On 25 April 2010 11:36, Claudius Henrichs <claudiu...@gmx.de> wrote:
> Am 25.04.2010 11:25, Alexander Sidorov:
>
> Hello!
>
> Is there any way to implement the following localization strategy:
> 1. display russian (russian is just for example) name if it exists
> 2. display english name if russian name doesn't exists
> 3. display local name otherwise
>
> The problem for me is to determine which language is used for some region
> by default. I have found this
> discussion<http://wiki.openstreetmap.org/wiki/Talk:Proposed_features/Language_of_this_element>but
> without solution.
>
> There's currently no way to determine what language the "name"-tag is in.
>
> Your question doesn't seem to be a tagging one though, because as I
> understand you want to have different rendered outputs, correct?
> Like John Smith explained this has to be done with some clever JOIN SQL
> statements.
>
> Some further readings here [1] and an example of a rendering of the world
> map in different languages here [2].
>
> Claudius
>
Part of the problem will be solved with the hstore patch I suspect. But
else, John is pretty much right, the SQL can be interesting to write. I
think a basic SQL statement would start like this. This is not real SQL code
since I don't have the name of the table in mind. Also I am keeping in mind
more of a simple osmosis schema for a just a russian extraction of name.
This is not meant to be used to extract all countries in the world. If I was
do that I would then use a slightly different approach:
SELECT ( CASE
WHEN nt1.v IS NOT NULL THEN nt1.v
WHEN nt2.v IS NOT NULL THEN nt2.v
ELSE nt.v
END
) AS russianName
FROM node_tags AS nt
LEFT OUTER JOIN
node_tags AS nt1
ON nt1.id = nt.id AND k = 'name:ru'
LEFT OUTER JOIN
node_tags AS nt2
ON nt2.id = nt.id AND k = 'name:en'
WHERE nt.k = 'name'
Here we go, we have a simple query to do as you asked. However, do keep mind
that if the place is in Russia I wouldn't expect name:ru to actually exists
since name should be containing the local name anyway. Of course, they are
probably better way to do that. Among others things, creating a partial
index on k = 'name:ru' and k = 'name:en' would speed up the query quite
dramatically.
The other approach would be to retrieve all name elements with a condition
nt.k LIKE 'name%' and then perform a subquery on this afterwards. Similarly,
creating a view with that information would prove quite useful. It is just
down to imagination.
Emilie Laffray
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