打开 url.openstream 会引发无效的 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，我收到一个带有“无效 Http 响应”消息的 IOExeption

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.

温度模块似乎没有在回复中发送任何标题。 不幸的是，当我查看HttpURLConnection.java 时，如果没有响应代码，它会抛出 IOException。

我的问题是，有没有办法在不关心响应代码的情况下通过库或其他方法获取文件内容？

Answer 1

由于服务器是not HTTP compliant，因此您不应使用 URL 或 URLConnection。改用普通的 Socket：

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();
}