Script 'Create a file in S3 (Python)' for s3

Create a file in S3 (Python)

Details on S3 integration at https://www.windmill.dev/docs/core_concepts/persistent_storage/large_data_files

Created by henri186 353 days ago Viewed 0 times

Submitted by henri186 Python3

Created 353 days ago

#requirements:
wmill>=1.251.7

# Details on S3 integration at https://www.windmill.dev/docs/core_concepts/persistent_storage/large_data_files

import wmill
from wmill import S3Object

def main(s3_file_path: str):
    s3_file_output = S3Object(s3=s3_file_path)

    file_content = b"Hello Windmill!"
	# file_content can be either bytes or a BufferedReader
    file_content = wmill.write_s3_file(s3_file_output, file_content)
    return s3_file_output

`1`	`#requirements:`
`2`	`wmill>=1.251.7`
`3`
`4`	`# Details on S3 integration at https://www.windmill.dev/docs/core_concepts/persistent_storage/large_data_files`
`5`
`6`	`import wmill`
`7`	`from wmill import S3Object`
`8`
`9`	`def main(s3_file_path: str):`
`10`	`s3_file_output = S3Object(s3=s3_file_path)`
`11`
`12`	`file_content = b"Hello Windmill!"`
`13`	`# file_content can be either bytes or a BufferedReader`
`14`	`file_content = wmill.write_s3_file(s3_file_output, file_content)`
`15`	`return s3_file_output`