java - TCP/IP 打开连接

Question

目前我正在使用以下代码与服务器交互

  public String connectToserverforincomingmsgs(String phonurl, String phno)
        throws IOException {
    URL url = new URL(phonurl);
    HttpURLConnection con = (HttpURLConnection) url.openConnection();
    con.setDoInput(true);

    // Allow Outputs
    con.setDoOutput(true);
    con.connect();
    BufferedWriter writer = null;
    writer = new BufferedWriter(new OutputStreamWriter(
            con.getOutputStream(), "UTF-8"));
    // give server your all parameters and values (replace param1 with you
    // param1 name and value with your one's)
    writer.write("sender_no=" + phno);

    writer.flush();

    String responseString = "";

    BufferedReader reader = null;
    reader = new BufferedReader(new InputStreamReader(con.getInputStream()));
    String line;

    while ((line = reader.readLine()) != null) {
        responseString = responseString.concat(line);
    }
    con.disconnect();
    return responseString;

}

我怎样才能建立tcp连接。现在我不知道。我对 android 和 java 都很陌生，所以任何有关 tcp 连接的示例代码将不胜感激

Answer 1

要创建 TCP 连接，您需要使用套接字:

Socket socket = new Socket(host_name_or_ip_address, port_no);

要发送数据，请使用socket.getOutputStream()

要接收数据，请使用socket.getInputStream()