aws-amplify / aws-amplify/amplify-cli

Amplify function specify source location

Open
#2,231 7 comments 10 reactions 0 assignees View on GitHub
feature-request functions p4
Dominant language
TypeScript
Stars
2.9k
Forks
825
Avg merge
11d 23h
Merged PRs (30d)
2

Description

>**Note**: If your question is regarding the AWS Amplify Console service, please log it in the
[official AWS Amplify Console forum](https://forums.aws.amazon.com/forum.jspa?forumID=314&start=0)

** Which Category is your question related to? **
Amplify Cli functions
** What AWS Services are you utilizing? **
Appysync, dynamo, lambda
** Provide additional details e.g. code snippets **
I couldn't find my answer in the docs for amplify or in the more general aws docs for functions.

I have an amplify project with a function that I originally created with all the code in `src/` but I have now added webpack to package up the function code which outputs to `src/bin`. How do I tell amplify to target that specific directory to create the zip that it puts in the `dist/` dir?
i.e I would like to zip the output of webpack not the src.

It looks like it runs my script `"amplify:my-function": "npm run lint && npm run webpack"`

I tried changing the lambda function aws:asset:path in the medatadate section in cloud formation template but I don't think its used here

Contributor guide

Open the contributing guide

Assessment

This issue has not been assessed yet.

Get new issues in your inbox

A short digest of beginner-friendly GitHub issues.