s3 urls – get bucket name and path

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("https://stackoverflow.com/")

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("https://stackoverflow.com/") + '?' + self._parsed.query
        else:
            return self._parsed.path.lstrip("https://stackoverflow.com/")

    @property
    def url(self):
        return self._parsed.geturl()