JAX-RS使用JSON并使用javax.json读取它

Question

如何从post中使用json并使用javax.json读取它

如何在不添加更多依赖项或将json与实体bean关联的情况下完成此操作？

json (application/json)

{"employees":[
   {"firstName":"John", "lastName":"Doe"}, 
   {"firstName":"Anna", "lastName":"Smith"},
   {"firstName":"Peter", "lastName":"Jones"}
]}

服务器

    @POST
    @Consumes(MediaType.APPLICATION_JSON)
    @Produces("text/plain")
    public String receiveJSON( whatShouldIput data ){

      JsonReader reader = Json.createReader(data );
      JsonObject obj = reader.readObject();
      JsonArray results = obj.getJsonArray("data");
      for (JsonObject result : results.getValuesAs(JsonObject.class)) {
          System.out.print(result.getJsonObject("employees").getString("firstName"));
          System.out.print(" ");
          System.out.println(result.getString("lastName", ""));
          System.out.println("-----------");
     }

        return "ok";
    }

Answer 1

使用Apache CXF API可以很容易地做到这一点。您必须提供直接将json数据转换为对象的json提供程序。Ideally, we should not write marshalling or unmarshalling code service implementation.

http://cxf.apache.org/docs/jax-rs.html

或者，如果您想解封json数据，请使用ObjectMapper of jackson framework。

http://wiki.fasterxml.com/JacksonInFiveMinutes

Answer 2

"whatShouldIput“应该是"String”类型。还需要做一些其他的调整。下面是一个使用您的输入的示例程序。

    public static void main(String[] args) {
    String input = "{\"employees\":["
            + "{\"firstName\":\"John\", \"lastName\":\"Doe\"},"
            + "{\"firstName\":\"Anna\", \"lastName\":\"Smith\"},"
            + "{\"firstName\":\"Peter\", \"lastName\":\"Jones\"}" + "]}";
    System.out.println(receiveJSON(input));
}

public static String receiveJSON(String data) {

    try (StringReader sReader = new StringReader(data);
            JsonReader reader = Json.createReader(sReader)) {
        JsonObject obj = reader.readObject();
        JsonArray results = obj.getJsonArray("employees");
        for (JsonObject result : results.getValuesAs(JsonObject.class)) {
            System.out.print(result.getString("firstName"));
            System.out.print(" ");
            System.out.println(result.getString("lastName", ""));
            System.out.println("-----------");
        }
        return "ok";
    }
}

输出：

John Doe
-----------
Anna Smith
-----------
Peter Jones
-----------
ok

Answer 3

至少在泽西，只需将其声明为JsonObject：

public String receiveJSON( JsonObject data )

它会自动为你解析

还请参见https://jersey.java.net/documentation/latest/media.html#json.json-p来设置依赖项。