Java 如何使用 Jersey 客户端 POST 方法提交数据

How to submit data with Jersey client POST method

提问by Li'

I am new to Jersey. I need to implement a Jersey client to submit data with POST method. The curl command is:

我是泽西岛的新手。我需要实现一个 Jersey 客户端来使用 POST 方法提交数据。curl 命令是：

curl -d '{"switch": "00:00:00:00:00:00:00:01", "name":"flow-mod-1", "priority":"32768", "ingress-port":"1","active":"true", "actions":"output=2"}' http://localhost:8080/wm/staticflowentrypusher/json

So I am trying to figure out how to use Jersey client to implement the above curl command.

所以我想弄清楚如何使用 Jersey 客户端来实现上面的 curl 命令。

So far I have done:

到目前为止，我已经做了：

public class FLClient {

private static Client client;
private static WebResource webResource;
private static String baseuri = "http://localhost:8080/wm/staticflowentrypusher/json";
private static ClientResponse response;
private static String output = null;

public static void main(String[] args) {
    try {

        client = Client.create();

        webResource = client.resource(baseuri);

                    // implement POST data 

    } catch (Exception e) {
        e.printStackTrace();
    }
}

}

Can someone help me with it?

有人可以帮我吗？

采纳答案by Li'

Now I figure it out. Here is my solution:

现在我想通了。这是我的解决方案：

public static void main(String[] args) {

    try {
        Client client = Client.create();

        WebResource webResource = client.resource(baseuri);

        String input = "{\"switch\": \"00:00:00:00:00:00:00:01\", "
                + "\"name\":\"flow-mod-1\", \"priority\":\"32768\", "
                + "\"ingress-port\":\"1\",\"active\":\"true\", "
                + "\"actions\":\"output=2\"}";

        // POST method
        ClientResponse response = webResource.accept("application/json")
                .type("application/json").post(ClientResponse.class, input);

        // check response status code
        if (response.getStatus() != 200) {
            throw new RuntimeException("Failed : HTTP error code : "
                    + response.getStatus());
        }

        // display response
        String output = response.getEntity(String.class);
        System.out.println("Output from Server .... ");
        System.out.println(output + "\n");
    } catch (Exception e) {
        e.printStackTrace();
    }

}

回答by Willy

If you want to post within a JSON body, here is a better approach.

如果您想在 JSON 正文中发布，这里有一个更好的方法。

ClientConfig clientConfig = new DefaultClientConfig();              
clientConfig.getFeatures().put(JSONConfiguration.FEATURE_POJO_MAPPING, Boolean.TRUE);     
client = Client.create(clientConfig);

WebResource webResource = client.resource(baseuri);

Map<String,Object> postBody = new HashMap<String,Object>();
//put switch, name,priority....
ClientResponse response = webResource.accept("application/json")
                .type("application/json").post(ClientResponse.class, postBody);

Remember you have to include jersey-json

记住你必须包括 jersey-json

回答by Ohad Eytan

For future users, with the new version of jerseythings has changed, so the way of doing such POST method is:

对于未来的用户来说，随着新版本的jersey事情发生了变化，所以做这样的POST方法的方式是：

• For versions 1.x(@Li'answer) :

  WebResource webResource = client.resource(baseuri);
  
  String input = "...";
  
  ClientResponse response = webResource.accept("application/json")
          .type("application/json").post(ClientResponse.class, input);
  

• For versions 2.x:

  WebTarget webTarget = client.target(baseuri);
  
  String input = "...";
  
  Response response = webTarget.request("application/json").post(Entity.json(input));
  

• 对于版本1.x（@Li'回答）：

  WebResource webResource = client.resource(baseuri);
  
  String input = "...";
  
  ClientResponse response = webResource.accept("application/json")
          .type("application/json").post(ClientResponse.class, input);
  

• 对于版本2.x：

  WebTarget webTarget = client.target(baseuri);
  
  String input = "...";
  
  Response response = webTarget.request("application/json").post(Entity.json(input));
  

Based on the Migration Guide

基于迁移指南