this post was submitted on 10 Aug 2023
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I'm a little confused as to how the "report" function works. I'm an instance admin, but if I report a post on another instance, that report shows up in MY report queue. If I mark that report as "resolved" to get rid of the notification count, does it resolve it on the home instance?

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[–] Crul@lemm.ee 2 points 1 year ago* (last edited 1 year ago) (2 children)

EDIT: This is wrong, see OP's comment with the right answer

~~Not an expert, those who know more please correct me.~~

~~I will explain what I understand with an example; let's say I want to report your comment (the one I'm replying to).~~

~~Because I'm reading it from https://lemm.ee/post/3809349 , if I report it here, I would be reporting it to lemm.ee admins. Now I need to decide if I want to report it to:~~

  • ~~lemmy.ml admins: the instance of the community in which we are commenting~~

~~or~~

  • ~~tkohhh.social: the instance of your user~~

~~Then, because there is no way to "translate" a post / comment link from one instance to another, I need to go to !lemmy_support@lemmy.ml through the instance I want to report it. Let's say tkohhh.social~~

~~So I go to: https://tkohhh.social/c/lemmy_support@lemmy.ml and log in~~

~~Then I look for the post:~~ https://tkohhh.social/post/10726

~~And if I report it there, the report should be sent to tkohhh.social's admins.~~

~~Note that (if this is correct) it requires an account on the instance you want to report to... which is why I did an unofficial report when I found a spammer on another instance, using the Meta community of that instance:~~

[–] Blaze@discuss.tchncs.de 3 points 1 year ago
[–] tko@tkohhh.social 2 points 1 year ago (3 children)

If this is true, what is the admin of the reporting user even supposed to do? Surely I'm not expected (or even able?) to go Bigfoot on somebody else's community because one of my users reported some content over there. I feel like I must be missing something.

[–] Crul@lemm.ee 2 points 1 year ago* (last edited 1 year ago) (1 children)

I have 0 experience with the admin side of Lemmy, so I don't know what actions an Admin can take over a user from another instance. If I had to guess:

  • In the post I linked before there is a comment talking about a "purge feature", but I don't know how that works.
  • Community Mods may be able (?) to ban the user in that specific community
  • Instance Admins may be able (?) to ban / block the replication of that user content in their instance

But again, not an expert, so I don't really know what is the intended way to handle this.

[–] Blaze@discuss.tchncs.de 2 points 1 year ago (1 children)

You can have a look at https://join-lemmy.org/docs/users/04-moderation.html to see a table with the different actions and the role required, as well as some explanation of how it works.

Personally, I received reports as a non-admin mod for a community from users from other instances, so that should be possible

[–] Crul@lemm.ee 2 points 1 year ago

What I don't see is how it works accross instances. If we were talking about a single instance, then everything is very easy.

But, I have some questions:
... which I don't expect you to answer :)

  • An admin from instance X purges a user from that instance, does the purge propagate to all other instances? Or maybe each instance's admins need to purge that user.
  • What happens if an admin from instance X purges a user from instance Y?
[–] PriorProject@lemmy.world 2 points 1 year ago

If this is true, what is the admin of the reporting user even supposed to do

Decide what instances to defederate. They can check up on:

  • Mods of other instances to see if they take appropriate action on reports.
  • Admins of other instances to see if they take action against bad-faith mods.
  • Admins of other instances who generate a disproportionate number of reports per capital, to address structural, cultural, or policy issues that lead the offending instance to be a bigger source of pain/reports than others.

And finally they can defederate if they don't like what they find.