service call returns me
{"param1":"value1", "param2":"value2",...."paramN":"valueN"}
但是,我希望它返回
["param1":"value1", "param2":"value2",...."paramN":"valueN"]
我需要对下面的代码进行哪些更改?
What are the changes I need to make in the code below?
@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
Part of my web.xml file looks like this
<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!
推荐答案
你可以定义你的服务方法如下,使用Person POJO:
You can define your service method as follows, using Person POJO:
@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 类:
@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;
}
}
}
结果将如下所示:
[{"fn":"1","ln":"2"}, {"fn":"3","ln":"4"}, {"fn":"5","ln":"6"}]
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