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Could not parse rfc1738 URL from string #2

@foocp

Description

@foocp

I have all the requirements, running python3.5.3 in virtualenvwrapper, on the most recent Raspbian Stretch (Jan2018), but I always get this. Any hints?

Traceback (most recent call last):
  File "scraper.py", line 8, in <module>
    db = dataset.connect(settings.CONNECTION_STRING)
  File "/home/pi/.virtualenvs/testtwitter_scrape/lib/python3.5/site-packages/dataset/__init__.py", line 41, in connect
    ensure_schema=ensure_schema, row_type=row_type)
  File "/home/pi/.virtualenvs/testtwitter_scrape/lib/python3.5/site-packages/dataset/database.py", line 53, in __init__
    self.engine = create_engine(url, **engine_kwargs)
  File "/home/pi/.virtualenvs/testtwitter_scrape/lib/python3.5/site-packages/sqlalchemy/engine/__init__.py", line 419, in create_engine
    return strategy.create(*args, **kwargs)
  File "/home/pi/.virtualenvs/testtwitter_scrape/lib/python3.5/site-packages/sqlalchemy/engine/strategies.py", line 50, in create
    u = url.make_url(name_or_url)
  File "/home/pi/.virtualenvs/testtwitter_scrape/lib/python3.5/site-packages/sqlalchemy/engine/url.py", line 205, in make_url
    return _parse_rfc1738_args(name_or_url)
  File "/home/pi/.virtualenvs/testtwitter_scrape/lib/python3.5/site-packages/sqlalchemy/engine/url.py", line 254, in _parse_rfc1738_args
    "Could not parse rfc1738 URL from string '%s'" % name)
sqlalchemy.exc.ArgumentError: Could not parse rfc1738 URL from string ''

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