Hi André, Is it possible to refer to a list of layers in the project, or does
your code run without the QGIS project being opened?Because, theoretically, you
need to point the layer or feature class you will use to extract the selected
items, and in this case, you can save the name in a separate variable to use
lately to create a name for the output.
I hope I helped.
José Roberto Ribeiro Filho - Geógrafo
Em terça-feira, 3 de dezembro de 2024 às 13:24:42 BRT, Andre Kotze via
QGIS-User <[email protected]> escreveu:
Dear all,
I have a processing algorithm that names its output based on the input layer's
name, e.g. if buffering a layer called 'powerlines_110kW' with 40m the output
layer would be called 'powerlines_110kW+40m'. For the input I use
QgsProcessingParameterFeatureSource (instead of VectorLayer), so that the
"Selected features only" option is available.
My problem is, when "Selected features" is unchecked, I can get the input
layer's name with the 'parameterAsVectorLayer(parameters, 'INPUT',
context).name()' method. However, when "Selected features" is checked,
'parameterAsVectorLayer()' returns None. This makes sense, as I believe the
input becomes wrapped in a QgsProcessingFeatureSourceDefinition. Is there any
way to get the layer's name before the selected features are extracted/filtered
from the selected vector layer (the one given by the user in the ComboBox)?
Many thanks
André
_______________________________________________
QGIS-User mailing list
[email protected]
List info: https://lists.osgeo.org/mailman/listinfo/qgis-user
Unsubscribe: https://lists.osgeo.org/mailman/listinfo/qgis-user
_______________________________________________
QGIS-User mailing list
[email protected]
List info: https://lists.osgeo.org/mailman/listinfo/qgis-user
Unsubscribe: https://lists.osgeo.org/mailman/listinfo/qgis-user
