Using AWS Secrets Manager with Python (Lambda Console)

Question

I am attempting to use Secrets Manager a Lambda function in AWS. Secrets a manager is used to store database credentials to Snowflake (username, password).

I managed to set up a secret in Secrets Manager which contains several key/value pairs (e.g. one for username, another for password).

Now I am trying to refer to these values in my Python function code. AWS documentation kindly provides the following snippet:

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.

Later in my def lambda_handler(event, context) function, I have the following snippet to establish a connection to my database:

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

However, I am unable to figure out how to use the get_secret() function to return values for parameters like USERNAME or PASSWORD.

How can this be accomplished? Thank you for the help!

Calling the function inside the handler function isn't working? — Munavir Chavody
– Munavir Chavody, Commented Sep 19, 2019 at 6:20

Ron Marcelino · Accepted Answer · 2019-11-08 10:18:04Z

30

update the last part of get_secret() to:

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

This will return a dictionary where you'll have the keys you specified in AWS Secret Manager console.

answered Nov 8, 2019 at 10:18

Ron Marcelino

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5 Comments

django-unchained Over a year ago

@Prashanthkumar Did you get dictionary just by adding the return? I am struggling to get my credentials.

Prashanth kumar Over a year ago

@django-unchained, hope you got it covered already, but otherwise, I just enclosed the base64.b64decode(get_secret_value_response['SecretBinary']) inside json.loads when assinging it to variable "secret", after that I could access the credentials as secret["username"] secret["password"], or whatever your variables are inside the secrets manager secret.

Denise Over a year ago

can u provide your code ? im stuck in this situation for days

Sarde Over a year ago

I am getting following error [ERROR] UnboundLocalError: local variable 'secret' referenced before assignment

gcj Over a year ago

@Sarde Hello I am getting same error, do you manage to sort it out? thanks

amittn · Accepted Answer · 2019-09-19 13:20:55Z

Here is how i have used it using arn, following this bloc hope that helps you.
Worth checking what you have used to store and accordingly use one SecretString or 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 Retrieves the contents of the encrypted fields SecretString or SecretBinary from the specified version of a secret, whichever contains content.
Your lambda role should have the following permissions depending on what is used
- secretsmanager:GetSecretValue
- kms:Decrypt required only if you use a customer-managed AWS KMS key to encrypt the secret. You do not need this permission to use the account's default AWS managed CMK for Secrets Manager.

gogogadgetprime · Accepted Answer · 2024-03-15 20:42:35Z

The code below worked well for me! Make sure your lambda function has the permissions to "listsecrets", and "GetsecretValue":

"{ "Sid": "Statement1", "Effect": "Allow", "Action": [ "secretsmanager:ListSecrets", "secretsmanager:GetSecretValue" ], "Resource": [ "arn:aws:secretsmanager:........" ] } ] }"

Lambda code:

import boto3 import json def lambda_handler(event, context): secrets_client = boto3.client('secretsmanager') secret_name = 'Insert_secret_name_here' secret_response = secrets_client.get_secret_value(SecretId=secret_name).get('SecretString') print(secret_response) secret_list = json.loads(secret_response) username = secret_list.get('username') print(username)

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