Try-catch旨在帮助进行异常处理。这意味着它将以某种方式帮助我们的系统更加健壮：尝试从意外事件中恢复。
我们怀疑在执行和指令（发送消息）时可能会发生某些事情，因此它会被包含在try中。如果发生几乎意外的事情，我们可以做点什么：我们写下了捕获。我认为我们没有打电话来记录异常。我认为catch块意味着让我们有机会从错误中恢复过来。
现在，假设我们可以从错误中恢复，因为我们可以解决错误。重做是非常好的：
try{ some_instruction(); }
catch (NearlyUnexpectedException e){
fix_the_problem();
retry;
}
这会很快落入永恒循环，但是假设fix_the_problem返回true，那么我们重试。鉴于Java中没有这样的东西，你将如何解决这个问题？解决这个问题的最佳设计代码是什么？
这就像一个哲学问题，因为我已经知道我所要求的并不是Java直接支持的。
答案 0 :(得分：257)
您需要将try-catch封闭在while循环中，如下所示： -
int count = 0;
int maxTries = 3;
while(true) {
try {
// Some Code
// break out of loop, or return, on success
} catch (SomeException e) {
// handle exception
if (++count == maxTries) throw e;
}
}
我已经采用count和maxTries来避免遇到无限循环，以防异常继续发生在try block。
答案 1 :(得分：48)
强制性的“企业家”解决方案：
public abstract class Operation {
abstract public void doIt();
public void handleException(Exception cause) {
//default impl: do nothing, log the exception, etc.
}
}
public class OperationHelper {
public static void doWithRetry(int maxAttempts, Operation operation) {
for (int count = 0; count < maxAttempts; count++) {
try {
operation.doIt();
count = maxAttempts; //don't retry
} catch (Exception e) {
operation.handleException(e);
}
}
}
}
致电：
OperationHelper.doWithRetry(5, new Operation() {
@Override public void doIt() {
//do some stuff
}
@Override public void handleException(Exception cause) {
//recover from the Exception
}
});
答案 2 :(得分：31)
与往常一样，最佳设计取决于具体情况。通常，我写的是：
for (int retries = 0;; retries++) {
try {
return doSomething();
} catch (SomeException e) {
if (retries < 6) {
continue;
} else {
throw e;
}
}
}
答案 3 :(得分：19)
虽然try/catch while是众所周知的好策略我想建议你递归调用：
void retry(int i, int limit) {
try {
} catch (SomeException e) {
// handle exception
if (i >= limit) {
throw e; // variant: wrap the exception, e.g. throw new RuntimeException(e);
}
retry(i++, limit);
}
}
答案 4 :(得分：14)
您可以使用jcabi-aspects中的AOP和Java注释（我是开发人员）：
@RetryOnFailure(attempts = 3, delay = 5)
public String load(URL url) {
return url.openConnection().getContent();
}
您还可以使用@Loggable和@LogException注释。
答案 5 :(得分：11)
您的确切方案是通过Failsafe处理的：
RetryPolicy retryPolicy = new RetryPolicy()
.retryOn(NearlyUnexpectedException.class);
Failsafe.with(retryPolicy)
.onRetry((r, f) -> fix_the_problem())
.run(() -> some_instruction());
非常简单。
答案 6 :(得分：6)
这些答案大多基本相同。我也是，但这是我喜欢的形式
boolean completed = false;
Throwable lastException = null;
for (int tryCount=0; tryCount < config.MAX_SOME_OPERATION_RETRIES; tryCount++)
{
try {
completed = some_operation();
break;
}
catch (UnlikelyException e) {
lastException = e;
fix_the_problem();
}
}
if (!completed) {
reportError(lastException);
}
答案 7 :(得分：3)
使用带有本地while标记的status循环。将标记初始化为false并在操作成功时将其设置为true，例如下面：
boolean success = false;
while(!success){
try{
some_instruction();
success = true;
} catch (NearlyUnexpectedException e){
fix_the_problem();
}
}
这将继续重试，直到成功。
如果您只想重试一定次数，请使用计数器：
boolean success = false;
int count = 0, MAX_TRIES = 10;
while(!success && count++ < MAX_TRIES){
try{
some_instruction();
success = true;
} catch (NearlyUnexpectedException e){
fix_the_problem();
}
}
if(!success){
//It wasn't successful after 10 retries
}
如果不成功，这将尝试最多10次，直到那时如果事先成功将退出。
答案 8 :(得分：2)
Spring AOP和基于注释的解决方案：
用法（{@RetryOperation是我们对作业的自定义注释）：
@RetryOperation(retryCount = 1, waitSeconds = 10)
boolean someMethod() throws Exception {
}
我们需要完成两件事：1.注释界面，和2. spring方面。这是实现这些目的的一种方法：
注释界面：
import java.lang.annotation.*;
@Target(ElementType.METHOD)
@Retention(RetentionPolicy.RUNTIME)
public @interface RetryOperation {
int retryCount();
int waitSeconds();
}
春季看点：
import org.aspectj.lang.ProceedingJoinPoint;
import org.aspectj.lang.annotation.Around;
import org.aspectj.lang.annotation.Aspect;
import org.aspectj.lang.reflect.MethodSignature;
import org.slf4j.Logger;
import org.slf4j.LoggerFactory;
import org.springframework.stereotype.Component;
import java.lang.reflect.Method;
@Aspect @Component
public class RetryAspect {
private static final Logger LOGGER = LoggerFactory.getLogger(RetryAspect.class);
@Around(value = "@annotation(RetryOperation)")
public Object retryOperation(ProceedingJoinPoint joinPoint) throws Throwable {
Object response = null;
Method method = ((MethodSignature) joinPoint.getSignature()).getMethod();
RetryOperation annotation = method.getAnnotation(RetryOperation.class);
int retryCount = annotation.retryCount();
int waitSeconds = annotation.waitSeconds();
boolean successful = false;
do {
try {
response = joinPoint.proceed();
successful = true;
} catch (Exception ex) {
LOGGER.info("Operation failed, retries remaining: {}", retryCount);
retryCount--;
if (retryCount < 0) {
throw ex;
}
if (waitSeconds > 0) {
LOGGER.info("Waiting for {} second(s) before next retry", waitSeconds);
Thread.sleep(waitSeconds * 1000l);
}
}
} while (!successful);
return response;
}
}
答案 9 :(得分：2)
如果不是所有例外都需要重试，则只有一部分例外。如果必须至少尝试一次，则可以使用另一种实用方法：
void runWithRetry(Runnable runnable, Class<Exception> exClass, int maxRetries) {
Exception err = null;
do {
maxRetries--;
try {
runnable.run();
err = null;
} catch (Exception e) {
if(exClass.isAssignableFrom(e.getClass())){
err = e;
}else {
throw e;
}
}
} while (err != null && maxRetries > 0);
if (err != null) {
throw err;
}
}
用法：
runWithRetry(() -> {
// do something
}, TimeoutException.class, 5)
答案 10 :(得分：2)
解决问题的一种简单方法是将try / catch包装在while循环中并保持计数。这样，您可以通过在保持故障日志的同时检查其他变量的计数来防止无限循环。它不是最精致的解决方案，但它会起作用。
答案 11 :(得分：1)
如果它有用，还需要考虑几个选项，所有选项（停止文件而不是重试，睡眠，继续更大的循环）都可能有用。
bigLoop:
while(!stopFileExists()) {
try {
// do work
break;
}
catch (ExpectedExceptionType e) {
// could sleep in here, too.
// another option would be to "restart" some bigger loop, like
continue bigLoop;
}
// ... more work
}
答案 12 :(得分：1)
将@ach先前的解决方案简化为一个文件并使用功能接口。
public class OperationHelper {
public static void doWithRetry(int maxAttempts, Runnable operation, Consumer<Exception> handle) {
for (int count = 0; count < maxAttempts; count++) {
try {
operation.run();
count = maxAttempts; //don't retry
} catch (Exception e) {
handle.accept(e);
}
}
}
}
答案 13 :(得分：1)
您可以使用https://github.com/bnsd55/RetryCatch
示例：
RetryCatch retryCatchSyncRunnable = new RetryCatch();
retryCatchSyncRunnable
// For infinite retry times, just remove this row
.retryCount(3)
// For retrying on all exceptions, just remove this row
.retryOn(ArithmeticException.class, IndexOutOfBoundsException.class)
.onSuccess(() -> System.out.println("Success, There is no result because this is a runnable."))
.onRetry((retryCount, e) -> System.out.println("Retry count: " + retryCount + ", Exception message: " + e.getMessage()))
.onFailure(e -> System.out.println("Failure: Exception message: " + e.getMessage()))
.run(new ExampleRunnable());
您可以传递自己的匿名函数来代替new ExampleRunnable()。
答案 14 :(得分：0)
这是我的解决方案，类似于其他一些可以包装函数的解决方案，但如果成功，则允许您获取函数的返回值。
/**
* Wraps a function with retry logic allowing exceptions to be caught and retires made.
*
* @param function the function to retry
* @param maxRetries maximum number of retires before failing
* @param delay time to wait between each retry
* @param allowedExceptionTypes exception types where if caught a retry will be performed
* @param <V> return type of the function
* @return the value returned by the function if successful
* @throws Exception Either an unexpected exception from the function or a {@link RuntimeException} if maxRetries is exceeded
*/
@SafeVarargs
public static <V> V runWithRetriesAndDelay(Callable<V> function, int maxRetries, Duration delay, Class<? extends Exception>... allowedExceptionTypes) throws Exception {
final Set<Class<? extends Exception>> exceptions = new HashSet<>(Arrays.asList(allowedExceptionTypes));
for(int i = 1; i <= maxRetries; i++) {
try {
return function.call();
} catch (Exception e) {
if(exceptions.contains(e.getClass())){
// An exception of an expected type
System.out.println("Attempt [" + i + "/" + maxRetries + "] Caught exception [" + e.getClass() + "]");
// Pause for the delay time
Thread.sleep(delay.toMillis());
}else {
// An unexpected exception type
throw e;
}
}
}
throw new RuntimeException(maxRetries + " retries exceeded");
}
答案 15 :(得分：0)
下面的代码段执行一些代码段。如果在执行代码片段时遇到任何错误，请睡眠M毫秒，然后重试。参考link。
public void retryAndExecuteErrorProneCode(int noOfTimesToRetry, CodeSnippet codeSnippet, int sleepTimeInMillis)
throws InterruptedException {
int currentExecutionCount = 0;
boolean codeExecuted = false;
while (currentExecutionCount < noOfTimesToRetry) {
try {
codeSnippet.errorProneCode();
System.out.println("Code executed successfully!!!!");
codeExecuted = true;
break;
} catch (Exception e) {
// Retry after 100 milliseconds
TimeUnit.MILLISECONDS.sleep(sleepTimeInMillis);
System.out.println(e.getMessage());
} finally {
currentExecutionCount++;
}
}
if (!codeExecuted)
throw new RuntimeException("Can't execute the code within given retries : " + noOfTimesToRetry);
}
答案 16 :(得分：0)
尝试使用springs @ Retryable注释，当RuntimeException发生时，以下方法将重试3次
@Retryable(maxAttempts=3,value= {RuntimeException.class},backoff = @Backoff(delay = 500))
public void checkSpringRetry(String str) {
if(StringUtils.equalsIgnoreCase(str, "R")) {
LOGGER.info("Inside retry.....!!");
throw new RuntimeException();
}
}
答案 17 :(得分：0)
这是一个古老的问题，但解决方案仍然有意义。这是我在Java 8中不使用任何第三方库的通用解决方案：
public interface RetryConsumer<T> {
T evaluate() throws Throwable;
}
public interface RetryPredicate<T> {
boolean shouldRetry(T t);
}
public class RetryOperation<T> {
private RetryConsumer<T> retryConsumer;
private int noOfRetry;
private int delayInterval;
private TimeUnit timeUnit;
private RetryPredicate<T> retryPredicate;
