java - 打开 url.openstream 抛出无效的 http 响应

Question

我正在尝试从温度模块读取文件，当我在 url 上调用 openStream() 时，我收到一个 IOExeption，其中包含消息“无效的 Http 响应”

SEVERE: null
java.io.IOException: Invalid Http response
at sun.net.www.protocol.http.HttpURLConnection.getInputStream0(HttpURLConnection.java:1555)
at sun.net.www.protocol.http.HttpURLConnection.getInputStream(HttpURLConnection.java:1441)
at java.net.URL.openStream(URL.java:1038)
at thermometerpoller.Poller.poll(Poller.java:38)

我可以远程登录到温度模块:

telnet 192.168.142.55 80
Trying 192.168.142.55...
Connected to 192.168.142.55.
Escape character is '^]'.
GET /state.xml HTTP/1.1

<?xml version='1.0' encoding='utf-8'?>
<datavalues>
<units>F</units>
<sensor1temp>74.0</sensor1temp>
<sensor2temp>67.0</sensor2temp>
<sensor3temp>xx.x</sensor3temp>
<sensor4temp>xx.x</sensor4temp>
<relay1state>0</relay1state>
<relay2state>0</relay2state>
</datavalues>

Connection closed by foreign host.

温度模块似乎没有随回复发送任何 header 。不幸的是，当我查看HttpURLConnection.java时如果没有响应代码，则会抛出 IOException。

我的问题是，有没有办法通过库或其他方法获取文件内容而不关心响应代码是什么？

Answer 1

由于服务器是 not HTTP compliant ，您不应该使用 URL 或 URLConnection。使用普通的 Socket 代替:

Document doc;

final Socket connection = new Socket("192.168.142.55", 80);

try (final OutputStream out = connection.getOutputStream();
     InputStream in = connection.getInputStream()) {

    Callable<Void> requestSender = new Callable<Void>() {
        @Override
        public Void call()
        throws IOException {
            String request = "GET /state.xml HTTP/1.1\n\n";
            out.write(request.getBytes(StandardCharsets.US_ASCII));
            return null;
        }
    };
    ExecutorService background = Executors.newSingleThreadExecutor();
    Future<?> request = background.submit(requestSender);

    doc = DocumentBuilderFactory.newInstance().newDocumentBuilder().parse(in);

    request.get();
    background.shutdown();
}