Stabilise filename parsing from url.

Whenever special sign ? is found in url analyse it in order to have
correct filename.

In case of http://host/path/filename/? use "filename" as part to
generate filename.

This fuzzy logic allows to generate quite sane filenames.

src/zc/buildout/networkcache.py

@@ -18,6 +18,7 @@ import os
 import posixpath
 import re
 import shutil
+import urlparse
 
 try:
     from slapos.libnetworkcache import NetworkcacheClient, UploadError, \
@@ -136,14 +137,15 @@ def upload_network_cached(dir_url, cache_url, external_url, path, logger,
 def get_filename_from_url(url):
     """Inspired how pip get filename from url.
     """
-    url = url.split('#', 1)[0]
-    url = url.rstrip('/')
-    name = posixpath.basename(url)
-    name_split_list = name.split('?', 1)
-    if name_split_list[0]:
-      name = name_split_list[0]
+    parsed_url = urlparse.urlparse(url)
+    if parsed_url.query and parsed_url.path.endswith('/'):
+      name = parsed_url.query.split('?', 1)[0]
+    elif parsed_url.path.endswith('/') and not parsed_url.query:
+      name = parsed_url.path.split('/')[-2]
     else:
-      name = name_split_list[1]
+      name = posixpath.basename(parsed_url.path)
+
+    name = name.split('#', 1)[0]
     assert name, (
            'URL %r produced no filename' % url)
     return name