从 SOAPException 中抛出超时异常
Throw timeout exception from SOAPException
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;
}
}
以下几行来自call
方法的公开jdk源代码。在代码中,他们只使用 Exception
(也使用链接?评论)。我认为没有其他方法,除非 Oracle jdk 以不同的方式处理此问题。
您仍然可以尝试 if(soapEx.getCause() instanceof SomeTimeoutException)
(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);
}
如果要查看源码HttpSoapConnection
经过几个小时的测试,我找到了将 SOAPException 与超时相关异常区分开来的正确方法。所以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;
}
}
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;
}
}
以下几行来自call
方法的公开jdk源代码。在代码中,他们只使用 Exception
(也使用链接?评论)。我认为没有其他方法,除非 Oracle jdk 以不同的方式处理此问题。
您仍然可以尝试 if(soapEx.getCause() instanceof SomeTimeoutException)
(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);
}
如果要查看源码HttpSoapConnection
经过几个小时的测试,我找到了将 SOAPException 与超时相关异常区分开来的正确方法。所以解决方案是获取异常的父原因字段并检查它是否是 SocketTimeoutException
.
try {
response = connection.call(request, endpoint);
} catch (SOAPException soapEx) {
if(soapEx.getCause().getCause() instanceof SocketTimeoutException) {
throw new TimeoutExpirationException(); //custom exception
} else {
throw soapEx;
}
}