从Java中的HTTP POST请求中读取JSON消息

Question

我是Java和客户端 - 服务器编程的新手。

我正在使用嵌入式Jetty，我正在尝试将JSON字符串发送到某个地址（ http：// localhost：7070 / json ），然后在该地址中显示JSON字符串。

我尝试了以下代码，但我得到的只是null。

嵌入式Jetty代码：

public static void main(String[] args) throws Exception {
    Server server = new Server(7070);
    ServletContextHandler handler = new ServletContextHandler(server, "/json");
    handler.addServlet(ExampleServlet.class, "/");
    server.start();
}

用于发送Http POST的客户端功能：

public static void sendHttp(){
    HttpClient httpClient = HttpClientBuilder.create().build(); //Use this instead 

    try {
        HttpPost request = new HttpPost("http://localhost:7070/json");

        JSONObject object = new JSONObject();
        try {
            object.put("name", "MyName");
            object.put("age", "26");
        } catch (Exception ex) {
            System.out.println("Error: " + ex.getMessage());
        }

        String message = object.toString();
        request.setEntity(new StringEntity(message, "UTF8"));
        request.setHeader("Content-type", "application/json");

        HttpResponse response = httpClient.execute(request);

        // handle response here...
    }catch (Exception ex) {
        // handle exception here
    } finally {
    }
}

和Servlet功能：

public class ExampleServlet extends HttpServlet{

    @Override
    protected void doGet(HttpServletRequest req, HttpServletResponse resp)
            throws ServletException, IOException {
        //System.out.println("test get\n");
        doPost(req, resp);
    }

    @Override
    protected void doPost(HttpServletRequest req, HttpServletResponse resp)
            throws ServletException, IOException {
        //System.out.println("test post\n");
        PrintWriter out = resp.getWriter();
        String json_str = req.getParameter("name");
        out.print(json_str);
    }
}

在运行嵌入式Jetty服务器代码（如果这很重要）之后，我从测试类调用sendHttp（）方法。

Answer 1

您需要阅读原始请求正文，如下所示。 将它放在servlet的doPost方法中，以便从请求中读取json ：

StringBuilder jsonBuff = new StringBuilder();
String line = null;
try {
    BufferedReader reader = req.getReader();
    while ((line = reader.readLine()) != null)
        jsonBuff.append(line);
} catch (Exception e) { /*error*/ }

System.out.println("Request JSON string :" + jsonBuff.toString());
//write the response here by getting JSON from jasonBuff.toString()

try {
    JSONObject jsonObject = JSONObject.fromObject(jb.toString());

    out.print(jsonObject.get("name"));//writing output as you did

} catch (ParseException e) {
    throw new IOException("Error parsing JSON ");
}

注意：您可以访问req.getParameter("name"); 只有当你的标题是这样的时候：

content type: "application/x-www-form-urlencoded"

在正常的html表单提交中。

Answer 2

要从Post请求中获取数据，您需要获取内容。 尝试这个：

String data = IOUtils.toString(req.getInputStream(), "UTF-8");

Answer 3

我没有使用过jetty但是我用这个代码做了类似的通信（PUT，而不是POST）：

URL url = new URL(desturl);
HttpURLConnection huc = (HttpURLConnection) url.openConnection();
huc.setRequestMethod("PUT");
byte[] postData = null; 
int postDataLength; 
huc.setDoOutput(true);
postData = data.getBytes( StandardCharsets.UTF_8 );
postDataLength = postData.length;
huc.setRequestProperty( "Content-Type", "application/json"); 
huc.setRequestProperty( "charset", "utf-8");
huc.setRequestProperty( "Content-Length", Integer.toString( postDataLength ));
huc.setUseCaches( false );
huc.connect();
huc.setConnectTimeout(10000);
DataOutputStream wr = new DataOutputStream( huc.getOutputStream());
wr.write( postData );
rd = new BufferedReader(new InputStreamReader(huc.getInputStream()));
retcode = huc.getResponseCode();

Answer 4

这是我的代码，这工作正常

    String data = "";   
    StringBuilder builder = new StringBuilder();
    BufferedReader reader = request.getReader();
    String line;
    while ((line = reader.readLine()) != null) {
        builder.append(line);
    }
    data = builder.toString();
    JSONObject object = new JSONObject(data); 
   //or JSONArray array = new JSONArray(data); which ever the one you want

祝好运.....