Java中如何处理超时异常?
这是我的密码:Java中如何处理超时异常?,java,timeout,Java,Timeout,这是我的密码: private void synCampaign() { List<Campaign> campaigns; try { campaigns = AdwordsCampaign.getAllCampaign(); for(Campaign c : campaigns) CampaignDao.save(c); } catch (ApiException e) { try
private void synCampaign() {
List<Campaign> campaigns;
try {
campaigns = AdwordsCampaign.getAllCampaign();
for(Campaign c : campaigns)
CampaignDao.save(c);
} catch (ApiException e) {
try {
Thread.sleep(5000);
} catch (InterruptedException e1) {
e1.printStackTrace();
}
synCampaign();
e.printStackTrace();
} catch (RemoteException e) {
try {
Thread.sleep(5000);
} catch (InterruptedException e1) {
e1.printStackTrace();
}
synCampaign();
e.printStackTrace();
}
}
private void syncamp(){
列出运动;
试一试{
活动=AdwordsCampaign.getAllCampaign();
对于(活动c:活动)
3.save(c);
}捕获(APIE){
试一试{
睡眠(5000);
}捕捉(中断异常e1){
e1.printStackTrace();
}
synCampaign();
e、 printStackTrace();
}捕获(远程异常){
试一试{
睡眠(5000);
}捕捉(中断异常e1){
e1.printStackTrace();
}
synCampaign();
e、 printStackTrace();
}
}
AdwordsCampaign.getAllCampaign()
尝试获取一些远程资源。由于internet连接超时,这可能引发RemoteException
。当捕获到异常时,我只希望线程休眠一段时间,然后再次尝试获取远程资源
我的代码有问题吗?或者有更好的方法吗?除了在catch块中重复代码之外,这看起来还可以(确保您想要的重试次数)。您可能希望创建一个私有方法来处理异常,如下所示:
private void synCampaign() {
List<Campaign> campaigns;
try {
campaigns = AdwordsCampaign.getAllCampaign();
for(Campaign c : campaigns)
CampaignDao.save(c);
} catch (ApiException e) {
e.printStackTrace();
waitAndSync();
} catch (RemoteException e) {
e.printStackTrace();
waitAndSync();
}
}
private void waitAndSync(){
try {
Thread.sleep(5000);
} catch (InterruptedException e1) {
e1.printStackTrace();
}
synCampaign();
}
private void syncamp(){
列出运动;
试一试{
活动=AdwordsCampaign.getAllCampaign();
对于(活动c:活动)
3.save(c);
}捕获(APIE){
e、 printStackTrace();
waitAndSync();
}捕获(远程异常){
e、 printStackTrace();
waitAndSync();
}
}
私有void waitAndSync(){
试一试{
睡眠(5000);
}捕捉(中断异常e1){
e1.printStackTrace();
}
synCampaign();
}
没有什么真正的错误,但是(可能是无限的)递归重试循环(以及堆栈增长)让我有点紧张。我会写:
private void synCampaignWithRetries(int ntries, int msecsRetry) {
while(ntries-- >=0 ) {
try {
synCampaign();
return; // no exception? success
}
catch (ApiException e ) {
// log exception?
}
catch (RemoteException e ) {
// log exception?
}
try {
Thread.sleep(msecsRetry);
} catch (InterruptedException e1) {
// log exception?
}
}
// no success , even with ntries - log?
}
private void synCampaign() throws ApiException ,RemoteException {
List<Campaign> campaigns = AdwordsCampaign.getAllCampaign();
for(Campaign c : campaigns)
CampaignDao.save(c);
}
private void synCampaignWithRetries(int ntries,int msecsRetry){
而(ntries-->=0){
试一试{
synCampaign();
return;//无异常?成功
}
捕获(APIE){
//日志异常?
}
捕获(远程异常){
//日志异常?
}
试一试{
睡眠(msecsRetry);
}捕捉(中断异常e1){
//日志异常?
}
}
//没有成功,即使使用ntries-log?
}
private void synCampaign()引发ApiException、RemoteException{
列出活动=AdwordsCampaign.getAllCampaign();
对于(活动c:活动)
3.save(c);
}
您确实无法将其作为SocketTimeoutException捕获。可以捕获RemoteException,检索其原因并检查这是否是SocketTimeoutException的实例
try{
// Your code that throws SocketTimeoutException
}catch (RemoteException e) {
if(e.getCause().getClass().equals(SocketTimeoutException.class)){
System.out.println("It is SocketTimeoutException");
// Do handling for socket exception
}else{
throw e;
}
}catch (Exception e) {
// Handling other exception. If necessary
}
你的代码有效吗?或者你只是问你这样做是否合适?我只是觉得我的代码有点奇怪,所以我想确定我这样做是否合适@ClarkTo要解决infgoax提到的问题,我需要一个作业调度程序类@约根德拉·辛格