Question

我试图了解如何向服务器发送REST请求。如果我必须使用httpconnections或任何其他连接在java中将其实现为请求，我该怎么做？

    POST /resource/1
    Host: myownHost
    DATE: date
    Content-Type: some standard type

这应该如何以标准方式构建？

    URL url= new URL("http://myownHost/resource/1");
    HttpsURLConnection connect= (HttpsURLConnection) url.openConnection();
    connect.setRequestMethod("POST");
    connect.setRequestProperty("Host", "myOwnHost");
    connect.setRequestProperty("Date","03:14:15 03:14:15 GMT");
    connect.setRequestProperty("Content-Type","application/x-www-form-urlencoded");

Answer 1

有很多选项，Apache HTTP客户端（http://hc.apache.org/httpcomponents-client-4.4.x/index.html）就是其中之一（并且让事情变得非常简单）

创建REST请求可以像这样简单（在这种情况下使用JSON）：

    DefaultHttpClient httpClient = new DefaultHttpClient();
    HttpGet getRequest = new HttpGet(
        "http://localhost:8080/RESTfulExample/json/product/get");
    getRequest.addHeader("accept", "application/json");

    HttpResponse response = httpClient.execute(getRequest);

    if (response.getStatusLine().getStatusCode() != 200) {
        throw new RuntimeException("Failed : HTTP error code : "
           + response.getStatusLine().getStatusCode());
    }

    BufferedReader br = new BufferedReader(
                     new   InputStreamReader((response.getEntity().getContent())));

更新：抱歉，文档的链接已更新。发布了新文档。

Answer 2

你应该在这里使用json

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.net.HttpURLConnection;
import java.net.MalformedURLException;
import java.net.URL;

public class NetClientPost {

// http://localhost:8080/RESTfulExample/json/product/post
public static void main(String[] args) {

  try {

    URL url = new URL("http://myownHost/resource/1");
    HttpURLConnection conn = (HttpURLConnection) url.openConnection();
    conn.setDoOutput(true);
    conn.setRequestMethod("POST");
    conn.setRequestProperty("Content-Type", "application/json");

    String input = "{\"DATE\":\"03:14:15 03:14:15 GMT\",\"host\":\"myownhost\"}";

    OutputStream os = conn.getOutputStream();
    os.write(input.getBytes());
    os.flush();

    if (conn.getResponseCode() != HttpURLConnection.HTTP_CREATED) {
        throw new RuntimeException("Failed : HTTP error code : "
            + conn.getResponseCode());
    }

    BufferedReader br = new BufferedReader(new InputStreamReader(
            (conn.getInputStream())));

    String output;
    System.out.println("Output from Server .... \n");
    while ((output = br.readLine()) != null) {
        System.out.println(output);
    }

    conn.disconnect();

  } catch (MalformedURLException e) {

    e.printStackTrace();

   } catch (IOException e) {

    e.printStackTrace();

  }

 }

}

更多内容浏览link

Answer 3

有几种方法可以使用Java调用RESTful服务，但不需要使用原始级API; - ）

它存在一些RESTful框架，如Restlet或JAX-RS。它们针对客户端和服务器端，旨在隐藏此类调用的技术管道。下面是一个代码示例，描述了如何使用Restlet和JSON解析器进行处理：

JSONObject jsonObj = new JSONObject();
jsonObj.put("host", "...");
ClientResource cr = new Client("http://myownHost/resource/1");
cr.post(new JsonRepresentation(jsonObject);

// In the case of form
// Form form = new Form ();
// form.set("host", "...");
// cr.post(form);

您可以注意到，在上一个代码段中，根据您发送的内容（表单，JSON，...）自动为您设置标题Content-type，Date

除非是一个小注释，要添加元素，您应该在元素列表资源（POST）或方法http://myownHost/resources/上使用方法PUT，如果您有所需的唯一标识符用来识别它（http://myownHost/resources/1）。此链接可能对您有用：https://templth.wordpress.com/2014/12/15/designing-a-web-api/。

希望它可以帮到你，亨利

如何在java中构建RESTful请求

3 个答案: