在收到POST请求的Java的httpserver崩溃

Question

我试图用Java编写一个简单的HTTP服务器，它可以处理POST请求。虽然我的服务器成功接收的GET，它崩溃的讯息。

这里是服务器

public class RequestHandler {
    public static void main(String[] args) throws Exception {
        HttpServer server = HttpServer.create(new InetSocketAddress(8080), 0);
        server.createContext("/requests", new MyHandler());
        server.setExecutor(null); // creates a default executor
        server.start();
    }

    static class MyHandler implements HttpHandler {
        public void handle(HttpExchange t) throws IOException {
            String response = "hello world";
            t.sendResponseHeaders(200, response.length());
            System.out.println(response);
            OutputStream os = t.getResponseBody();
            os.write(response.getBytes());
            os.close();
        }
    }
}

这里是我用来发送POST Java代码

// HTTP POST request
private void sendPost() throws Exception {

    String url = "http://localhost:8080/requests";
    URL obj = new URL(url);
    HttpsURLConnection con = (HttpsURLConnection) obj.openConnection();

    //add reuqest header
    con.setRequestMethod("POST");
    con.setRequestProperty("User-Agent", USER_AGENT);
    con.setRequestProperty("Accept-Language", "en-US,en;q=0.5");

    String urlParameters = "sn=C02G8416DRJM&cn=&locale=&caller=&num=12345";

    // Send post request
    con.setDoOutput(true);
    DataOutputStream wr = new DataOutputStream(con.getOutputStream());
    wr.writeBytes(urlParameters);
    wr.flush();
    wr.close();

    int responseCode = con.getResponseCode();
    System.out.println("\nSending 'POST' request to URL : " + url);
    System.out.println("Post parameters : " + urlParameters);
    System.out.println("Response Code : " + responseCode);

    BufferedReader in = new BufferedReader(
            new InputStreamReader(con.getInputStream()));
    String inputLine;
    StringBuffer response = new StringBuffer();

    while ((inputLine = in.readLine()) != null) {
        response.append(inputLine);
    }
    in.close();

    //print result
    System.out.println(response.toString());
}

每次POST请求在此行崩溃

HttpsURLConnection con = (HttpsURLConnection) obj.openConnection();

但是当我改变的URL在我发现这它的工作原理所提供的示例之一。

Answer 1

代替

HttpsURLConnection con = (HttpsURLConnection) obj.openConnection();

采用

HttpURLConnection con = (HttpURLConnection) obj.openConnection();

您要连接到这不是一个HTTPS URL。当你调用obj.openConnection()，它决定连接是HTTP或HTTPS，并返回相应的对象。当它的http，它不会返回HttpsURLConnection，所以你不能转换到它。

然而，由于HttpsURLconnection延伸HttpURLConnection，使用HttpURLConnection将两个http和https网址。你是在你的代码中调用的所有方法中存在的int HttpURLConnection类。