从SOAPException抛出超时异常

Question

我试图在下面的代码中抛出一个超时异常。我尝试了一个简单的条件，但这不是正确的方法。我的问题是如何区分超时异常和SOAPException？

URL endpoint = new URL(null,
    urlStr,
    new URLStreamHandler() {
      // The url is the parent of this stream handler, so must create clone
      protected URLConnection openConnection(URL url) throws IOException {
        URL cloneURL = new URL(url.toString());
        HttpURLConnection cloneURLConnection = (HttpURLConnection) cloneURL.openConnection();
        // TimeOut settings
        cloneURLConnection.setConnectTimeout(10000);
        cloneURLConnection.setReadTimeout(10000);
        return cloneURLConnection;
      }
    });

try {
  response = connection.call(request, endpoint);
} catch (SOAPException soapEx) {
  if(soapEx.getMessage().contains("Message send failed")) {
    throw new TimeoutExpirationException();
  } else {
    throw soapEx;
  }
}

Answer 1

经过几个小时的测试，我找到了解决SOAPException超时相关异常的正确方法。因此，解决方案是接受异常的父原因字段，并检查它是否是SocketTimeoutException的实例。

try {
  response = connection.call(request, endpoint);
} catch (SOAPException soapEx) {
  if(soapEx.getCause().getCause() instanceof SocketTimeoutException) {
    throw new TimeoutExpirationException(); //custom exception
  } else {
    throw soapEx;
  }
}

Answer 2

下面的行来自call方法的开放jdk源代码。在代码中，它们只使用Exception (也是使用链接)捕获？评论)。我不认为有其他方法，除非Oracle jdk对此有不同的处理方式。

您仍然可以尝试类似if(soapEx.getCause() instanceof SomeTimeoutException)的方法(不确定这是否有效)。

            try {
                SOAPMessage response = post(message, (URL)endPoint);
                return response;
            } catch (Exception ex) {
                // TBD -- chaining?
                throw new SOAPExceptionImpl(ex);
            } 

如果您想检查源代码HttpSoapConnection