1. How to upload a folder to S3 without adding the bucket name prefix?


    Sure, I'll help you write a Pulumi program for this.

    We can use the aws.s3.BucketObject resource from the Pulumi AWS package to upload a file to an S3 bucket. However, uploading a folder requires handling multiple files. For this, we'll need to use Python's built-in os library to list all files in the folder and create a BucketObject for each file.

    Here's how you can accomplish this:

    import pulumi import pulumi_aws as aws import os # Define the S3 bucket bucket = aws.s3.Bucket("my-bucket") # Directory to upload directory_path = "path/to/your/directory" # iterate through all files in the directory for file in os.listdir(directory_path): # ensure we're dealing with a file and not a subdirectory if os.path.isfile(os.path.join(directory_path, file)): # Create an S3 object for each file obj = aws.s3.BucketObject(file, bucket=bucket.id, # reference to the S3 bucket source=pulumi.FileAsset(os.path.join(directory_path, file)), # path to file key=file, # use file name as s3 object key (removes directory prefix) ) # Export the name of the bucket pulumi.export("bucket", bucket.id)

    Note: This program uploads all files in the specified directory to the root of the S3 bucket, maintaining the filename but omitting the directory prefix entirely. Individual files will have their original names, but the bucket won't contain the directory structure from your local filesystem.

    For more details on how to use aws.s3.BucketObject and pulumi.FileAsset, refer to the Pulumi AWS Package documentation and Pulumi Programming Model.