我试图在下面的代码中抛出超时异常。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SOAPException的超时异常?
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;
}
}
答案 0 :(得分:0)
以下几行来自call
方法的open jdk源代码。在代码中,他们只使用Exception
(也包含链接?注释)。除非Oracle jdk以不同的方式处理这个问题,否则我认为还有其他方法
您仍然可以尝试if(soapEx.getCause() instanceof SomeTimeoutException)
之类的内容(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);
}
如果您想查看源代码HttpSoapConnection
答案 1 :(得分:0)
经过几个小时的测试后,我找到了从Timeout相关异常中分离SOAPException的正确方法。因此,解决方案是获取异常的父原因字段,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;
}
}