Antonin:
Thanks for share your query. I want to create a row with name, year of
creation, area protected (in hectare), a photo, the Commons category (if
exists) and the legal decree that creates the protected area, so I see that
seems impossible to do it with a few concats :)
I already have the query and the data from Wikidata, so this isn't the key
problem.
On Mon, Apr 16, 2018 at 8:40 AM, Antonin Delpeuch (lists) <
lists(a)antonin.delpeuch.eu> wrote:
Hi,
It is possible to use templates in Listeria table cells, by tweaking
your SPARQL query so that it returns the appropriate wikicode:
https://www.wikidata.org/wiki/User:Pintoch/orgid
If you have templates which depend on multiple variables in your SPARQL
query, I suppose you could take advantage of the fact that the pipe
characters in wiki tables can also be used to separate template
arguments: you could generate parts of the template in various
consecutive SPARQL variables and they would all be rendered as one
template in a cell (but the headers would go out of sync).
But that's a ugly hack - I would be interested in a cleaner way too.
Antonin
On 16/04/2018 13:30, Dennis Tobar wrote:
Hi:
I'm working on protected areas in Chile and I want to get the list using
Listeriabot but in a "special format" using templates. I tried to find
some example without luck.
Does anyone know a working example of Listeriabot + template to render
the rows?.
Kindly,
--
Dennis Tobar Calderón
Ingeniero en Informática UTEM
_______________________________________________
Wikidata mailing list
Wikidata(a)lists.wikimedia.org
https://lists.wikimedia.org/mailman/listinfo/wikidata
_______________________________________________
Wikidata mailing list
Wikidata(a)lists.wikimedia.org
https://lists.wikimedia.org/mailman/listinfo/wikidata
--
Dennis Tobar Calderón
Ingeniero en Informática UTEM