This example of API
service = lp.load('/+services/sharing')
service.getPillarGranteeData(pillar=project)
Raise a TypeError error because wadllib resource.url returned a URI instead a string
Traceback (most recent call last):
File "./share-projects-with-team.py", line 106, in <module>
sys.exit(main(sys.argv[1:]))
File "./share-projects-with-team.py", line 102, in main
migrator.share()
File "./share-projects-with-team.py", line 39, in share
self.unshare_with_team_members()
File "./share-projects-with-team.py", line 68, in unshare_with_team_members
grantees = self.service.getPillarGranteeData(pillar=self.project)
File "/usr/lib/python2.7/dist-packages/lazr/restfulclient/resource.py", line 563, in __call__
url = self.wadl_method.build_request_url(**args)
File "/usr/lib/python2.7/dist-packages/wadllib/application.py", line 615, in build_request_url
return self.request.build_url(param_values, **kw_param_values)
File "/usr/lib/python2.7/dist-packages/wadllib/application.py", line 731, in build_url
if '?' in url:
TypeError: argument of type 'URI' is not iterable
This is the hack in my script to allow me to use this method:
# Every normal man must be tempted at times to spit on his hands,
# hoist the black flag, and begin to slit throats.
def safe_build_
"""Return the request URL to use to invoke this method.
The unsafe method raises an error for
lp.
because the wadl resource.uri it gets a URI instead of a str.
"""
try:
from urllib.parse import urlencode
urlencode
except ImportError:
from urllib import urlencode
validated_
url = str(self.
if len(validated_
if '?' in url:
append = '&'
else:
append = '?'
url += append + urlencode(
return url