Apply fixes to s3:// URL parsing under Python 2.6.

The urlparse module starting in Python 2.6 appears to be compliant with RFC
3986, while previous versions were not.  This causes a change in the
behaviro of parsing s3:// URLs, however, resulting in breakage with Python
2.6.

Try to fix this by adjusting our code so that it works with either the old
or the new behavior.

diff --git a/python/cumulus/store/s3.py b/python/cumulus/store/s3.py
index 65884ea..52f03dd 100644 (file)
--- a/python/cumulus/store/s3.py
+++ b/python/cumulus/store/s3.py
@@ -1,3 +1,4 @@
+# Amazon S3 storage backend.  Uses a URL of the form s3://BUCKET/PATH/.
 import os, sys, tempfile
 import boto
 from boto.s3.bucket import Bucket
@@ -7,10 +8,21 @@ import cumulus.store
 
 class S3Store(cumulus.store.Store):
     def __init__(self, url, **kw):
-        (bucket, prefix) = self.path.lstrip("/").split("/", 1)
+        # Old versions of the Python urlparse library will take a URL like
+        # s3://bucket/path/ and include the bucket with the path, while new
+        # versions (2.6 and later) treat it as the netloc (which seems more
+        # correct).
+        #
+        # But, so that we can work with either behavior, for now just combine
+        # the netloc and path together before we do any further processing
+        # (which will then split the combined path apart into a bucket and path
+        # again).  If we didn't want to support Python 2.5, this would be
+        # easier as we could just use the netloc as the bucket directly.
+        path = self.netloc + '/' + self.path
+        (bucket, prefix) = path.lstrip("/").split("/", 1)
         self.conn = boto.connect_s3(is_secure=False)
         self.bucket = self.conn.create_bucket(bucket)
-        self.prefix = prefix.rstrip ("/")
+        self.prefix = prefix.strip("/")
         self.scan_cache = {}
 
     def _get_key(self, type, name):