Python-从路径获取AWS Bucket密钥

Question

鉴于我有这样的东西：

"s3://folder1/folder2/folder3/folder4/folder4.5/folder4.6/folder5/folder6/folder7/file_name.csv"

我该如何解析它，以使我得到Bucket，然后得到Key，因为它减去文件名的last /？

我正在尝试获取路径，以便可以列出路径中的所有文件。

Answer 1

一种方法是使用简单的split和join组合：

s3_path = "s3://folder1/folder2/folder3/folder4/folder4.5/folder4.6/folder5/folder6/folder7/file_name.csv"

s3_path_split = s3_path.split('/')

bucket_name = s3_path_split[2]

# 'folder1'

key_name = '/'.join(s3_path_split[3:])

# 'folder2/folder3/folder4/folder4.5/folder4.6/folder5/folder6/folder7/file_name.csv'

key_name_without_file = '/'.join(s3_path_split[3:-1])

# folder2/folder3/folder4/folder4.5/folder4.6/folder5/folder6/folder7'