Mediawiki-api June 2011

mediawiki-api@lists.wikimedia.org

11 participants
9 discussions

Need to extract abstract of a wikipedia page
by aditya srinivas 23 Nov '23

23 Nov '23

Hello, I am writing a Java program to extract the abstract of the wikipedia page given the title of the wikipedia page. I have done some research and found out that the abstract with be in rvsection=0 So for example if I want the abstract of 'Eiffel Tower" wiki page then I am querying using the api in the following way. http://en.wikipedia.org/w/api.php?action=query&prop=revisions&titles=Eiffel… and parse the XML data which we get and take the wikitext in the tag <rev xml:space="preserve"> which represents the abstract of the wikipedia page. But this wiki text also contains the infobox data which I do not need. I would like to know if there is anyway in which I can remove the infobox data and get only the wikitext related to the page's abstract Or if there is any alternative method by which I can get the abstract of the page directly. Looking forward to your help. Thanks in Advance Aditya Uppu

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api.php file not found
by Erel Segal 01 Aug '11

01 Aug '11

Hi, I have Mediwiki 1.13.1: http://irsrv2.cs.biu.ac.il/wiki and I cannot find the api.php page. When I open the url: http://irsrv2.cs.biu.ac.il/w/api.php I get a "File not found" error. How can I enable the api.php?

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installing action modules
by Erel Segal 23 Jun '11

23 Jun '11

My API contains only 3 modules: action - What action you would like to perform One value: login, logout, help Default: help How can I install the other modules (specifically the query module)? I have MediaWiki 1.13.1 - should I upgrade it to 1.17? Or is it possible to upgrade the API only?

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description page for math images
by Bagrecha, Priyank 22 Jun '11

22 Jun '11

Hi I have been trying to find the description page for latex math images. Here is an example of such an image: http://upload.wikimedia.org/math/f/6/a/f6ac8632c237011599f300e62d916859.png obtained from the page http://en.wikipedia.org/wiki/Acid However I am unable to find the description page for such images. I tried http://en.wikipedia.org/w/api.php?action=parse&format=xml&page=File:f6ac863… and that resulted in an xml with error code="missingtitle" info="The page you specified doesn't exist" I have also tried http://en.wikipedia.org/wiki/File:F6ac8632c237011599f300e62d916859.png Which says the file does not exist. I am trying to get the licensing information for images, and then try to download them. Any suggestions or pointers are welcome. Thanks Priyank

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Get number of views
by Christian Koncilia 11 Jun '11

11 Jun '11

Hi, I'm looking for a way to get an indicator for how relevant a given Wiki-Page is. Something like "number of views". I found the "clicktracking" action but I'm not sure if this would really be useful. Thanks Christian

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Letting other websites use wiki content
by Eric K 07 Jun '11

07 Jun '11

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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UNSUBSCRIBE
by Craig Surtees 07 Jun '11

