sam local invoke always use same image even I run sam build first when I update the source code.

[0417taehyun]

sam local invoke always use same image even I run sam build first when I update the source code.

• Mac OS: Pro 2019, Intel Core i7, Ventuar 13.3.1
• Docker: Docker Desktop for Mac 23.0.5
• AWS SAM CLI: 1.82.0
• Runtime: Python3.9

Here is my template.yaml file.

Transform: AWS::Serverless-2016-10-31
Description: A simple AWS Lambda

Resources:
  SimpleAWSLambda:
    Type: AWS::Serverless::Function
    Properties:
      Runtime: python3.9
      PackageType: Image
      ImageUri: simpleawslambda
      Timeout: 600
      MemorySize: 1024
      Environment:
        Variables:
          SLACK_BOT_TOKEN: "SLACK_BOT_TOKEN"
    Metadata:
      Dockerfile: Dockerfile
      DockerContext: ./
      DockerTag: v1

I run like the one below.

sam build -t template.yaml --use-container && \
sam local invoke -t template.yaml -e events/event.json --env-vars env.json

sam build create a new image when I change the source code. However, sam local invoke build a image which tag is rapid-x86_64, and I understand that this image is for lambda execution, which is made by sam local invoke command. But, I think if the sam build builds a new image, a followed sam local invoke builds a new rapid-x86_64 image. However, it didn't in my local environment.

Did I miss something? I want to invoke the AWS Lambda function on my local environment with my local Docker when I change my source code.

[0417taehyun]

It works if I add the command --force-image-build on sam local invoke. Is there any related information that I should know with Docker? I thought, it would be updated automatically when I push the new image.

[Tigatok]

@0417taehyun I think this thread answers the reason why it does that #5910