如何在不手动转换为 JSON 的情况下使用 Jersey 客户端发布 Pojo?

2022-01-21 00:00:00 json client java jackson jersey

我有一个可用的 json 服务，如下所示:

I have a working json service which looks like this:

@POST
@Path("/{id}/query")
@Consumes(MediaType.APPLICATION_JSON)
@Produces(JSON)
public ListWrapper query(@Context SecurityContext sc, @PathParam("id") Integer projectId, Query searchQuery) {
    ...
    return result
}

查询对象看起来像这样，当发布该查询对象的 json 表示时，效果很好.

The query object looks like this and when posting a json representation of that Query object it works out nice. 

@XmlRootElement
public class Query {
    Integer id;
    String query;
    ... // Getters and Setters etc..
}

现在我想从客户端填充该对象并使用 Jersey 客户端将该 Query 对象发布到服务并获得 JSONObject 作为结果.我的理解是，它可以在不先将其转换为 json 对象然后作为字符串发布的情况下完成.

Now I want to fill that object from a client and use Jersey client to post that Query object to the service and get an JSONObject as a result. My understanding is that it could be done without converting it to a json object first and then posted as a String.

我尝试过这样的事情，但我想我错过了一些东西.

I have tried something like this but I think I miss something.

public static JSONObject query(Query searchQuery){
    String url = baseUrl + "project/"+searchQuery.getProjectId() +"/query";
    WebResource webResource = client.resource(url);
    webResource.entity(searchQuery, MediaType.APPLICATION_JSON_TYPE);
    JSONObject response = webResource.post(JSONObject.class);
    return response;
}

我使用的是 Jersey 1.12.

I'm using Jersey 1.12.

任何正确方向的帮助或指示将不胜感激.

Any help or pointer in the right direction would be much appreciated.

推荐答案

如果您的 Web 服务生成 JSON，您必须在客户端使用 accept() 方法处理它:

If your web-service produces a JSON you must handle that in your client by using an accept() method:

ClientResponse response = webResource.accept(MediaType.APPLICATION_JSON).post(searchQuery, MediaType.APPLICATION_JSON);
ListWrapper listWrapper = response.getEntity(ListWrapper.class);

试试这个并给出你的结果.

Try this and give your results.

相关文章