KafkaConsumer分析

一 重要的字段

String clientId：Consumer惟一標識

ConsumerCoordinator coordinator： 控制Consumer與服務器端GroupCoordinator之間的通訊邏輯

Fetcher<K, V> fetcher： 負責從服務器端獲取消息的組件，而且更新partition的offset

ConsumerNetworkClient client: 負責和服務器端通訊

SubscriptionState subscriptions: 便於快速獲取topic partition等狀態，維護了消費者消費狀態

Metadata metadata: 集羣元數據信息

AtomicLong currentThread: 當前使用KafkaConsumer的線程id

AtomicInteger refcount: 重入次數

二 核心的方法2.1 subscribe 訂閱主題

訂閱給定的主題列表，以得到動態分配的分區

主題的訂閱不是增量的，這個列表將會代替當前的分配。注意，不可能將主題訂閱與組管理與手動分區分配相結合

做爲組管理的一部分，消費者將會跟蹤屬於某一個特殊組的消費者列表，若是知足在下列條件，將會觸發再平衡操做:1 訂閱的主題列表的那些分區數量的改變2 主題建立或者刪除3 消費者組的成員掛了4 經過join api將一個新的消費者添加到一個存在的消費者組public void subscribe(Collection<String> topics, ConsumerRebalanceListener listener) {

// 取得一把鎖    acquire();
try {
    if (topics == null) { // 主題列表爲null，拋出異常            throw new IllegalArgumentException("Topiccollection to subscribe to cannot be null");
    } else if (topics.isEmpty()) {// 主題列表爲空，取消訂閱            this.unsubscribe();
    } else {
        for (String topic : topics) {
            if (topic == null || topic.trim().isEmpty())
                throw new IllegalArgumentException("Topic collection to subscribe to cannot contain null or emptytopic");
        }
        log.debug("Subscribed to topic(s):{}", Utils.join(topics, ", "));
        this.subscriptions.subscribe(new HashSet<>(topics), listener);
        // 用新提供的topic集合替換當前的topic集合，若是啓用了主題過時，主題的過時時間將在下一次更新中從新設置。            metadata.setTopics(subscriptions.groupSubscription());
    }
} finally {
    // 釋放鎖        release();
}

}2.2 assign 手動分配分區

對於用戶手動指定topic的訂閱模式，經過此方法能夠分配分區列表給一個消費者：public void assign(Collection<TopicPartition> partitions) {

acquire();
try {
    if (partitions == null) {
        throw new IllegalArgumentException("Topic partition collection to assign to cannot be null");
    } else if (partitions.isEmpty()) {// partition爲空取消訂閱            this.unsubscribe();
    } else {
        Set<String> topics = new HashSet<>();
        // 遍歷TopicPartition，把topic添加到一個集合裏            for (TopicPartition tp : partitions) {
            String topic = (tp != null) ? tp.topic() : null;
            if (topic == null || topic.trim().isEmpty())
                throw new IllegalArgumentException("Topic partitions to assign to cannot have null or empty topic");
            topics.add(topic);
        }

        // 進行一次自動提交            this.coordinator.maybeAutoCommitOffsetsNow();

        log.debug("Subscribed to partition(s): {}", Utils.join(partitions, ", "));
        // 根據用戶提供的指定的partitions 改變assignment            this.subscriptions.assignFromUser(new HashSet<>(partitions));
        metadata.setTopics(topics);// 更新metatdata topic        }
} finally {
    release();
}

}2.3 commitSync & commitAsync 提交消費者已經消費完的消息的offset,爲指定已訂閱的主題和分區列表返回最後一次poll返回的offsetpublic void commitSync(final Map<TopicPartition, OffsetAndMetadata> offsets) {

acquire();
try {
    coordinator.commitOffsetsSync(offsets);
} finally {
    release();
}

}
public void commitAsync(final Map<TopicPartition, OffsetAndMetadata> offsets, OffsetCommitCallback callback) {

acquire();
try {
    log.debug("Committing offsets: {} ", offsets);
    coordinator.commitOffsetsAsync(new HashMap<>(offsets), callback);
} finally {
    release();
}

}2.4 seek 指定消費者消費的起始位置public void seek(TopicPartition partition, long offset) {

if (offset < 0) {
    throw new IllegalArgumentException("seek offset must not be a negative number");
}
acquire();
try {
    log.debug("Seeking to offset {} for partition {}", offset, partition);
    this.subscriptions.seek(partition, offset);
} finally {
    release();
}

}// 爲指定的分區查找第一個offsetpublic void seekToBeginning(Collection<TopicPartition> partitions) {

acquire();
try {
    Collection<TopicPartition> parts = partitions.size() == 0 ? this.subscriptions.assignedPartitions() : partitions;
    for (TopicPartition tp : parts) {
        log.debug("Seeking to beginning of partition {}", tp);
        subscriptions.needOffsetReset(tp, OffsetResetStrategy.EARLIEST);
    }
} finally {
    release();
}

