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

Question

Pulumi · Accepted Answer

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:

```python
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](https://www.pulumi.com/registry/packages/aws/api-docs/s3/bucketobject/) and [Pulumi Programming Model](https://www.pulumi.com/docs/intro/concepts/programming-model/#fileasset).