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jersey获取json参数,该如何解决

热度:287   发布时间:2016-04-17 10:16:46.0
jersey获取json参数
谁给个demo啊?
我这样写获取不到客户端的json数据

前台请求:
JScript code
$.ajax( {        url : '/Test/services/update',        type : 'post',        data : '{"1":"1","2":"2"}',        dataType : 'json',        contentType:'application/json',        async : false,        success : function(data) {            alert("success");            alert(data.name);            alert(data.age);        },        error : function() {            alert("ajax error");        }    });


后台的jsrsey处理,如何获取前台的json数据:data : '{"1":"1","2":"2"}',

Java code
@POST    @Path("/update")    @Produces("application/json")        //@Produces(MediaType.APPLICATION_JSON)    public Person update(HashMap map) {        System.out.println("aaaaaaaaaaa");        System.out.println(map.get("1"));        System.out.println(map.get("2"));//这里取不到,而且还报错                //        Map<String,Person> aa = new HashMap<String,Person>();//        aa.put("1", new Person("aaa","123"));//        aa.put("2", new Person("BBB","456"));        System.out.println("bbbbbbbbbbb");        //return  new GenericEntity<Map<String, Integer>> (aa){};        return new Person("test","test");    }



报错:SEVERE: A message body reader for Java type, class java.util.HashMap, and MIME media type, application/json;charset=UTF-8, was not found

415 Unsupported Media Type
61ms

------解决方案--------------------
Java code
@GET    @Path("/requestTest")    @Produces("application/json")    public Person update(@Context HttpServletRequest request) {        System.out.println("request:" + request.getParameter("a11"));        System.out.println("request:" + request.getParameter("a22"));        System.out.println("request:" + request.getQueryString());        return new Person("testPerson", "33");    }
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