Question

我正在尝试在AWS中使用Secrets Manager的Lambda函数。管理员的秘密用于将数据库凭据存储到Snowflake（用户名，密码）。

我设法在Secrets Manager中设置了一个秘密，其中包含几个键/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文档请提供以下代码段：

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，我具有以下代码段可建立与数据库的连接：

        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()函数返回诸如USERNAME或PASSWORD之类的参数的值。

这如何完成？谢谢您的帮助！

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，您将在其中拥有在AWS Secret Manager控制台中指定的密钥。

Answer 2

hi @Prashanth kumar你的意思是这样的： def get_secret（）：

secret_name = "MySecretForRedshift"
region_name = "us-east-1"

# 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 = json.loads(get_secret_value_response['SecretString'])
    else:
        secret = json.loads(base64.b64decode(get_secret_value_response['SecretBinary']))
return secret

Answer 3

这是我使用arn的方式，following this bloc希望对您有所帮助。
值得检查一下您曾经存储过的内容，并相应地使用了一个 SecretString或SecretBinary

    secrets_client = boto3.client('secretsmanager')
    secret_arn = 'arn:aws:secretsmanager:eu-west-2:xxxxxxxxxxxx:secret:dashboard/auth_token'
    auth_token = secrets_client.get_secret_value(SecretId=secret_arn).get('SecretString')

boto3 docs
get_secret_value从机密的指定版本中检索加密字段SecretString或SecretBinary的内容，以任何内容为准。
根据使用的内容，您的lambda角色应具有以下权限
- secretsmanager:GetSecretValue
- kms:Decrypt仅在您使用客户管理的AWS KMS密钥加密机密时才需要。您不需要此权限即可将账户的默认AWS管理CMK用于Secrets Manager。

Answer 4

我创建了一个名为pysecret的开源库，这是AWS Secret Manager集成的文档：https://github.com/MacHu-GWU/pysecret-project#aws-key-management-service-and-secret-manager-integration

我可以引导您完成最简单的方法：

手动将您的秘密值放入json或使用 pysecret创建一个。

from pysecret import AWSSecret

aws_profile = "my_aws_profile"
aws = AWSSecret(profile_name=aws_profile)

secret_id = "my-example-secret"
secret_data = {
    "host": "www.example.com",
    "port": 1234,
    "database": "mydatabase",
    "username": "admin",
    "password": "mypassword",
    "metadata": {
        "creator": "Alice"
    }
}
aws.deploy_secret(name=secret_id, secret_data=secret_data) # or you can pass kms_key_id if you created a custom kms key

然后，您应该可以看到在aws控制台中创建的机密。

在lambda函数或任何python代码中读取您的秘密值。

aws = AWSSecret(profile_name=aws_profile) # in lambda code, don't need ``profile_name=aws_profile``
password = aws.get_secret_value(secret_id="my-example-secret", key="password") # mypassword
creator = aws.get_secret_value(secret_id="my-example-secret", key="metadata.creator") # Alice

注意，Lambda功能IAM角色要求访问机密

您的Lambda IAM角色必须具有Secret Manager的读取访问权限。 aws内置策略arn：aws：iam :: aws：policy / SecretsManagerReadWrite具有读写功能，如果您很懒，可以使用它。但是我建议您创建仅具有“读取”访问权限的自定义策略。
如果您使用自动生成的kms密钥作为秘密，则必须使用aws kms create-grant命令授予Lambda Function IAM角色访问kms密钥进行加密，这就是https://docs.aws.amazon.com/cli/latest/reference/kms/create-grant.html
如果您使用自定义kms密钥，则应该能够编辑kms密钥的用户，请在aws控制台中选择Lambda Function IAM角色。

希望这能回答您的问题。

如果有此帮助，请为我的项目https://github.com/MacHu-GWU/pysecret-project加上星号。

Answer 5

AWS为某些受支持的数据库引擎（例如MySQL等）提供了一些模板，请查看以下template：

对于不受支持的数据库，请检查this

上面提供的模板将为您提供一个自定义功能的示例。

结合使用AWS Secrets Manager和Python（Lambda Console）

5 个答案: