
Hi Mariatta, people contributing to Python will also have to sign up for maintaining the code they contribute to a certain extent, so why would they not need a bpo account ? Also note that bpo is easy for us to customize, so the natural place to maintain such flags. That said, the CLA process is manual (at least AFAIK), so if there's a way to set a flag in Github, perhaps this could be integrated into CLA process. Thanks, -- Marc-Andre Lemburg eGenix.com Professional Python Services directly from the Experts (#1, Jun 15 2018)
Python Projects, Coaching and Consulting ... http://www.egenix.com/ Python Database Interfaces ... http://products.egenix.com/ Plone/Zope Database Interfaces ... http://zope.egenix.com/
::: We implement business ideas - efficiently in both time and costs ::: eGenix.com Software, Skills and Services GmbH Pastor-Loeh-Str.48 D-40764 Langenfeld, Germany. CEO Dipl.-Math. Marc-Andre Lemburg Registered at Amtsgericht Duesseldorf: HRB 46611 http://www.egenix.com/company/contact/ http://www.malemburg.com/ On 15.06.2018 02:30, Mariatta Wijaya wrote:
Currently, people wanting to contribute to projects like peps, devguide, core-workflow, bedevere, need to sign CLA, which requires a bpo account, and then they need to add their GitHub account to their bpo account.
I find this quite complicated process and a barrier of entry.
I think it will be great to have this process simplified. I think people should not need to have bpo account unless they're actually participating in the bug tracker for CPython.
1. Can we have people sign CLA and not require their bpo account?
2. Other big issue with bpo as CLA host is we don't have easy way that can let the-knights-who-say-ni update the label in the PR once the contributor has signed the CLA.
I've brought this up at the language summit, as one of the necessary step before we can start using GitHub issues. (that is not yet decided, and this is not the thread to argue about it).
But whether we are using GitHub issues or not, the current process needs improvement anyway.
Mariatta ᐧ
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