Question

我想使用SDK Amazon的batchWriteItem方法将很多项放入表中。我从Kinesis那里收回物品，并且它有很多碎片。我将这种方法用于一个项目：

public static void addSingleRecord(Item thingRecord) {


    // Add an item
    try
    {

        DynamoDB dynamo = new DynamoDB(dynamoDB); 
        Table table = dynamo.getTable(dataTable);
        table.putItem(thingRecord);

    } catch (AmazonServiceException ase) {
        System.out.println("addThingsData request  "
                + "to AWS was rejected with an error response for some reason.");
        System.out.println("Error Message:    " + ase.getMessage());
        System.out.println("HTTP Status Code: " + ase.getStatusCode());
        System.out.println("AWS Error Code:   " + ase.getErrorCode());
        System.out.println("Error Type:       " + ase.getErrorType());
        System.out.println("Request ID:       " + ase.getRequestId());
    } catch (AmazonClientException ace) {
        System.out.println("addThingsData - Caught an AmazonClientException, which means the client encountered "
                + "a serious internal problem while trying to communicate with AWS, "
                + "such as not being able to access the network.");
        System.out.println("Error Message: " + ace.getMessage());
    }
}

public static void addThings(String thingDatum) {
    Item itemJ2;
    itemJ2 = Item.fromJSON(thingDatum);
    addSingleRecord(itemJ2);

}

该项目来自：

private void processSingleRecord(Record record) {
    // TODO Add your own record processing logic here


    String data = null;
    try {


        // For this app, we interpret the payload as UTF-8 chars.
        data = decoder.decode(record.getData()).toString();
        System.out.println("**processSingleRecord - data " + data);
        AmazonDynamoDBSample.addThings(data);

    } catch (NumberFormatException e) {
        LOG.info("Record does not match sample record format. Ignoring record with data; " + data);
    } catch (CharacterCodingException e) {
        LOG.error("Malformed data: " + data, e);
    }
}

现在，如果我想要记录很多，我将使用：

    public static void writeMultipleItemsBatchWrite(Item thingRecord) {
    try {                    


            dataTableWriteItems.addItemToPut(thingRecord);
        System.out.println("Making the request.");
        BatchWriteItemOutcome outcome = dynamo.batchWriteItem(dataTableWriteItems);
        do {

                // Check for unprocessed keys which could happen if you exceed provisioned throughput

                Map<String, List<WriteRequest>> unprocessedItems = outcome.getUnprocessedItems();

                if (outcome.getUnprocessedItems().size() == 0) {
                    System.out.println("No unprocessed items found");
                } else {
                    System.out.println("Retrieving the unprocessed items");
                    outcome = dynamo.batchWriteItemUnprocessed(unprocessedItems);
                }

            } while (outcome.getUnprocessedItems().size() > 0);

    }  catch (Exception e) {
        System.err.println("Failed to retrieve items: ");
        e.printStackTrace(System.err);
    }  

}

但是如何发送最后一组呢？因为只有当我有25件物品时我才发送，但最后这个号码更低。

Answer 1

您可以使用PutItem或UpdateItem附加到Document SDK的Lambda function中的Kinesis Stream一次一个地向DynamoDB表中写入项目。这样，您可以对流记录中显示的流记录作出反应，而无需担心是否还有其他记录需要处理。在幕后，BatchWriteItem消耗与相应的PutItem调用相同的写入容量单位。 BatchWriteItem将与批量中的PUT一样潜伏，耗时最长。因此，使用BatchWriteItem，您可能会遇到比并行PutItem / UpdateItem调用更高的平均延迟。

通过Java代码将多个项目放入DynamoDB

1 个答案: