如何使用 Jersey API 从 RESTful Web 服务发送和接收 JSON 数据

2022-01-21 00:00:00 json java jersey 
@Path("/hello")
public class Hello {

    @POST
    @Path("{id}")
    @Produces(MediaType.APPLICATION_JSON)
    @Consumes(MediaType.APPLICATION_JSON)
    public JSONObject sayPlainTextHello(@PathParam("id")JSONObject inputJsonObj) {

        String input = (String) inputJsonObj.get("input");
        String output="The input you sent is :"+input;
        JSONObject outputJsonObj = new JSONObject();
        outputJsonObj.put("output", output);

        return outputJsonObj;
    }
}

这是我的网络服务(我正在使用 Jersey API).但是我想不出一种从java rest客户端调用这个方法来发送和接收json数据的方法.我尝试了以下方式写客户端

This is my webservice (I am using Jersey API). But I could not figure out a way to call this method from a java rest client to send and receive the json data. I tried the following way to write the client

ClientConfig config = new DefaultClientConfig();
Client client = Client.create(config);
WebResource service = client.resource(getBaseURI());
JSONObject inputJsonObj = new JSONObject();
inputJsonObj.put("input", "Value");
System.out.println(service.path("rest").path("hello").accept(MediaType.APPLICATION_JSON).entity(inputJsonObj).post(JSONObject.class,JSONObject.class));

但这显示以下错误

Exception in thread "main" com.sun.jersey.api.client.ClientHandlerException: com.sun.jersey.api.client.ClientHandlerException: A message body writer for Java type, class java.lang.Class, and MIME media type, application/octet-stream, was not found

推荐答案

您对@PathParam 的使用不正确.它不遵循 javadoc here.我相信您只想发布 JSON 实体.您可以在资源方法中修复此问题以接受 JSON 实体.

Your use of @PathParam is incorrect. It does not follow these requirements as documented in the javadoc here. I believe you just want to POST the JSON entity. You can fix this in your resource method to accept JSON entity.

@Path("/hello")
public class Hello {

  @POST
  @Produces(MediaType.APPLICATION_JSON)
  @Consumes(MediaType.APPLICATION_JSON)
  public JSONObject sayPlainTextHello(JSONObject inputJsonObj) throws Exception {

    String input = (String) inputJsonObj.get("input");
    String output = "The input you sent is :" + input;
    JSONObject outputJsonObj = new JSONObject();
    outputJsonObj.put("output", output);

    return outputJsonObj;
  }
}

而且，您的客户端代码应如下所示:

And, your client code should look like this:

  ClientConfig config = new DefaultClientConfig();
  Client client = Client.create(config);
  client.addFilter(new LoggingFilter());
  WebResource service = client.resource(getBaseURI());
  JSONObject inputJsonObj = new JSONObject();
  inputJsonObj.put("input", "Value");
  System.out.println(service.path("rest").path("hello").accept(MediaType.APPLICATION_JSON).post(JSONObject.class, inputJsonObj));

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