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ValueError: url cannot be parsed. #191

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Ly0n opened this issue Nov 23, 2020 · 1 comment
Open

ValueError: url cannot be parsed. #191

Ly0n opened this issue Nov 23, 2020 · 1 comment

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@Ly0n
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Ly0n commented Nov 23, 2020

File "/usr/local/lib/python3.7/site-packages/libreselery-0.0.3.dev13-py3.7.egg/libreselery/github_connector.py", line 39, in parseRemoteProjectId
repo = self.github.get_repo(self.parseRemoteToOwnerProjectName(url))
File "/usr/local/lib/python3.7/site-packages/libreselery-0.0.3.dev13-py3.7.egg/libreselery/github_connector.py", line 33, in parseRemoteToOwnerProjectName
raise ValueError("url cannot be parsed. (url: %s)" % (url))
ValueError: url cannot be parsed. (url: https://github.com/cityflow-project/CityFlow/)

Appeared with @kikass13 new architecture. Other URLs worked perfectly. I do not understand how this URL differs from others.

@Ly0n
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Ly0n commented Nov 23, 2020

ValueError: url cannot be parsed. (url: https://github.com/IAMconsortium/)

In this case the error is simple to understand. We also should parse links in the style github.com/organization/projectname
When just the organization page is been linked we need to skip that.

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