python - 将 AWS Secrets Manager 与 Python 结合使用(Lambda 控制台)

Question

我正在尝试在 AWS 中使用 Secrets Manager 和 Lambda 函数。管理器用于将数据库凭据存储到 Snowflake 的 secret (用户名、密码)。

我设法在 Secrets Manager 中设置了一个 secret ，其中包含多个键/值对(例如，一个用于用户名，另一个用于密码)。

现在我试图在我的 Python 函数代码中引用这些值。 AWS 文档提供了以下片段:

import boto3
import base64
from botocore.exceptions import ClientError


def get_secret():

    secret_name = "MY/SECRET/NAME"
    region_name = "us-west-2"

    # Create a Secrets Manager client
    session = boto3.session.Session()
    client = session.client(
        service_name='secretsmanager',
        region_name=region_name
    )

    # In this sample we only handle the specific exceptions for the 'GetSecretValue' API.
    # See https://docs.aws.amazon.com/secretsmanager/latest/apireference/API_GetSecretValue.html
    # We rethrow the exception by default.

    try:
        get_secret_value_response = client.get_secret_value(
            SecretId=secret_name
        )
    except ClientError as e:
        if e.response['Error']['Code'] == 'DecryptionFailureException':
            # Secrets Manager can't decrypt the protected secret text using the provided KMS key.
            # Deal with the exception here, and/or rethrow at your discretion.
            raise e
        elif e.response['Error']['Code'] == 'InternalServiceErrorException':
            # An error occurred on the server side.
            # Deal with the exception here, and/or rethrow at your discretion.
            raise e
        elif e.response['Error']['Code'] == 'InvalidParameterException':
            # You provided an invalid value for a parameter.
            # Deal with the exception here, and/or rethrow at your discretion.
            raise e
        elif e.response['Error']['Code'] == 'InvalidRequestException':
            # You provided a parameter value that is not valid for the current state of the resource.
            # Deal with the exception here, and/or rethrow at your discretion.
            raise e
        elif e.response['Error']['Code'] == 'ResourceNotFoundException':
            # We can't find the resource that you asked for.
            # Deal with the exception here, and/or rethrow at your discretion.
            raise e
    else:
        # Decrypts secret using the associated KMS CMK.
        # Depending on whether the secret is a string or binary, one of these fields will be populated.
        if 'SecretString' in get_secret_value_response:
            secret = get_secret_value_response['SecretString']
        else:
            decoded_binary_secret = base64.b64decode(get_secret_value_response['SecretBinary'])

    # Your code goes here.

稍后在我的 def lambda_handler(event, context) 函数中，我有以下片段来建立与我的数据库的连接:

        conn = snowflake.connector.connect(
            user=USERNAME,
            password=PASSWORD,
            account=ACCOUNT,
            warehouse=WAREHOUSE,
            role=ROLE
            )

但是，我无法弄清楚如何使用 get_secret() 函数返回参数值，例如 USERNAME 或 PASSWORD。

如何做到这一点？感谢您的帮助!

Answer 1

将 get_secret() 的最后一部分更新为:

else:
        # Decrypts secret using the associated KMS CMK.
        # Depending on whether the secret is a string or binary, one of these fields will be populated.
        if 'SecretString' in get_secret_value_response:
            secret = get_secret_value_response['SecretString']
        else:
            secret = base64.b64decode(get_secret_value_response['SecretBinary'])

return json.loads(secret)  # returns the secret as dictionary

这将返回一个字典，其中包含您在 AWS Secret Manager 控制台中指定的 key 。