private List<Class<? extends Throwable>> exceptionList;
public static class OperationBuilder<T> {
private RetryConsumer<T> iRetryConsumer;
private int iNoOfRetry;
private int iDelayInterval;
private TimeUnit iTimeUnit;
private RetryPredicate<T> iRetryPredicate;
private Class<? extends Throwable>[] exceptionClasses;
private OperationBuilder() {
}
public OperationBuilder<T> retryConsumer(final RetryConsumer<T> retryConsumer) {
this.iRetryConsumer = retryConsumer;
return this;
}
public OperationBuilder<T> noOfRetry(final int noOfRetry) {
this.iNoOfRetry = noOfRetry;
return this;
}
public OperationBuilder<T> delayInterval(final int delayInterval, final TimeUnit timeUnit) {
this.iDelayInterval = delayInterval;
this.iTimeUnit = timeUnit;
return this;
}
public OperationBuilder<T> retryPredicate(final RetryPredicate<T> retryPredicate) {
this.iRetryPredicate = retryPredicate;
return this;
}
@SafeVarargs
public final OperationBuilder<T> retryOn(final Class<? extends Throwable>... exceptionClasses) {
this.exceptionClasses = exceptionClasses;
return this;
}
public RetryOperation<T> build() {
if (Objects.isNull(iRetryConsumer)) {
throw new RuntimeException("'#retryConsumer:RetryConsumer<T>' not set");
}
List<Class<? extends Throwable>> exceptionList = new ArrayList<>();
if (Objects.nonNull(exceptionClasses) && exceptionClasses.length > 0) {
exceptionList = Arrays.asList(exceptionClasses);
}
iNoOfRetry = iNoOfRetry == 0 ? 1 : 0;
iTimeUnit = Objects.isNull(iTimeUnit) ? TimeUnit.MILLISECONDS : iTimeUnit;
return new RetryOperation<>(iRetryConsumer, iNoOfRetry, iDelayInterval, iTimeUnit, iRetryPredicate, exceptionList);
}
}
public static <T> OperationBuilder<T> newBuilder() {
return new OperationBuilder<>();
}
private RetryOperation(RetryConsumer<T> retryConsumer, int noOfRetry, int delayInterval, TimeUnit timeUnit,
RetryPredicate<T> retryPredicate, List<Class<? extends Throwable>> exceptionList) {
this.retryConsumer = retryConsumer;
this.noOfRetry = noOfRetry;
this.delayInterval = delayInterval;
this.timeUnit = timeUnit;
this.retryPredicate = retryPredicate;
this.exceptionList = exceptionList;
}
public T retry() throws Throwable {
T result = null;
int retries = 0;
while (retries < noOfRetry) {
try {
result = retryConsumer.evaluate();
if (Objects.nonNull(retryPredicate)) {
boolean shouldItRetry = retryPredicate.shouldRetry(result);
if (shouldItRetry) {
retries = increaseRetryCountAndSleep(retries);
} else {
return result;
}
} else {
// no retry condition defined, no exception thrown. This is the desired result.
return result;
}
} catch (Throwable e) {
retries = handleException(retries, e);
}
}
return result;
}
private int handleException(int retries, Throwable e) throws Throwable {
if (exceptionList.contains(e.getClass()) || (exceptionList.isEmpty())) {
// exception is excepted, continue retry.
retries = increaseRetryCountAndSleep(retries);
if (retries == noOfRetry) {
// evaluation is throwing exception, no more retry left. Throw it.
throw e;
}
} else {
// unexpected exception, no retry required. Throw it.