}// 爲指定的分區查找最後的offsetpublic void seekToEnd(Collection<TopicPartition> partitions) {

acquire();
try {
    Collection<TopicPartition> parts = partitions.size() == 0 ? this.subscriptions.assignedPartitions() : partitions;
    for (TopicPartition tp : parts) {
        log.debug("Seeking to end of partition {}", tp);
        subscriptions.needOffsetReset(tp, OffsetResetStrategy.LATEST);
    }
} finally {
    release();
}

}2.5 poll方法 獲取消息

從指定的主題或者分區獲取數據，在poll以前，你沒有訂閱任何主題或分區是不行的,每一次poll,消費者都會嘗試使用最後一次消費的offset做爲接下來獲取數據的start offset，最後一次消費的offset也能夠經過seek(TopicPartition, long)設置或者自動設置public ConsumerRecords<K, V> poll(long timeout) {

acquire();
try {
    if (timeout < 0)
        throw new IllegalArgumentException("Timeout must not be negative");
    // 若是沒有任何訂閱，拋出異常        if (this.subscriptions.hasNoSubscriptionOrUserAssignment())
        throw new IllegalStateException("Consumer is not subscribed to any topics or assigned any partitions");

    // 一直poll新數據直到超時        long start = time.milliseconds();
    // 距離超時還剩餘多少時間        long remaining = timeout;
    do {
        // 獲取數據，若是自動提交，則進行偏移量自動提交，若是設置offset重置，則進行offset重置            Map<TopicPartition, List<ConsumerRecord<K, V>>> records = pollOnce(remaining);
        if (!records.isEmpty()) {
            // 再返回結果以前，咱們能夠進行下一輪的fetch請求，避免阻塞等待                fetcher.sendFetches();
            client.pollNoWakeup();
            // 若是有攔截器進行攔截，沒有直接返回                if (this.interceptors == null)
                return new ConsumerRecords<>(records);
            else                    return this.interceptors.onConsume(new ConsumerRecords<>(records));
        }

        long elapsed = time.milliseconds() - start;
        remaining = timeout - elapsed;
    } while (remaining > 0);

    return ConsumerRecords.empty();
} finally {
    release();
}

}private Map<TopicPartition, List<ConsumerRecord<K, V>>> pollOnce(long timeout) {

// 輪詢coordinator事件，處理週期性的offset提交    coordinator.poll(time.milliseconds());

// fetch positions if we have partitions we're subscribed to that we    // don't know the offset for    // 判斷上一次消費的位置是否爲空,若是不爲空，則    if (!subscriptions.hasAllFetchPositions())
    // 更新fetch position        updateFetchPositions(this.subscriptions.missingFetchPositions());

// 數據你準備好了就當即返回，也就是還有可能沒有準備好    Map<TopicPartition, List<ConsumerRecord<K, V>>> records = fetcher.fetchedRecords();
if (!records.isEmpty())
    return records;

// 咱們須要發送新fetch請求    fetcher.sendFetches();

long now = time.milliseconds();
long pollTimeout = Math.min(coordinator.timeToNextPoll(now), timeout);

client.poll(pollTimeout, now, new PollCondition() {
    @Override        public boolean shouldBlock() {
        // since a fetch might be completed by the background thread, we need this poll condition            // to ensure that we do not block unnecessarily in poll()            return !fetcher.hasCompletedFetches();
    }
});
// 早長時間的poll以後，咱們應該在返回數據以前檢查是否這個組須要從新平衡，以致於這個組可以迅速的穩定    if (coordinator.needRejoin())
    return Collections.emptyMap();
// 獲取返回的消息    return fetcher.fetchedRecords();

}2.6 pause 暫停消費者，暫停後poll返回空public void pause(Collection<TopicPartition> partitions) {

acquire();
try {
    for (TopicPartition partition: partitions) {
        log.debug("Pausing partition {}", partition);
        subscriptions.pause(partition);
    }
} finally {
    release();
}

}// 返回暫停的分區public Set<TopicPartition> paused() {

acquire();
try {
    return Collections.unmodifiableSet(subscriptions.pausedPartitions());
} finally {
    release();
}

}2.7 resume 恢復消費者public void resume(Collection<TopicPartition> partitions) {

acquire();
try {
    for (TopicPartition partition: partitions) {
        log.debug("Resuming partition {}", partition);
        subscriptions.resume(partition);
    }
} finally {
    release();
}

}2.8 position方法 獲取下一個消息的offset// 獲取下一個record的offsetpublic long position(TopicPartition partition) {

acquire();
try {
    if (!this.subscriptions.isAssigned(partition))
        throw new IllegalArgumentException("You can only check the position for partitions assigned to this consumer.");
    Long offset = this.subscriptions.position(partition);
    if (offset == null) {
        updateFetchPositions(Collections.singleton(partition));
        offset = this.subscriptions.position(partition);
    }
    return offset;
} finally {
    release();
}

}