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Sent: 06 June 2011 17:54
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Subject: Mediawiki-api Digest, Vol 48, Issue 1
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Today's Topics:
1. Re: Return old_revid in watchlist results (Jim Safley)
2. Re: Return old_revid in watchlist results (Jim Safley)
3. Blocked user can logon via api (- xarax- )
4. Re: Blocked user can logon via api (Roan Kattouw)
5. Re: Blocked user can logon via api (Roan Kattouw)
6. Re: Blocked user can logon via api (- xarax- )
7. Letting other websites use wiki content (Eric K)
8. Re: Letting other websites use wiki content
(Mohamed Mahir Ahamed Ibrahim)
9. [Mediawiki-api-announce] Fwd: [Wikitech-l] BREAKING CHANGE:
action=watch now requires token (and API requires token and POST)
(Roan Kattouw)
10. Re: Letting other websites use wiki content (Eric K)
----------------------------------------------------------------------
Message: 1
Date: Tue, 31 May 2011 16:59:14 -0400
From: Jim Safley <jimsaf...@gmail.com>
Subject: Re: [Mediawiki-api] Return old_revid in watchlist results
To: "MediaWiki API announcements & discussion"
<mediawiki-api@lists.wikimedia.org>
Message-ID: <banlktins_c06bmaxnjcgjvnzahqvjyk...@mail.gmail.com>
Content-Type: text/plain; charset=ISO-8859-1
> This is not implemented right now, but should be easy to add. I've
> filed it at https://bugzilla.wikimedia.org/29221
Thanks, Roan. It will be a helpful addition.
------------------------------
Message: 2
Date: Tue, 31 May 2011 18:22:11 -0400
From: Jim Safley <jimsaf...@gmail.com>
Subject: Re: [Mediawiki-api] Return old_revid in watchlist results
To: "MediaWiki API announcements & discussion"
<mediawiki-api@lists.wikimedia.org>
Message-ID: <BANLkTimPrRk6ZWqvsE6a3RjLPC=aaok...@mail.gmail.com>
Content-Type: text/plain; charset=ISO-8859-1
In addition to old_revid, would it make sense to add type, logid, logtype,
and logaction to the watchlist response? This would bring the watchlist in
line with recentchanges, rcprop=loginfo
Jim
------------------------------
Message: 3
Date: Fri, 3 Jun 2011 14:38:08 +0200
From: "- xarax- " <m...@xarax.eu>
Subject: [Mediawiki-api] Blocked user can logon via api
To: <mediawiki-api@lists.wikimedia.org>
Message-ID:
<!&!AAAAAAAAAAAYAAAAAAAAANPZL5CFR+RGoyDtNjycpefCgAAAEAAAABvhrXUYbFNMjxtaFmao
9jcBAAAAAA==@xarax.eu>
Content-Type: text/plain; charset="iso-8859-1"
Hi all,
Why are locked users can log in via the API? Would have expected that the
users get back the status blocked. It says so in any case in the
documentation. I have a standard SVN installation without extensions or
other modifications. Can anyone confirm this? Below the query and the result
comes from unit testing. It looks like get, but in reality it is a post
request.
http://unittest.xarax.eu/w/api.php?format=xml&action=login&lgname=unittestin
g02&lgpassword=xxxxx&lgtoken=05b7b7fc4d5d18e389520d5e3ba7f0c2
<?xml version="1.0"?><api xmlns="http://www.mediawiki.org/xml/api/";><login
result="Success" lguserid="3" lgusername="Unittesting02"
lgtoken="ef8fc0d6e4f463e3b2c8d1716d306f5f" cookieprefix="mw_ut_19a"
sessionid="2bce1ed6a491fe957894bab610b77ae4" /></api>
The user is really blocked, like the log says.
12:01, 3. Jun. 2011 Unittesting02 (Diskussion | Beitr?ge) unbegrenzt
(Freigeben | Sperre ?ndern) Sysadmin (Diskussion | Beitr?ge | Sperren)
Erstellung von Benutzerkonten gesperrt, darf eigene Diskussionsseite nicht
bearbeiten (Ungeb?hrliches Verhalten)
I forgot to configure something or is this by design and the documentation
is wrong?
Best regards
- xarax -
------------------------------
Message: 4
Date: Fri, 3 Jun 2011 16:12:35 +0200
From: Roan Kattouw <roan.katt...@gmail.com>
Subject: Re: [Mediawiki-api] Blocked user can logon via api
To: "MediaWiki API announcements & discussion"
<mediawiki-api@lists.wikimedia.org>
Message-ID: <BANLkTi=ZhF2nV-=ya39myGmVf=krsz1...@mail.gmail.com>
Content-Type: text/plain; charset=ISO-8859-1
On Fri, Jun 3, 2011 at 2:38 PM, - xarax- <m...@xarax.eu> wrote:
> Hi all,
>
> Why are locked users can log in via the API?
Blocked users can log in just fine, both via the UI and the API.
However, they can't edit or move pages or change anything in any other
way, except if $wgBlockAllowsUTEdit is set to true, in which case
blocked users can edit their own user talk page (usually to appeal the
block and present arguments and evidence in their favor).
Roan Kattouw (Catrope)
------------------------------
Message: 5
Date: Fri, 3 Jun 2011 16:13:56 +0200
From: Roan Kattouw <roan.katt...@gmail.com>
Subject: Re: [Mediawiki-api] Blocked user can logon via api
To: "MediaWiki API announcements & discussion"
<mediawiki-api@lists.wikimedia.org>
Message-ID: <BANLkTimZrUOFRM+=oafuq73q6pavwbn...@mail.gmail.com>
Content-Type: text/plain; charset=ISO-8859-1
On Fri, Jun 3, 2011 at 4:12 PM, Roan Kattouw <roan.katt...@gmail.com> wrote:
> Blocked users can log in just fine, both via the UI and the API.
...except if $wgBlockDisablesLogin is true, in that case blocked users
can't even log in. This is useful on private wikis where anonymous
users can't read pages.
Roan Kattouw (Catrope)
------------------------------
Message: 6
Date: Fri, 3 Jun 2011 17:00:00 +0200
From: "- xarax- " <m...@xarax.eu>
Subject: Re: [Mediawiki-api] Blocked user can logon via api
To: "'MediaWiki API announcements & discussion'"
<mediawiki-api@lists.wikimedia.org>
Message-ID:
<!&!AAAAAAAAAAAYAAAAAAAAANPZL5CFR+RGoyDtNjycpefCgAAAEAAAAPqfS59ooqhCiV7Kp0/U
5hkBAAAAAA==@xarax.eu>
Content-Type: text/plain; charset="iso-8859-1"
Thank you Roan for your clarification.
-----Urspr?ngliche Nachricht-----
Von: mediawiki-api-boun...@lists.wikimedia.org
[mailto:mediawiki-api-boun...@lists.wikimedia.org] Im Auftrag von Roan
Kattouw
Gesendet: Freitag, 3. Juni 2011 16:14
An: MediaWiki API announcements & discussion
Betreff: Re: [Mediawiki-api] Blocked user can logon via api
On Fri, Jun 3, 2011 at 4:12 PM, Roan Kattouw <roan.katt...@gmail.com> wrote:
> Blocked users can log in just fine, both via the UI and the API.
...except if $wgBlockDisablesLogin is true, in that case blocked users can't
even log in. This is useful on private wikis where anonymous users can't
read pages.
Roan Kattouw (Catrope)
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------------------------------
Message: 7
Date: Sun, 5 Jun 2011 19:36:44 -0700 (PDT)
From: Eric K <ek79...@yahoo.com>
Subject: [Mediawiki-api] Letting other websites use wiki content
To: mediawiki-api@lists.wikimedia.org
Message-ID: <843337.6100...@web39322.mail.mud.yahoo.com>
Content-Type: text/plain; charset="iso-8859-1"
I've been refered to this list after asking this question on the Mediawiki-l
list.
Suppose we have a wiki, and we want other websites (blogs and any other
kinds of sites) to use our wiki content. For example say the wiki has
different types of random facts about countries. When someone visits a blog,
they can see a box where that random fact is presented, and that data is
drawn from the wiki.
Whats the best way to do this? RSS? Is this kind of thing possible using
Mediawiki API?
?
thanks
Eric
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Message: 8
Date: Mon, 6 Jun 2011 10:46:59 +0530
From: Mohamed Mahir Ahamed Ibrahim <mahi...@gmail.com>
Subject: Re: [Mediawiki-api] Letting other websites use wiki content
To: "MediaWiki API announcements & discussion"
<mediawiki-api@lists.wikimedia.org>
Message-ID: <banlktinysqlt1z9rxvxeghahbmyib04...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
I think its possible calling a template through api.
Mahir
On Mon, Jun 6, 2011 at 8:06 AM, Eric K <ek79...@yahoo.com> wrote:
> I've been refered to this list after asking this question on the
> Mediawiki-l list.
> Suppose we have a wiki, and we want other websites (blogs and any other
> kinds of sites) to use our wiki content. For example say the wiki has
> different types of random facts about countries. When someone visits a
> blog, they can see a box where that random fact is presented, and that
> data is drawn from the wiki.
> Whats the best way to do this? RSS? Is this kind of thing possible using
> Mediawiki API?
>
> thanks
> Eric
>
> _______________________________________________
> Mediawiki-api mailing list
> Mediawiki-api@lists.wikimedia.org
> https://lists.wikimedia.org/mailman/listinfo/mediawiki-api
>
>
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Message: 9
Date: Mon, 6 Jun 2011 10:56:08 +0200
From: Roan Kattouw <roan.katt...@gmail.com>
Subject: [Mediawiki-api] [Mediawiki-api-announce] Fwd: [Wikitech-l]
BREAKING CHANGE: action=watch now requires token (and API requires
token and POST)
To: mediawiki-api-annou...@lists.wikimedia.org
Message-ID: <banlktik4avn4+p7weqtjhymm2rd90q1...@mail.gmail.com>
Content-Type: text/plain; charset="us-ascii"
---------- Forwarded message ----------
From: Krinkle <krinklem...@gmail.com>
Date: Mon, Jun 6, 2011 at 3:08 AM
Subject: [Wikitech-l] BREAKING CHANGE: action=watch now requires token
(and API requires token and POST)
To: MediaWiki announcements and site admin list
<mediawik...@lists.wikimedia.org>, Wikimedia developers
<wikitec...@lists.wikimedia.org>
Hi all,
As of MediaWiki 1.19 the action of watching or unwatching a page
requires a
token [1]. A similar measure was taken during the development of 1.17
for
the markpatrolled action, and the reason is the same: To prevent
third-party sites from executing write actions without the users'
permission.
The ApiWatch module must be posted and given a token. As with other
edittoken-based api actions, the token is salted but stays the same
throughout a session. Scripts may retrieve this token, as usual, through
the ApiQueryInfo (must be logged in, anon users don't have action-watch)
[4]
On a sidenote, recently the the mw.user.tokens resourceloader module [8]
has been created [9]. This, together with the mw.user.options module
introduced in 1.17, gadgets can do advanced actions without polling
the API
for common data. If you script is ran from a wiki, you can get the
tokens
from [5] this Map without an http request to the query info module. An
example has been made in the mediawiki.action.watch.ajax module [6].
This
(un)watches through the API.
The actual change in the WatchAction class was made in r89545 [3].
The ApiWatch module was changed in r88522 [7].
--
Krinkle
[1] https://bugzilla.wikimedia.org/27655 Require token for
(un)watching pages
[2] https://bugzilla.wikimedia.org/29070 Add token to action=watch API
[3] http://www.mediawiki.org/wiki/Special:Code/MediaWiki/89545
[4]
http://yourdomain/w/api.php?action=query&prop=info&titles=Main+Page&intoken=
watch
[5] http://www.mediawiki.org/wiki/ResourceLoader/Default_modules#tokens
[6]
http://svn.wikimedia.org/viewvc/mediawiki/trunk/phase3/resources/mediawiki.a
ction/mediawiki.action.watch.ajax.js?view=markup&pathrev=88554#l117
[7] http://www.mediawiki.org/wiki/Special:Code/MediaWiki/88522
[8] https://bugzilla.wikimedia.org/29067 Expose user.tokens like we do
user.options in ResourceLoader
[9] http://www.mediawiki.org/wiki/Special:Code/MediaWiki/88553
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------------------------------
Message: 10
Date: Mon, 6 Jun 2011 09:54:05 -0700 (PDT)
From: Eric K <ek79...@yahoo.com>
Subject: Re: [Mediawiki-api] Letting other websites use wiki content
To: MediaWiki API announcements & discussion
<mediawiki-api@lists.wikimedia.org>
Message-ID: <94062.24463...@web39304.mail.mud.yahoo.com>
Content-Type: text/plain; charset="iso-8859-1"
Thanks! I'm completely new to the Mediawiki API system. Should I be look at
this page?
http://www.mediawiki.org/wiki/API:Parsing_wikitext
--- On Mon, 6/6/11, Mohamed Mahir Ahamed Ibrahim <mahi...@gmail.com> wrote:
From: Mohamed Mahir Ahamed Ibrahim <mahi...@gmail.com>
Subject: Re: [Mediawiki-api] Letting other websites use wiki content
To: "MediaWiki API announcements & discussion"
<mediawiki-api@lists.wikimedia.org>
Date: Monday, June 6, 2011, 12:16 AM
I think its possible calling a template through api.
Mahir
On Mon, Jun 6, 2011 at 8:06 AM, Eric K <ek79...@yahoo.com> wrote:
I've been refered to this list after asking this question on the Mediawiki-l
list.
Suppose we have a wiki, and we want other websites (blogs and any other
kinds of sites) to use our wiki content. For example say the wiki has
different types of random facts about countries. When someone visits a blog,
they can see a box where that random fact is presented, and that data is
drawn from the wiki.
Whats the best way to do this? RSS? Is this kind of thing possible using
Mediawiki API?
?
thanks
Eric
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https://lists.wikimedia.org/mailman/listinfo/mediawiki-api
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