07 Jun '11

-----Original Message----- From: mediawiki-api-bounces(a)lists.wikimedia.org [mailto:mediawiki-api-bounces@lists.wikimedia.org] On Behalf Of mediawiki-api-request(a)lists.wikimedia.org Sent: 06 June 2011 17:54 To: mediawiki-api(a)lists.wikimedia.org Subject: Mediawiki-api Digest, Vol 48, Issue 1 Send Mediawiki-api mailing list submissions to mediawiki-api(a)lists.wikimedia.org To subscribe or unsubscribe via the World Wide Web, visit https://lists.wikimedia.org/mailman/listinfo/mediawiki-api or, via email, send a message with subject or body 'help' to mediawiki-api-request(a)lists.wikimedia.org You can reach the person managing the list at mediawiki-api-owner(a)lists.wikimedia.org When replying, please edit your Subject line so it is more specific than "Re: Contents of Mediawiki-api digest..." 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 <jimsafley(a)gmail.com> Subject: Re: [Mediawiki-api] Return old_revid in watchlist results To: "MediaWiki API announcements & discussion" <mediawiki-api(a)lists.wikimedia.org> Message-ID: <BANLkTinS_C06BmAXnJcGjVNzaHqVjykCNA(a)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 <jimsafley(a)gmail.com> Subject: Re: [Mediawiki-api] Return old_revid in watchlist results To: "MediaWiki API announcements & discussion" <mediawiki-api(a)lists.wikimedia.org> Message-ID: <BANLkTimPrRk6ZWqvsE6a3RjLPC=aaOk44g(a)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- " <ms(a)xarax.eu> Subject: [Mediawiki-api] Blocked user can logon via api To: <mediawiki-api(a)lists.wikimedia.org> Message-ID: <!&!AAAAAAAAAAAYAAAAAAAAANPZL5CFR+RGoyDtNjycpefCgAAAEAAAABvhrXUYbFNMjxtaFmao 9jcBAAAAAA==(a)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.kattouw(a)gmail.com> Subject: Re: [Mediawiki-api] Blocked user can logon via api To: "MediaWiki API announcements & discussion" <mediawiki-api(a)lists.wikimedia.org> Message-ID: <BANLkTi=ZhF2nV-=ya39myGmVf=KRsz1Qqg(a)mail.gmail.com> Content-Type: text/plain; charset=ISO-8859-1 On Fri, Jun 3, 2011 at 2:38 PM, - xarax- <ms(a)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.kattouw(a)gmail.com> Subject: Re: [Mediawiki-api] Blocked user can logon via api To: "MediaWiki API announcements & discussion" <mediawiki-api(a)lists.wikimedia.org> Message-ID: <BANLkTimZrUOFRM+=oAFuq73q6PAvwbnaiA(a)mail.gmail.com> Content-Type: text/plain; charset=ISO-8859-1 On Fri, Jun 3, 2011 at 4:12 PM, Roan Kattouw <roan.kattouw(a)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- " <ms(a)xarax.eu> Subject: Re: [Mediawiki-api] Blocked user can logon via api To: "'MediaWiki API announcements & discussion'" <mediawiki-api(a)lists.wikimedia.org> Message-ID: <!&!AAAAAAAAAAAYAAAAAAAAANPZL5CFR+RGoyDtNjycpefCgAAAEAAAAPqfS59ooqhCiV7Kp0/U 5hkBAAAAAA==(a)xarax.eu> Content-Type: text/plain; charset="iso-8859-1" Thank you Roan for your clarification. -----Urspr?ngliche Nachricht----- Von: mediawiki-api-bounces(a)lists.wikimedia.org [mailto:mediawiki-api-bounces@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.kattouw(a)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) _______________________________________________ Mediawiki-api mailing list Mediawiki-api(a)lists.wikimedia.org https://lists.wikimedia.org/mailman/listinfo/mediawiki-api ------------------------------ Message: 7 Date: Sun, 5 Jun 2011 19:36:44 -0700 (PDT) From: Eric K <ek79501(a)yahoo.com> Subject: [Mediawiki-api] Letting other websites use wiki content To: mediawiki-api(a)lists.wikimedia.org Message-ID: <843337.6100.qm(a)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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[Mediawiki-api-announce] Fwd: [Wikitech-l] BREAKING CHANGE: action=watch now requires token (and API requires token and POST)
by Roan Kattouw 06 Jun '11

06 Jun '11

---------- Forwarded message ---------- From: Krinkle <krinklemail(a)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 <mediawiki-l(a)lists.wikimedia.org>, Wikimedia developers <wikitech-l(a)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… [5] http://www.mediawiki.org/wiki/ResourceLoader/Default_modules#tokens [6] http://svn.wikimedia.org/viewvc/mediawiki/trunk/phase3/resources/mediawiki.… [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 _______________________________________________ Wikitech-l mailing list Wikitech-l(a)lists.wikimedia.org https://lists.wikimedia.org/mailman/listinfo/wikitech-l _______________________________________________ Mediawiki-api-announce mailing list Mediawiki-api-announce(a)lists.wikimedia.org https://lists.wikimedia.org/mailman/listinfo/mediawiki-api-announce

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Blocked user can logon via api
by - xarax- 03 Jun '11

03 Jun '11

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 | Beiträge) unbegrenzt (Freigeben | Sperre ändern) Sysadmin (Diskussion | Beiträge | Sperren) Erstellung von Benutzerkonten gesperrt, darf eigene Diskussionsseite nicht bearbeiten (Ungebührliches Verhalten) I forgot to configure something or is this by design and the documentation is wrong? Best regards - xarax -

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Mediawiki-api June 2011