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It breaks with SERVER_NAME defined #36
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Please provide steps to reproduce, expected outcome, and actual outcome. |
Here, the steps to reproduce the issue: >>> import flask
>>> app = flask.Flask(__name__)
>>> @app.route('/')
... def index(): return 'index'
...
>>> from flask_frozen import Freezer
>>> f = Freezer(app)
>>> n = f._generate_all_urls()
>>> next(n)
(u'/', 'index')
>>> app.config['SERVER_NAME'] = 'www.example.com'
>>> n = f._generate_all_urls()
>>> next(n)
(u'://www.example.com/', 'index')
>>> f.freeze()
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "C:\Python27\lib\site-packages\flask_frozen\__init__.py", line 149, in freeze
new_filename = self._build_one(url)
File "C:\Python27\lib\site-packages\flask_frozen\__init__.py", line 263, in _build_one
% (response.status, url))
ValueError: Unexpected status '404 NOT FOUND' on URL ://www.example.com/ |
I’m not sure if this was always broken or if this something changed in recent Flask versions, but it should be fixed now. Please let me know if this fixes the issue for you. |
Thanks, it works for me. |
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I'm trying to freeze my app, I get a
ValueError
when SERVER_NAME is defined.SERVER_NAME = 'www.example.com'
The
url_for
try to generate the external URL without scheme (://www.example.com
). I'm using Flask 0.10.1 and last Frozen-Flask.The text was updated successfully, but these errors were encountered: