Question

我的服务结构：

-MyService
    -common
    -node_modules
    -functions_folder
        -Function1.js
        -Function2.js
        -Function3.js

yaml文件：

service: MyService

provider:
  name: aws
  runtime: nodejs6.10
  stage: dev

functions:
    Function1:
       handler: functions_folder/Function1.handler
       memorySize: 512
       timeout: 10

    Function2:
       handler: functions_folder/Function2.handler
       memorySize: 512
       timeout: 10

     Function2:
       handler: functions_folder/Function3.handler
       memorySize: 512
       timeout: 10

当我正在部署时，我有3个不同的lambda函数，但每个函数都包含Function1.js，Function2.js，Function3.js。

有人可以解释一下如何排除Lambda不需要的文件吗？

Answer 1

经过一段时间的研究，我找到了解决方案。所以这就是：

service: MyService

package:
  individually: true
  exclude:
    - ./**
  include:
    - common/**
    - node_modules/**

provider:
  name: aws
  runtime: nodejs6.10
  stage: dev
  memorySize: 512
  timeout: 10

functions:
    Function1:
       handler: functions_folder/Function1.handler
       package:
         include:
           - functions_folder/Function1.js

    Function2:
       handler: functions_folder/Function2.handler
       package:
         include:
           - functions_folder/Function2.js

     Function2:
       handler: functions_folder/Function3.handler
       package:
         include:
           - functions_folder/Function3.js

因此，您可以在包部分中看到我添加了包含/排除部分，首先，我排除了所有文件，然后我将包含2个所需的文件夹＆＃34; common＆＃34;和＆＃34; node_modules＆＃34;。在此之后，我还使用include命令来添加所需的文件。

无服务器框架。从包中排除不需要的功能

1 个答案: