为什么不成功的“ HTTP使用套接字获取请求”

Question

我想在我的Android应用程序上发送获取消息。在我想收到200个好的回复之后。但是我没有完成。我收到408要求超时或什么都没有。你能帮我吗？

String requestmsg = "GET / HTTP/1.1rn";
requestmsg += "Host: www.ktu.edu.trrn";
requestmsg += "Connection: keep-alivern";
requestmsg += "Accept: text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,*/*;q=0.8rn";
requestmsg += "User-Agent: Mozilla/5.0 (Windows NT 6.3; WOW 64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/42.0.2311.135 Safari/537.36rn";
requestmsg += "Accept-Encoding: gzip, deflate, sdchrn";
requestmsg += "Accept-Language: en-US,en;q=0.8,en-GB;q=0.6rn";

DataOutputStream dos = null;
BufferedReader dis = null;
try {
    Log.d("ClientActivity", "Connecting...");
    String addr = InetAddress.getByName("www.ktu.edu.tr").getHostAddress().toString();
    Socket socket = new Socket(addr, 80);
    String data = "";
    try {
        Log.d("ClientActivity", "C: Sending command.");
        dos = new DataOutputStream(socket.getOutputStream());
        dis = new BufferedReader(new InputStreamReader(socket.getInputStream()));
        dos.write(requestmsg.getBytes());
        Log.i("ClientActivity", "RequestMsg Sent");
        StringBuilder sb = new StringBuilder();
        while ((data = dis.readLine()) != null) {
            sb.append(data);
        }
        Log.i("ClientActivity", "C: Sent.");
        Log.i("ClientActivity", "C: Received " + sb.toString());
    } catch (Exception e) {
        Log.e("ClientActivity", "S: Error", e);
    }
    socket.close();
    Log.d("ClientActivity", "C: Closed.");
} catch (Exception e) {
    Log.e("ClientActivity", "C: Error", e);
}

Answer 1

• 您缺少Content-Length标头
• 您在请求标题末尾缺少空白行
• 您没有根据其内容长度或分支正确阅读响应。

简而言之，不要自己实施HTTP。使用URL和HttpURLConnection。它会照顾所有这些细节，还有更多。