Since it's just a normal URL, you can use urlparse
to get all the parts of the URL.
>>> from urlparse import urlparse
>>> o = urlparse('s3://bucket_name/folder1/folder2/file1.json', allow_fragments=False)
>>> o
ParseResult(scheme='s3', netloc='bucket_name', path='/folder1/folder2/file1.json', params='', query='', fragment='')
>>> o.netloc
'bucket_name'
>>> o.path
'/folder1/folder2/file1.json'
You may have to remove the beginning slash from the key as the next answer suggests.
o.path.lstrip('/')
With Python 3 urlparse
moved to urllib.parse
so use:
from urllib.parse import urlparse
Here's a class that takes care of all the details.
try:
from urlparse import urlparse
except ImportError:
from urllib.parse import urlparse
class S3Url(object):
"""
>>> s = S3Url("s3://bucket/hello/world")
>>> s.bucket
'bucket'
>>> s.key
'hello/world'
>>> s.url
's3://bucket/hello/world'
>>> s = S3Url("s3://bucket/hello/world?qwe1=3#ddd")
>>> s.bucket
'bucket'
>>> s.key
'hello/world?qwe1=3#ddd'
>>> s.url
's3://bucket/hello/world?qwe1=3#ddd'
>>> s = S3Url("s3://bucket/hello/world#foo?bar=2")
>>> s.key
'hello/world#foo?bar=2'
>>> s.url
's3://bucket/hello/world#foo?bar=2'
"""
def __init__(self, url):
self._parsed = urlparse(url, allow_fragments=False)
@property
def bucket(self):
return self._parsed.netloc
@property
def key(self):
if self._parsed.query:
return self._parsed.path.lstrip('/') + '?' + self._parsed.query
else:
return self._parsed.path.lstrip('/')
@property
def url(self):
return self._parsed.geturl()
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