How to get CodeBuild project name from buildspec file - amazon-web-services

Question

This is my buildspec file to build an Angular, React, Vue or similar project through AWS CodeCommit and publish the resulting artifact to an S3 bucket:
version: 0.2
env:
variables:
S3_BUCKET: "my-bucket"
phases:
install:
runtime-versions:
nodejs: 16
pre_build:
commands:
- echo Installing source NPM dependencies...
- npm install
build:
commands:
- echo Build started on `date`
- npm run build
post_build:
commands:
- aws s3 cp dist s3://${S3_BUCKET} --recursive
- echo Build completed on `date`
What I would like to do is to use a subfolder with the name of the project when publishing the result files in the bucket. Now all files go to my-bucket but I would like them to go to my-bucket/name-of-the-project
I could change the post-build command to something like
- aws s3 cp dist s3://${S3_BUCKET}/name-of-the-project --recursive
That way it would be always the same directory name. What I want is to get dynamically the name of the CodeBuild project or from the package.json or similar to make that directory match the project name.

Answer 1

Here are two ways to read a project identifier from the build context at run-time:
Option 1: Read the project name from package.json:
PROJECT_NAME=$(cat package.json | jq -r '.name')
echo $PROJECT_NAME # -> name-of-the-project
Option 2: Extract the CodeCommit repo name from the source URL. Each CodeBuild execution exposes several environment variables, including CODEBUILD_SOURCE_REPO_URL.
echo $CODEBUILD_SOURCE_REPO_URL # -> https://git-codecommit.us-east-1.amazonaws.com/v1/repos/my-repo
REPO_NAME=$(echo $CODEBUILD_SOURCE_REPO_URL | awk -F\"/\" '{print $NF}') # split the url at '/', return the last item
echo $REPO_NAME # -> my-repo
Pass one of the captured names to the S3 command:
aws s3 cp dist s3://${S3_BUCKET}/${PROJECT_NAME} --recursive