Vik*_*kas 10 amazon-web-services aws-lambda aws-lambda-layers
我正在开发的应用程序中有超过 20 个 lambda 函数。还有一个 lambda 层,其中包含大量通用代码。
Lambda 函数将其挂钩到该层的特定版本,每次更新层时,它都会生成一个新版本。由于它是一个正在开发的应用程序,我几乎每天都有该层的新版本。这会导致 lambda 函数变得混乱,而这些函数每天都必须接触——以升级层版本。
我知道在生产环境中冻结 lambda 函数的代码非常重要,并且必须将 lambda 函数的一个版本挂接到该层的一个版本。
但是,对于开发环境来说,是否可以防止每次更新图层时都生成新的图层版本呢?或者配置 lambda 函数,使最新的 lambda 版本始终引用最新的层版本?
从 @Chris 答案中得到增强,您还可以在堆栈中使用 lambda 支持的自定义资源,并使用此 lambda 使用新层 ARN 更新目标配置。我注意到这一点,以防几天前我发现这个帖子时有人有类似的需求。
此解决方案有一些注释:
这是示例代码
AWSTemplateFormatVersion: '2010-09-09'
Transform: AWS::Serverless-2016-10-31
Description: >
myshared-libraries layer
Resources:
LambdaLayer:
Type: AWS::Serverless::LayerVersion
Properties:
LayerName: !Sub MyLambdaLayer
Description: Shared library layer
ContentUri: my_layer/layerlib.zip
CompatibleRuntimes:
- python3.7
ConsumerUpdaterLambda:
Type: AWS::Serverless::Function
Properties:
FunctionName: consumer-updater
InlineCode: |
import os, boto3, json
import cfnresponse
def handler(event, context):
print('EVENT:[{}]'.format(event))
if event['RequestType'].upper() == 'UPDATE':
shared_layer = os.getenv("DB_LAYER")
lambda_client = boto3.client('lambda')
consumer_lambda_list = ["target_lamda"]
for consumer in consumer_lambda_list:
try:
lambda_name = consumer.split(':')[-1]
lambda_client.update_function_configuration(FunctionName=consumer, Layers=[shared_layer])
print("Updated Lambda function: '{0}' with new layer: {1}".format(lambda_name, shared_layer))
except Exception as e:
print("Lambda function: '{0}' has exception: {1}".format(lambda_name, str(e)))
responseValue = 120
responseData = {}
responseData['Data'] = responseValue
cfnresponse.send(event, context, cfnresponse.SUCCESS, responseData)
Handler: index.handler
Runtime: python3.7
Role: !GetAtt ConsumerUpdaterRole.Arn
Environment:
Variables:
DB_LAYER: !Ref LambdaLayer
ConsumerUpdaterRole:
Type: AWS::IAM::Role
Properties:
Path: /
AssumeRolePolicyDocument:
Version: '2012-10-17'
Statement:
- Effect: Allow
Principal:
Service: lambda.amazonaws.com
Action: sts:AssumeRole
ManagedPolicyArns:
- Fn::Sub: arn:${AWS::Partition}:iam::aws:policy/service-role/AWSLambdaBasicExecutionRole
Policies:
- PolicyName:
Fn::Sub: updater-lambda-configuration-policy
PolicyDocument:
Version: '2012-10-17'
Statement:
- Effect: Allow
Action:
- lambda:GetFunction
- lambda:GetFunctionConfiguration
- lambda:UpdateFunctionConfiguration
- lambda:GetLayerVersion
- logs:DescribeLogGroups
- logs:CreateLogGroup
Resource: "*"
ConsumerUpdaterMacro:
DependsOn: ConsumerUpdaterLambda
Type: Custom::ConsumerUpdater
Properties:
ServiceToken: !GetAtt ConsumerUpdaterLambda.Arn
DBLayer: !Ref LambdaLayer
Outputs:
SharedLayer:
Value: !Ref LambdaLayer
Export:
Name: MySharedLayer
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另一种选择是使用堆栈通知 ARN,它将所有堆栈事件发送到定义的 SNS,您将在其中使用它来触发更新 lambda。在 lambda 中,您将使用 AWS::Lambda::Layer 资源过滤 SNS 消息正文(这是一个可读的 json 格式字符串),然后获取层 ARN 的 PhysicalResourceId。如何将 SNS 主题纳入您的堆栈,请使用 CLI sam/cloudformation deploy --notification-arns 选项。不幸的是,CodePipeline 不支持此配置选项,因此您只能与 CLI 一起使用
用于使用资源数据提取/过滤 SNS 消息正文的 lambda 示例代码
import os, boto3, json
def handler(event, context):
print('EVENT:[{}]'.format(event))
resource_data = extract_subscription_msg(event['Records'][0]['Sns']['Message'])
layer_arn = ''
if len(resource_data) > 0:
if resource_data['ResourceStatus'] == 'CREATE_COMPLETE' and resource_data['ResourceType'] == 'AWS::Lambda::LayerVersion':
layer_arn = resource_data['PhysicalResourceId']
if layer_arn != '':
lambda_client = boto3.client('lambda')
consumer_lambda_list = ["target_lambda"]
for consumer in consumer_lambda_list:
lambda_name = consumer.split(':')[-1]
try:
lambda_client.update_function_configuration(FunctionName=consumer, Layers=[layer_arn])
print("Update Lambda: '{0}' to layer: {1}".format(lambda_name, layer_arn))
except Exception as e:
print("Lambda function: '{0}' has exception: {1}".format(lambda_name, str(e)))
return
def extract_subscription_msg(msg_body):
result = {}
if msg_body != '':
attributes = msg_body.split('\n')
for attr in attributes:
if attr != '':
items = attr.split('=')
if items[0] in ['PhysicalResourceId', 'ResourceStatus', 'ResourceType']:
result[items[0]] = items[1].replace('\'', '')
return result
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