Disputatio Vicipaediae:Imago

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E Vicipaedia

Two questions[fontem recensere]

Two questions (getting to 96.5%?):

1- Do we have a {{spoiler}} equivalent to add to our Latin pages? If so, what is it?

2- when I create a new Vicipaedia page, is there (I hope there is) a way I can add the link to my new Latin page link to all the equivalent pages that already exist in other Wiki-languages or do I have to add it to the other pages one by one? This is the way I'm currently doing it but I strongly suspect it can be done some other handier way if you read all the instructions before you have a go.

Animus gratissimus,

D Ambulans 17:43, 30 Ianuarii 2006 (UTC)[reply]

ad 1) If someone writes {{spoiler}} (see the use of "nowiki" in the source code of this talk page, which is just to make the construct inactive on this talk page) another page is included (we are using "{" instead of "["). Depending on the specific Wikipedia (en, de, la, ...) there is a sepcial namespace for such templates. In the English WP it is called ... yes! ... "Template". In the Latin WP it is "Formula". Like User/Usor. So {{spoiler}} looks for page Template:spoiler in the English WP and for Formula:spoiler in the Latin WP. It will not find a page Formula:spoiler. You can simply create the template in the Latin WP like every other page. However, it might be a better idea to find a Latin name for the template. Given the Latin name should be "xxx", you would copy "Template:spoiler" from the English WP to "Formula:xxx" in the Latin WP. Then you have to make some corrections in the content, because the template uses other things which are maybe just available on the English WP. Of course you can copy these other templates as well. Example: "Formula:Usor de" is a copy from "Template:User de". Then the names of the categories within the template text had to be corrected: You have to change - in this case - at least 4 times the word User to Usor. If you want a Latin text, you have to translate the content as well. Then you can use your template by writing {{xxx}} ... but then just in the Latin WP.

ad 2) Two answers: a) You have to do it by hand and b) if there is at least one interwiki-link, there are robots, which will add the other links (which can take weeks). See Specialis:Recentchanges and look e. g. for the line "Patavium; 15:42 . . FlaBot (Disputatio) (robot Adding: sv)". Then there is a trick for a) If you want to get most interwiki links, simply copy it from the source of any of the foreign pages. It is a good idea to choose a bigger language which has mostly more interwiki links. If you choose French, for example, you have to add just [[fr:xxxxxx]] by hand and you are done (or you leave French and wait for the robot). And maybe you have to remove the Latin link [[la:xxxxxx]] otherwise it will be shown on the Latin page.

--Roland2 18:39, 30 Ianuarii 2006 (UTC)[reply]

Descriptiones imaginum[fontem recensere]

Vide Vicipaedia:Taberna/Tabularium 3#Diagnoses (vel descriptiones) imaginum et Usor:IacobusAmor/Disputata anni 2006#Quid est imago?. --UV 14:59, 26 Octobris 2006 (UTC)[reply]

What about a template and/or category like "Imago sine descriptione"? This would help in case I can add an image but do not know a good (Latin) description for it. --Roland (disp.) 12:36, 27 Octobris 2006 (UTC)[reply]
Nunc est: {{Imago sine descriptione}} et Categoria:Imago sine descriptione. Vide etiam Disputatio Usoris:Fabullus#Asterix et Vicipaedia:Taberna#De imaginum explanationibus. --UV 17:24, 23 Maii 2009 (UTC)[reply]

Pinacotheca imaginum vs Porticus imaginum[fontem recensere]

What term is better? We have both ... --Rolandus 19:41, 3 Decembris 2006 (UTC)[reply]

I like pinacotheca for four reasons:
1. I have seen it in classical texts to refer to gallery, I have not seen porticus.
2. The string "~E gallery" at William Whitaker's Words yields (among other things):
porticus, porticus N C 4 1 C [XXXBO]
colonnade, covered walk; portico; covered gallery atop ampitheater/siege works;
pinacotheca/e, pinacothecae/es N F 1 1 F [XDXDO]
picture gallery;
3. Traupman says for gallery: "porticus, us, f; (open) perystylium, i, n; (for paintings) pinocatheca, ae, f.
4. It's greek and as UV and I discussed, we hardly use enough greek words around here.
--Ioshus (disp) 19:57, 3 Decembris 2006 (UTC)[reply]
Furthermore, I think pinocatheca imaginum is redundant. Wholly unnecessary genitive.--Ioshus (disp) 19:58, 3 Decembris 2006 (UTC)[reply]
Agree that imaginum is really redundant. Probably pinacotheca is better than porticus, for the reasons you have stated. --UV 20:37, 3 Decembris 2006 (UTC)[reply]

Wikimedia Foundation board passes resolution on licensing policy[fontem recensere]

Quoting from wikimedia:#Latest news:

Licensing policy is introduced

The Wikimedia Foundation board has now passed its resolution on the media licensing policy. This policy is intended to reflect the principles in the message posted earlier at mailing list and to make the guidelines for acceptable media licenses clearer across all projects.(27 March 2007)

--UV 23:11, 28 Martii 2007 (UTC)[reply]

De magnitudine imaginum[fontem recensere]

Vide Disputatio Usoris:UV#Cicero. --UV 14:15, 1 Iunii 2009 (UTC)[reply]