To convert all files in an AWS S3 bucket to zip files

Question

Accepted Answer

S3 is an object storage, it's not a file system. So you'll have to download the files first from S3, zip them and then upload again back to s3, if you don't do in-memory operation. Once you verify that zip upload is successful for all the objects, you can consider archiving/deleting the objects based on data criticality.

You can either do it through an EC2 instance or lambda function or may be your local machine(through CLI). With lambda there is a limitation of 900 seconds max time. Consider local machine for downloading/uploading as last option as that may add data transfer cost.

Hope this explanation helps and provide you a direction for how to move forward.

Comment here if you have additional questions.

Happy to help.

Abhishek

Answer

To convert all files in your s3 bucket into one single zip file you can use use AWS Lambda (Python) with the AWS SDK for Python (Boto3).

* The below is code to convert all content of bucket into one single zip file

```
import boto3
import zipfile
import io

s3 = boto3.client('s3')

def lambda_handler(event, context):
    source_bucket = 'your-source-bucket'
    target_bucket = 'your-target-bucket'
   
    # List objects in the source bucket
    response = s3.list_objects_v2(Bucket=source_bucket)
   
    if 'Contents' in response:
        objects = response.get('Contents', [])
       
        for obj in objects:
            key = obj.get('Key')
            if key:
                # Get object content
                response = s3.get_object(Bucket=source_bucket, Key=key)
                content = response['Body'].read()
               
                # Create a zip file in-memory
                zip_buffer = io.BytesIO()
                with zipfile.ZipFile(zip_buffer, 'w', zipfile.ZIP_DEFLATED) as zipf:
                    zipf.writestr(key, content)
               
                # Upload the zip file to the target bucket
                zip_buffer.seek(0)
                target_key = key + '.zip'
                s3.upload_fileobj(zip_buffer, target_bucket, target_key)
    else:
        print("No objects found in the source bucket.")

```

*  To covert each object into a zip file you can use this code. (If you have 10 objects, you will have 10 zip files)

```
import boto3
import zipfile
import io

s3 = boto3.client('s3')

def lambda_handler(event, context):
    source_bucket = 'your-source-bucket'
    target_bucket = 'your-target-bucket'
    zip_file_name = 'all_files.zip'  # Name of the zip file
    
    # List objects in the source bucket
    response = s3.list_objects_v2(Bucket=source_bucket)
   
    if 'Contents' in response:
        objects = response.get('Contents', [])
       
        # Create a zip file in-memory
        zip_buffer = io.BytesIO()
        with zipfile.ZipFile(zip_buffer, 'w', zipfile.ZIP_DEFLATED) as zipf:
            for obj in objects:
                key = obj.get('Key')
                if key:
                    # Get object content
                    response = s3.get_object(Bucket=source_bucket, Key=key)
                    content = response['Body'].read()
                    
                    # Write the file to the zip archive
                    zipf.writestr(key, content)
            
        # Upload the zip file to the target bucket
        zip_buffer.seek(0)
        s3.upload_fileobj(zip_buffer, target_bucket, zip_file_name)
    else:
        print("No objects found in the source bucket.")

```

To convert all files in an AWS S3 bucket to zip files

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