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Mediawiki-apiJune 2011

mediawiki-api@lists.wikimedia.org
  • 11 participants
  • 10 discussions
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20 Jan '26
Hi there,I'm using the API to extract the raw wiki text from my pages, usingthe "?action=query&titles=Main_Page&export&exportnowrap" syntax. Thatworks perfectly.Now I would like to get the templates expanded out in the result, so Iuse: "?action=query&titles=Main_Page&prop=revisions&rvlimit=1&rvprop=content&rvexpandtemplates",which does the job, as expected, but it also strips out the comments.My problem is that the comments are meaningful to me (I use them tohelp process the wiki text in subsequent steps).Is there a way to expand templates with the API, but leave thecomments intact?Thanks,Kevin
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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 pagegiven the title of the wikipedia page. I have done some research and foundout that the abstract with be in rvsection=0 So for example if I want the abstract of 'Eiffel Tower" wiki page then I amquerying 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 <revxml:space="preserve"> which represents the abstract of the wikipedia page.But this wiki text also contains the infobox data which I do not need. Iwould like to know if there is anyway in which I can remove the infobox dataand get only the wikitext related to the page's abstract Or if there is anyalternative method by which I can get the abstract of the page directly.Looking forward to your help.Thanks in AdvanceAditya 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/wikiand I cannot find the api.php page. When I open the url:http://irsrv2.cs.biu.ac.il/w/api.phpI 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: helpHow 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 toupgrade the API only?
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description page for math images
by Bagrecha, Priyank 22 Jun '11

22 Jun '11
HiI 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 pagehttp://en.wikipedia.org/wiki/AcidHowever I am unable to find the description page for such images. I triedhttp://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 triedhttp://en.wikipedia.org/wiki/File:F6ac8632c237011599f300e62d916859.pngWhich 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.ThanksPriyank
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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-Pageis. Something like "number of views". I found the "clicktracking" actionbut I'm not sure if this would really be useful. Thanks Christian
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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? thanksEric
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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 Ofmediawiki-api-request(a)lists.wikimedia.orgSent: 06 June 2011 17:54To: mediawiki-api(a)lists.wikimedia.orgSubject: Mediawiki-api Digest, Vol 48, Issue 1Send Mediawiki-api mailing list submissions tomediawiki-api(a)lists.wikimedia.orgTo subscribe or unsubscribe via the World Wide Web, visithttps://lists.wikimedia.org/mailman/listinfo/mediawiki-apior, via email, send a message with subject or body 'help' tomediawiki-api-request(a)lists.wikimedia.orgYou can reach the person managing the list atmediawiki-api-owner(a)lists.wikimedia.orgWhen 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: 1Date: Tue, 31 May 2011 16:59:14 -0400From: Jim Safley <jimsafley(a)gmail.com>Subject: Re: [Mediawiki-api] Return old_revid in watchlist resultsTo: "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 athttps://bugzilla.wikimedia.org/29221Thanks, Roan. It will be a helpful addition.------------------------------Message: 2Date: Tue, 31 May 2011 18:22:11 -0400From: Jim Safley <jimsafley(a)gmail.com>Subject: Re: [Mediawiki-api] Return old_revid in watchlist resultsTo: "MediaWiki API announcements & discussion"<mediawiki-api(a)lists.wikimedia.org>Message-ID: <BANLkTimPrRk6ZWqvsE6a3RjLPC=aaOk44g(a)mail.gmail.com>Content-Type: text/plain; charset=ISO-8859-1In addition to old_revid, would it make sense to add type, logid, logtype,and logaction to the watchlist response? This would bring the watchlist inline with recentchanges, rcprop=loginfoJim------------------------------Message: 3Date: Fri, 3 Jun 2011 14:38:08 +0200From: "- xarax- " <ms(a)xarax.eu>Subject: [Mediawiki-api] Blocked user can logon via apiTo: <mediawiki-api(a)lists.wikimedia.org>Message-ID:<!&!AAAAAAAAAAAYAAAAAAAAANPZL5CFR+RGoyDtNjycpefCgAAAEAAAABvhrXUYbFNMjxtaFmao9jcBAAAAAA==(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 theusers get back the status blocked. It says so in any case in thedocumentation. I have a standard SVN installation without extensions orother modifications. Can anyone confirm this? Below the query and the resultcomes from unit testing. It looks like get, but in reality it is a postrequest.http://unittest.xarax.eu/w/api.php?format=xml&action=login&lgname=unittesting02&lgpassword=xxxxx&lgtoken=05b7b7fc4d5d18e389520d5e3ba7f0c2<?xml version="1.0"?><api xmlns="http://www.mediawiki.org/xml/api/"><loginresult="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. 2011Unittesting02 (Diskussion | Beitr?ge)unbegrenzt(Freigeben | Sperre ?ndern)Sysadmin (Diskussion | Beitr?ge | Sperren)Erstellung von Benutzerkonten gesperrt, darf eigene Diskussionsseite nichtbearbeiten(Ungeb?hrliches Verhalten)I forgot to configure something or is this by design and the documentationis wrong?Best regards- xarax -------------------------------Message: 4Date: Fri, 3 Jun 2011 16:12:35 +0200From: Roan Kattouw <roan.kattouw(a)gmail.com>Subject: Re: [Mediawiki-api] Blocked user can logon via apiTo: "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-8859-1On 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 otherway, except if $wgBlockAllowsUTEdit is set to true, in which caseblocked users can edit their own user talk page (usually to appeal theblock and present arguments and evidence in their favor).Roan Kattouw (Catrope)------------------------------Message: 5Date: Fri, 3 Jun 2011 16:13:56 +0200From: Roan Kattouw <roan.kattouw(a)gmail.com>Subject: Re: [Mediawiki-api] Blocked user can logon via apiTo: "MediaWiki API announcements & discussion"<mediawiki-api(a)lists.wikimedia.org>Message-ID: <BANLkTimZrUOFRM+=oAFuq73q6PAvwbnaiA(a)mail.gmail.com>Content-Type: text/plain; charset=ISO-8859-1On 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 userscan't even log in. This is useful on private wikis where anonymoususers can't read pages.Roan Kattouw (Catrope)------------------------------Message: 6Date: Fri, 3 Jun 2011 17:00:00 +0200From: "- xarax- " <ms(a)xarax.eu>Subject: Re: [Mediawiki-api] Blocked user can logon via apiTo: "'MediaWiki API announcements & discussion'"<mediawiki-api(a)lists.wikimedia.org>Message-ID:<!&!AAAAAAAAAAAYAAAAAAAAANPZL5CFR+RGoyDtNjycpefCgAAAEAAAAPqfS59ooqhCiV7Kp0/U5hkBAAAAAA==(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 RoanKattouwGesendet: Freitag, 3. Juni 2011 16:14An: MediaWiki API announcements & discussionBetreff: Re: [Mediawiki-api] Blocked user can logon via apiOn 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'teven log in. This is useful on private wikis where anonymous users can'tread pages.Roan Kattouw (Catrope)_______________________________________________Mediawiki-api mailing listMediawiki-api(a)lists.wikimedia.orghttps://lists.wikimedia.org/mailman/listinfo/mediawiki-api------------------------------Message: 7Date: Sun, 5 Jun 2011 19:36:44 -0700 (PDT)From: Eric K <ek79501(a)yahoo.com>Subject: [Mediawiki-api] Letting other websites use wiki contentTo: mediawiki-api(a)lists.wikimedia.orgMessage-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-llist. Suppose we have a wiki, and we want other websites (blogs and any otherkinds of sites) to use our wiki content. For example say the wiki hasdifferent 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 isdrawn from the wiki.Whats the best way to do this? RSS? Is this kind of thing possible usingMediawiki API??thanksEric
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---------- Forwarded message ----------From: Krinkle <krinklemail(a)gmail.com>Date: Mon, Jun 6, 2011 at 3:08 AMSubject: [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 pagerequires atoken [1]. A similar measure was taken during the development of 1.17forthe markpatrolled action, and the reason is the same: To preventthird-party sites from executing write actions without the users'permission.The ApiWatch module must be posted and given a token. As with otheredittoken-based api actions, the token is salted but stays the samethroughout a session. Scripts may retrieve this token, as usual, throughthe 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 moduleintroduced in 1.17, gadgets can do advanced actions without pollingthe APIfor common data. If you script is ran from a wiki, you can get thetokensfrom [5] this Map without an http request to the query info module. Anexample 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 douser.options in ResourceLoader[9]http://www.mediawiki.org/wiki/Special:Code/MediaWiki/88553_______________________________________________Wikitech-l mailing listWikitech-l(a)lists.wikimedia.orghttps://lists.wikimedia.org/mailman/listinfo/wikitech-l_______________________________________________Mediawiki-api-announce mailing listMediawiki-api-announce(a)lists.wikimedia.orghttps://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 theusers get back the status blocked. It says so in any case in thedocumentation. I have a standard SVN installation without extensions orother modifications. Can anyone confirm this? Below the query and the resultcomes from unit testing. It looks like get, but in reality it is a postrequest.http://unittest.xarax.eu/w/api.php?format=xml&action=login&lgname=unittesting02&lgpassword=xxxxx&lgtoken=05b7b7fc4d5d18e389520d5e3ba7f0c2<?xml version="1.0"?><api xmlns="http://www.mediawiki.org/xml/api/"><loginresult="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. 2011Unittesting02 (Diskussion | Beiträge)unbegrenzt(Freigeben | Sperre ändern)Sysadmin (Diskussion | Beiträge | Sperren)Erstellung von Benutzerkonten gesperrt, darf eigene Diskussionsseite nichtbearbeiten(Ungebührliches Verhalten)I forgot to configure something or is this by design and the documentationis wrong?Best regards - xarax -
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