I don't see why it would be impossible… Could you share your query and
give the wikicode you would like to generate for a sample row?
Antonin
On 16/04/2018 14:19, Dennis Tobar wrote:
> 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)
> > Wikidata@lists.wikimedia.org <mailto:Wikidata@lists.> <lists@antonin.delpeuch.eu <mailto:lists@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
> <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
> >
> >
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>
> --
> Dennis Tobar Calderón
> Ingeniero en Informática UTEM
>
>
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