无服务器框架从单独的 function 获取 AWS“函数 url”

Question

I have a serverless application defined like this

service: hello_world

frameworkVersion: '3'

provider:
  name: aws
  runtime: python3.8

functions:
  hello:
    handler: handler.run
  world:
    handler: handler2.run
    url: true

Part of the function hello is to hand the function url of world to an external api.

How can I get the function url to hello , as an environment variable or otherwise?

Answer 1

Have not tested, but according to the following you can use what looks like cloudformation fn + backreferences

https://forum.serverless.com/t/how-do-i-get-the-url-for-a-function-in-my-serverless-yml-file/1386/4

custom:
  stage: ${opt:stage, self:provider.stage}
  region: ${opt:region, self:provider.region}

environment:
  URL: { "Fn::Join" : ["", [" https://", { "Ref" : "ApiGatewayRestApi" }, ".execute-api.${self:custom.region}.amazonaws.com/${self:custom.stage}/path/to/resource" ] ]  }

Answer 2

You need to know how Serverless is internally referencing the function URLs, which should be accessable in this format. You can check this by viewing the resource names in AWS.

service: hello_world

frameworkVersion: '3'

provider:
  name: aws
  runtime: python3.8

functions:
  fn1:
    handler: handler.run
    url: true
  fn2:
    handler: handler2.run
    environment:
      GRANT_URL_ENDPOINT: !Ref Fn1LambdaFunctionUrl