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你好,非常喜欢您的分享,在运行chapter22一章的demo时发现了两个问题,不知道是不是遗留的bug,如理解有问题,请指正😄
@Around("execution(public * *(..)) && @annotation(com.battcn.annotation.CacheLock)") public Object interceptor(ProceedingJoinPoint pjp) { MethodSignature signature = (MethodSignature) pjp.getSignature(); Method method = signature.getMethod(); CacheLock lock = method.getAnnotation(CacheLock.class); if (StringUtils.isEmpty(lock.prefix())) { throw new RuntimeException("lock key don't null..."); } final String lockKey = cacheKeyGenerator.getLockKey(pjp); String value = UUID.randomUUID().toString(); try { // 假设上锁成功,但是设置过期时间失效,以后拿到的都是 false final boolean success = redisLockHelper.lock(lockKey, value, lock.expire(), lock.timeUnit()); if (!success) { throw new RuntimeException("重复提交"); } try { return pjp.proceed(); } catch (Throwable throwable) { throw new RuntimeException("系统异常"); } } finally { // TODO 如果演示的话需要注释该代码;实际应该放开 redisLockHelper.unlock(lockKey, value); } }
这里的String value = UUID.randomUUID().toString();是不是有多线程问题,当第一个线程未处理完,第二个线程运行到这里是会刷新value,导致第一个线程unlock时会使用第二个线程的value
String value = UUID.randomUUID().toString();
public boolean lock(String lockKey, final String uuid, long timeout, final TimeUnit unit) { final long milliseconds = Expiration.from(timeout, unit).getExpirationTimeInMilliseconds(); boolean success = stringRedisTemplate.opsForValue().setIfAbsent(lockKey, (System.currentTimeMillis() + milliseconds) + DELIMITER + uuid); if (success) { stringRedisTemplate.expire(lockKey, timeout, TimeUnit.SECONDS); } else { // 这里直接更新这个值是不是有点问题,应该判断时间,超时再更新 String oldVal = stringRedisTemplate.opsForValue().getAndSet(lockKey, (System.currentTimeMillis() + milliseconds) + DELIMITER + uuid); final String[] oldValues = oldVal.split(Pattern.quote(DELIMITER)); if (Long.parseLong(oldValues[0]) + 1 <= System.currentTimeMillis()) { return true; } } return success; }
我认为这里应该这样写
public boolean lock(String lockKey, final String uuid, long timeout, final TimeUnit unit) { final long milliseconds = Expiration.from(timeout, unit).getExpirationTimeInMilliseconds(); boolean success = stringRedisTemplate.opsForValue().setIfAbsent(lockKey, (System.currentTimeMillis() + milliseconds) + DELIMITER + uuid); if (success) { stringRedisTemplate.expire(lockKey, timeout, TimeUnit.SECONDS); } else { String oldVal = stringRedisTemplate.opsForValue().get(lockKey); final String[] oldValues = oldVal.split(Pattern.quote(DELIMITER)); if (Long.parseLong(oldValues[0]) + 1 <= System.currentTimeMillis()) { stringRedisTemplate.opsForValue().set(lockKey, (System.currentTimeMillis() + milliseconds) + DELIMITER + uuid); return true; } } return success; }
如理解有问题请指正😄
The text was updated successfully, but these errors were encountered:
恩,不过现在有更简单的方案了,springboot2.1.x 支持setex 直接的写法了,不用这么麻烦了
/** * 获取锁 * * @param lockKey lockKey * @param uuid UUID * @param timeout 超时时间 * @param unit 过期单位 * @return true or false */ public boolean lock(String lockKey, final String uuid, long timeout, final TimeUnit unit) { // 新版本 API 中已经直接支持 set nx 方式 Boolean success = stringRedisTemplate.opsForValue().setIfAbsent(lockKey, uuid, timeout, unit); return success == null ? false : success; }
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你好,非常喜欢您的分享,在运行chapter22一章的demo时发现了两个问题,不知道是不是遗留的bug,如理解有问题,请指正😄
这里的
String value = UUID.randomUUID().toString();
是不是有多线程问题,当第一个线程未处理完,第二个线程运行到这里是会刷新value,导致第一个线程unlock时会使用第二个线程的value我认为这里应该这样写
如理解有问题请指正😄
The text was updated successfully, but these errors were encountered: