s3 urls - get bucket name and path

Lijju Mathew picture Lijju Mathew · Mar 7, 2017 · Viewed 45.1k times · Source

I have a variable which has the aws s3 url

s3://bucket_name/folder1/folder2/file1.json

I want to get the bucket_name in a variables and rest i.e /folder1/folder2/file1.json in another variable. I tried the regular expressions and could get the bucket_name like below, not sure if there is a better way.

m = re.search('(?<=s3:\/\/)[^\/]+', 's3://bucket_name/folder1/folder2/file1.json')
print(m.group(0))

How do I get the rest i.e - folder1/folder2/file1.json ?

I have checked if there is a boto3 feature to extract the bucket_name and key from the url, but couldn't find it.

Answer

kichik picture kichik · Mar 7, 2017

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()