[Wikimedia-l] Under block threat on fr.wp because of request on meta

Thomas Dalton thomas.dalton at gmail.com
Sat Nov 3 12:31:49 UTC 2012

You're taking about a whistleblower policy[1], essentially. Normally, they
are restricted to reporting violations off the law, rather than internal
policies (see the Foundation's policy[2] for example) but there is no
reason we couldn't have a broader one.

It would need to be quite limited in scope to avoid it being too open to
abuse, though.

1. http://en.wikipedia.org/wiki/Whistleblower
2. http://wikimediafoundation.org/wiki/Whistleblower_policy
On Nov 3, 2012 11:01 AM, "Teofilo" <teofilowiki at gmail.com> wrote:

> A group of French admins is threatening me of what they call a "block
> with consequences" in the case I would perform any "similar move", a
> move similar with what I did which they interpret as "disrupting
> Wikipedia to illustrate a point" (1).
> As the wording is totally vague ("similar move") this deprives me of
> the right to express myself on community matters. My freedom of speech
> on community matters is being denied.
> What I did, was a request to stewards on meta to remove access for all
> current French Checkusers as a consequence of the French Wikipedia
> switching from the "wiki with arbcom" to the "wiki without arbcom"
> status (2).
> So I am under threat, because I tried to enforce the checkuser policy,
> which provides different access procedures according to whether the
> wiki is with or without arbcom (3).
> Would it be possible to provide some kind of protection to users
> making requests on meta in reference to WMF policies ?
> Would it be possible to have some kind of "meta-arbcom" that would be
> a supreme court responsible for guaranteeing a set of fundamental
> principles, such as freedom of speech ?
> References:
> (1)
> http://fr.wikipedia.org/w/index.php?title=Discussion_utilisateur%3ATeofilo&diff=84877524&oldid=84615519
> (2)
> http://meta.wikimedia.org/w/index.php?title=Steward_requests/Permissions&oldid=4347135#per_CheckUser_policy.23Checkuser_access.2C_all_current_checkusers_on_fr.Wikipedia.org_.28wiki_without_an_Arbitration_Committee.29
> (3) http://meta.wikimedia.org/wiki/CheckUser_policy#Access_to_CheckUser
> See also:
> http://fr.wikipedia.org/wiki/Wikip%C3%A9dia:Prise_de_d%C3%A9cision/Checkuser
>  [The community vote in 2005 where checkusers where agreed by only a
> very short majority (52.4%)]
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