amazon-web-services - CloudFormation模板中的错误到底在哪里？

Question

我面临的事实是我的 CloudFormation 模板大于 51200 字节。所以我决定把这个主模板分成几个模板。当我尝试构建堆栈时，收到错误消息，指出嵌套模板中的资源属性不正确。

问题:

嵌套模板是否可能看不到主模板中声明的参数值？

错误:

Invalid template resource property 'CreateWidgetRole'

template.yaml(片段):

AWSTemplateFormatVersion: '2010-09-09'
Transform: 'AWS::Serverless-2016-10-31'
Parameters:
  EnvironmentName:
    Type: String
  SettingsServiceStackName:
    Type: String
  S3Bucket:
    Type: String
Conditions:
  IsProductionEnvironment:
    'Fn::Equals':
      - Ref: EnvironmentName
      - Prod
Resources:
  SettingsServiceResources:
    'Fn::Transform':
      Name: 'AWS::Include'
      Parameters:
        Location:
          'Fn::Sub': >-
            s3://${S3Bucket}/included_templates/settings_service_resources_template.yaml

settings_service_resources_template.yaml:

CreateWidgetRole:
  Type: 'AWS::IAM::Role'
  Properties:
    AssumeRolePolicyDocument:
      Version: '2012-10-17'
      Statement:
        - Effect: Allow
          Principal:
            Service: appsync.amazonaws.com
          Action: 'sts:AssumeRole'
    Policies:
      - PolicyName: CreateWidgetFunctionAccess
        PolicyDocument:
          Version: '2012-10-17'
          Statement:
            - Effect: Allow
              Action: 'lambda:InvokeFunction'
              Resource:
                'Fn::ImportValue':
                  'Fn::Sub': >-
                    ${EnvironmentName}${SettingsServiceStackName}:${EnvironmentName}CreateWidgetARN
CreateWidgetDataSource:
  Type: 'AWS::AppSync::DataSource'
  Properties:
    ApiId:
      'Fn::GetAtt': GraphQLApi.ApiId
    Name:
      'Fn::Sub': '${EnvironmentName}CreateWidget'
    Type: AWS_LAMBDA
    LambdaConfig:
      LambdaFunctionArn:
        'Fn::ImportValue':
          'Fn::Sub': >-
            ${EnvironmentName}${SettingsServiceStackName}:${EnvironmentName}CreateWidgetARN
    ServiceRoleArn:
      'Fn::GetAtt': CreateWidgetRole.Arn
CreateWidgetResolver:
  Type: 'AWS::AppSync::Resolver'
  Properties:
    ApiId:
      'Fn::GetAtt': GraphQLApi.ApiId
    FieldName: createWidget
    TypeName: Mutation
    DataSourceName:
      'Fn::GetAtt': CreateWidgetDataSource.Name

Answer 1

问题出在 template.yaml 中的 SettingsServiceResources 行。嵌套模板中的所有资源都有逻辑 ID(例如 CreateWidgetRole)，因此无需在 Transform block 前面放置另一个 ID。以下内容对我有用:

Resources:
  'Fn::Transform':
    Name: 'AWS::Include'
    Parameters:
      Location:
        'Fn::Sub': >-
          s3://${S3Bucket}/included_templates/settings_service_resources_template.yaml

您还可以检查docs中的语法。 .