Question

我正在使用Python和AWS Lambda检索DynamoDB项目。我无法解析响应包含的值。

例如，这会返回以回复[＆＃39;项目＆＃39;] [＆＃39; GUID＆＃39;]

{u＆＃39;：＆＃39; 8898f389-c282-4c4f-952a-87a0fbbb6d70＆＃39;}

最后，我只想要没有DynamoDB插入的无关信息的实际值。您如何使用Python 2.7最好地处理DynamoDB的JSON格式？

Answer 1

这是一个例子：

def lambda_handler(event, context):

        print("Received event: " + json.dumps(event, indent=2))

        if 'Records' not in event:
            print ('records not in event')
            return

        for record in event['Records']:
            if record['eventName'] == 'INSERT':
                print 'do something on insert'

            your_integer_hash_key = record['dynamodb']['Keys']['your_hash_key']['N']

            # if all image is stream
            if not 'NewImage' in record['dynamodb']:
                continue

            # get new image    
            new_image = record['dynamodb']['NewImage']

AWS Lambda

1 个答案: