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
Hello. I'm looking for a way to return the previous revision ID
(identical to old_revid in list=recentchanges) when making a watchlist
request:
?action=query&list=watchlist[&wlallrev]
Thanks,
Jim
Hi,
What query should I use to find to which Wiki an image belongs ? Most of the
images in Wikipedia are stored in Wikimedia Commons, but there may be some
stored on Local Wiki.
Right now I'm using this query to fetch image list for a given title:
http://en.wikipedia.org/w/api.php?action=parse&page=Pune&prop=images&format…
Regards,
Siteshwar Vashisht
Hello list,
I am trying to get all images in a page but I want absolute paths of those
images not just image name (aka image titles).
e.g.
To get all images in article "Pakistan International Airlines", I call
http://commons.wikimedia.org/w/api.php?action=query&prop=images&titles=Paki…
but then I am returned a image titles. I have to call this for each image to
get obsoulte URL:
http://commons.wikimedia.org/w/api.php?action=query&titles=File:Pakistan%20…
This means that I have to make many (order of 2) calls to get absolute paths
of all images in an article. Is there any way this all can be squeezed into
just one API call?
Regards
Hasanat Kazmi
+923464362473
Is there a way to hide the "You are looking at the HTML representation
of the ??? format." message that shows at the top of most of the *fm
formats?
A flag that can be used as needed in the url would be great.
Hacking the source to remove it would also be acceptable, if someone
would point me to the correct place.
I understand the reason for it, and the links it gives are useful for
new people, but the thousandth time I've seen it, I just starts to irk
me.
I would accept this showing on the xmlfm format if no format were
passed, and it was assuming xmlfm as the default, but it would be nice
if it didnt show it if I explicitly say that format.
(sorry to the mailing gods for the accidental sending to the
announce-only list. clicked the wrong bookmarked mailman page)
---------- Forwarded message ----------
From: Brion Vibber <brion(a)pobox.com>
Date: Tue, May 3, 2011 at 8:19 PM
Subject: API YAML output change heads-up
To: Roan Kattouw <roan(a)wikimedia.org>
One of the MediaWiki API's output formats for some time has been YAML
<http://www.yaml.org> -- while it doesn't seem to be widely used, it
is there and I assume someone, somewhere is pulling data in this
format. :)
The spyc library we've used to generate YAML output has had a number
of bugs; recently, MediaWiki development trunk has resolved this by
replacing spyc with the existing JSON output module:
https://bugzilla.wikimedia.org/show_bug.cgi?id=28591
Since YAML is a superset of JSON, this should be a transparent change
-- YAML clients should interpret the new output into the same data
structures as before.
Be aware though that there will be differences where previously the
formatting was affected by bugs in the YAML outputter, such as this
example:
http://www.mediawiki.org/wiki/Special:Code/MediaWiki/86302#c16513
If your code relied on particular structures in YAML output, it could
fail with the corrected structures; please double-check that your code
continues to work if switched over to format=json.
-- brion vibber (brion @ pobox.com / brion @ wikimedia.org)
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