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<p>Hi Richard,</p>
<p>your last question first: Nope, there is definitely no WFS (would
be too easy). I am living in Bavaria, the home of "Laptop and
Lederhosen". Our administrations are among the greediest and most
retarded (offering hardly any WFS which then mostly have to paid
as well) in Germany.</p>
<p>How would I know this is a Geoserver instance? The url of the WMS
is <code><a class="moz-txt-link-freetext" href="https://www.lfu.bayern.de/gdi/wms/natur/oefk">https://www.lfu.bayern.de/gdi/wms/natur/oefk</a>?</code></p>
<p>My "python" is actually nonexistant, so I was hoping that there
was something "prebuilt" in someone's drawer already. But as my
web research showed, there does seem to be much on this topic
besides of "not possible".<br>
</p>
<p>It's a bit mystical to me why QGIS can show the result of a click
on a layer, but not store the data by any means ...</p>
<p>Mhm ...</p>
<p>Bernd<br>
</p>
<div class="moz-cite-prefix">On 19.10.20 14:07, Richard Duivenvoorde
wrote:<br>
</div>
<blockquote type="cite"
cite="mid:3ed9e71d-0b47-07ec-1d4f-5c5f722da391@duif.net">
<pre class="moz-quote-pre" wrap="">Hi Jorge, Bernd,
I thought to play with that, but ...
although I get a valid result, I cannot get to the actual features in python:
TypeError: unable to convert a C++ 'QgsFeatureStoreList' instance to a Python object
So I'm not sure if you could (currently) easily come to the actual feature...
What I did (working with a national buildings WMS, and a (to me) known valid x,y,width/height etc:
l=qgis.utils.iface.addRasterLayer(
"crs=EPSG:28992&layers=Bebouwingvlak&styles=&format=image/png&url=<a class="moz-txt-link-freetext" href="https://geodata.nationaalgeoregister.nl/kadastralekaart/wms/v4_0">https://geodata.nationaalgeoregister.nl/kadastralekaart/wms/v4_0</a>?", # uri
"buildings", # name for layer (as seen in QGIS)
"wms" # dataprovider key
)
result = l.dataProvider().identify( QgsPointXY(104606,490213),
QgsRaster.IdentifyFormatFeature,
QgsRectangle(104391.1406077263819,490161.0749502293766,104899.6563292678911,490283.3410253541078),
1465,
352,
96)
print(result.isValid()) # prints True
print(result.results())
# prints TypeError: unable to convert a C++ 'QgsFeatureStoreList' instance to a Python object
Regards,
Richard Duivenvoorde
PS @Bernd: you are sure there is no WFS running there (if it is a Geoserver instance, it often has (silently) also a WFS....)?
On 10/19/20 1:36 PM, Jorge Gustavo Rocha wrote:
</pre>
<blockquote type="cite">
<pre class="moz-quote-pre" wrap="">Hi Bernd,
Try rlayer.dataProvider().identify()
Checks the docs at <a class="moz-txt-link-freetext" href="https://docs.qgis.org/3.10/en/docs/pyqgis_developer_cookbook/raster.html#query-values">https://docs.qgis.org/3.10/en/docs/pyqgis_developer_cookbook/raster.html#query-values</a>
Good luck,
Jorge
Às 12:25 de 19/10/20, Bernd Vogelgesang escreveu:
</pre>
<blockquote type="cite">
<pre class="moz-quote-pre" wrap="">Hi there,
unfortunately I am provided with data only in form of a WMS layer
representing areas as polygons.
These areas match with the cadastral boundaries.
I am looking for a way how to hand over the data from the WMS to my
cadastral units to be able to really work with these informations.
I created a "Point on surface" layer of the cadastre.
Is there any pythonic or other way to "mimic" the "Identify features"
click on that layer with the coordinates of those points and store the
results?
I assume, that each click is a server request, so running these (in my
case) 8000 "cĺicks" should not be sent in a millisecond, but with some
pause.
It doesn't really matter how long this takes, cause it only has to be
done once and would be in any case faster than manually clicking and
noting the data.
Obviously, this is not what the provider of the data intends, but it's
like an act of self-defence against a stubborn and bureaucratic
administration.
So, data rebels, come to help please ;)
Cheers,
Bernd
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