throw e;
}
return retries;
}
private int increaseRetryCountAndSleep(int retries) {
retries++;
if (retries < noOfRetry && delayInterval > 0) {
try {
timeUnit.sleep(delayInterval);
} catch (InterruptedException ignore) {
Thread.currentThread().interrupt();
}
}
return retries;
}
}
让我们有一个测试用例，例如：
@Test
public void withPredicateAndException() {
AtomicInteger integer = new AtomicInteger();
try {
Integer result = RetryOperation.<Integer>newBuilder()
.retryConsumer(() -> {
int i = integer.incrementAndGet();
if (i % 2 == 1) {
throw new NumberFormatException("Very odd exception");
} else {
return i;
}
})
.noOfRetry(10)
.delayInterval(10, TimeUnit.MILLISECONDS)
.retryPredicate(value -> value <= 6)
.retryOn(NumberFormatException.class, EOFException.class)
.build()
.retry();
Assert.assertEquals(8, result.intValue());
} catch (Throwable throwable) {
Assert.fail();
}
}
答案 18 :(得分：0)
简单
ReferenceError: XMLHttpRequest is not defined
at loadBooks (C:\Users\Hp\book-list.js:38:19)
at Object.<anonymous> (C:\Users\Hp\book-list.js:73:1)
答案 19 :(得分：0)
其余解决方案的问题在于，相应的函数连续尝试，而没有中间的时间间隔，从而导致栈泛滥。
为什么不只是每秒try和一个末尾？
以下是使用setTimeout和递归函数的解决方案：
(function(){
try{
Run(); //tries for the 1st time, but Run() as function is not yet defined
}
catch(e){
(function retry(){
setTimeout(function(){
try{
console.log("trying...");
Run();
console.log("success!");
}
catch(e){
retry(); //calls recursively
}
}, 1000); //tries every second
}());
}
})();
//after 5 seconds, defines Run as a global function
var Run;
setTimeout(function(){
Run = function(){};
}, 5000);
每秒用您要重新使用的功能或代码Run()替换功能try。
答案 20 :(得分：0)
这是一种可重用且更通用的Java 8+方法，不需要外部库：
public interface IUnreliable<T extends Exception>
{
void tryRun ( ) throws T;
}
public static <T extends Exception> void retry (int retryCount, IUnreliable<T> runnable) throws T {
for (int retries = 0;; retries++) {
try {
runnable.tryRun();
return;
} catch (Exception e) {
if (retries < retryCount) {
continue;
} else {
throw e;
}
}
}
}
用法：
@Test
public void demo() throws IOException {
retry(3, () -> {
new File("/tmp/test.txt").createNewFile();
});
}
答案 21 :(得分：0)
https://github.com/tusharmndr/retry-function-wrapper/tree/master/src/main/java/io
int MAX_RETRY = 3;
RetryUtil.<Boolean>retry(MAX_RETRY,() -> {
//Function to retry
return true;
});
答案 22 :(得分：0)
我不确定这是否是“专业”方式，我不完全确定它是否适用于所有事情。
boolean gotError = false;
do {
try {
// Code You're Trying
} catch ( FileNotFoundException ex ) {
// Exception
gotError = true;
}
} while ( gotError = true );
答案 23 :(得分：0)
以下是我的解决方案，方法非常简单！
while (true) {
try {
/// Statement what may cause an error;
break;
} catch (Exception e) {
}
}
答案 24 :(得分：0)
我知道这里已经有很多类似的答案，而且我的情况并没有太大的不同，但无论如何我会发布它，因为它涉及特定的案例/问题。
当处理facebook Graph API中的PHP时，您有时会收到错误，但立即重新尝试相同的操作会产生正面结果（因为各种魔法互联网原因超出了这个问题的范围）。在这种情况下，不需要修复任何错误，但只需再次尝试，因为有某种＆＃34; facebook错误＆＃34;。
创建Facebook会话后立即使用此代码：
//try more than once because sometimes "facebook error"
$attempt = 3;
while($attempt-- > 0)
{
// To validate the session:
try
{
$facebook_session->validate();
$attempt = 0;
}
catch (Facebook\FacebookRequestException $ex)
{
// Session not valid, Graph API returned an exception with the reason.
if($attempt <= 0){ echo $ex->getMessage(); }
}
catch (\Exception $ex)
{
// Graph API returned info, but it may mismatch the current app or have expired.
if($attempt <= 0){ echo $ex->getMessage(); }
}
}
此外，通过将for循环计数降至零（$attempt--），可以很容易地更改将来的尝试次数。
答案 25 :(得分：0)
使用do-while设计重试块。
boolean successful = false;
int maxTries = 3;
do{
try {
something();
success = true;
} catch(Me ifUCan) {
maxTries--;
}
} while (!successful || maxTries > 0)
答案 26 :(得分：0)
所有Try-Catch都会让您的程序正常失败。在catch语句中，您通常会尝试记录错误，如果需要，可以回滚更改。
bool finished = false;
while(finished == false)
{
try
{
//your code here
finished = true
}
catch(exception ex)
{
log.error("there was an error, ex");
}
}