The more relevant question, to me, is *why* you would want to do any of this. The whole point of a wiki is to be able to collaboratively create content, where it doesn't really matter who added what. So why you need to query on this data?


On Sun, Apr 21, 2013 at 1:03 PM, Marcelo Chiaradía <chiaradiamarcelo@gmail.com> wrote:
Yeah, I know. I've been checking the  Semantic Extra Special Properties extension, but it only solves half of my problem.

This is my situation:

I need to save somehow the author that made a contribution to a page, and somehow classify it regarding the information modified by him. First I thought an easy solution including a checkbox in the semantic form for the classification problem (like "is minor edit"), but it looks a lot more complex.

So now Im considering the develop of my own extension to semantify this information, following the Semantic Extra Special Properties and s13n approach.

I guess there's no easier way to do it, rigth?

Cheers,
Marcelo.


2013/4/21 Yaron Koren <yaron@wikiworks.com>
Hi,

This seems like a bad hack. I assume this is part of your effort to have the page store the usernames of all its contributors? If so, the Semantic Extra Special Properties extension may be the better approach - I believe it stores that information using the "Page author" property:


-Yaron



On Sat, Apr 20, 2013 at 3:05 PM, Marcelo Chiaradía <chiaradiamarcelo@gmail.com> wrote:
Hi everyone,

I have saved an internal object which has several properties:

{{#set_internal:isInternalObjectOf
|property1=value1
|property2=value2
|name=object_name
}}

I have a situation where eventually I need to define a new internal object using the properties of the object saved before as values.

The way I found to do this is through semantic queries:

{{#set_internal:isInternalObjectOf
|property1={{#ask: [[-property1::[[name::object_name]] ]]}}
|property2={{#ask: [[-property2::[[name::object_name]] ]]}}
}}

The sintax may be not quite correct, just to show the idea.

The problem with this approach is that for every property, I have to make a new query. Is there some way to do this with only one query?

Thanks in advance.

Marcelo.

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