AWS山姆构建而不是构建lambda函数

Question

我刚开始用AWS SAM做建筑。当我执行sam build时，它显示生成成功，但我没有在build目录中看到我的函数。这是我的目录结构

Folder PATH listing for volume Code
Volume serial number is B243-6647
D:.
ª   .gitignore
ª   template.yaml
ª   tree.txt
ª   
+---.aws-sam
ª   ª   build.toml
ª   ª   
ª   +---build
ª           template.yaml
ª           
+---events
ª       testHelloWorld.json
ª       
+---lambdas
ª   +---HelloWorld
ª           index.py
ª           __init__.py
ª           

这是template.yaml的内容

AWSTemplateFormatVersion: '2010-09-09'
Transform: AWS::Serverless-2016-10-31
Globals:
  Function:
    Runtime: python3.9
    Handler: index.handler

Resources:
  HelloWorldFunction:
    Properties:
      Type: AWS::Serverless::Function
      CodeUri: lambdas/HelloWorld/

    Policies:
      - AWSLambdaExecute
      - Version: '2012-10-17'

这是lambdas/HelloWorld/index.py

import json

def handler(event, context):
  return json.dumps(event, indent=2)

这是sam build之后构建文件夹的状态

Folder PATH listing for volume Code
Volume serial number is B243-6647
D:\WEB\URLSHORTENER\.AWS-SAM
│   build.toml
│
└───build
        template.yaml

有人能帮我弄清楚为什么这些功能不可用吗？

Answer 1

我几乎把一切都做好了。只需在Type之外添加Properties :/ Updatedtemplate.yaml

AWSTemplateFormatVersion: '2010-09-09'
Transform: AWS::Serverless-2016-10-31
Description: Url Shortener

Globals:
  Function:
    Handler: index.handler
    Runtime: python3.9

Resources:
  HelloWorldFunction:
    Type: AWS::Serverless::Function
    Properties:
      CodeUri: lambdas/HelloWorld/
    Policies:
      - AWSLambdaExecute
      - Version: '2012-10-17'