[Qgis-user] Get QgsProcessingParameterFeatureSource's parent layer

[José Roberto Ribeiro Filho]

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 <qgis-user at lists.osgeo.org> 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
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