On Fri, 7 Aug 2020 at 02:57, C Hamilton <[email protected]> wrote: > > How do I access a layer's style information in the processing algorithm. If > it was outside processing it would be layer.renderer(). As far as I can tell > I cannot access it from a QgsProcessingParameterFeatureSource.
You can't, if it's a feature source. A feature source isn't necessarily a vector layer, and accordingly doesn't have a renderer. If you want to write an algorithm which modifies a renderer then the algorithm should defined a QgsProcessingParameterVectorLayer instead. Nyall _______________________________________________ QGIS-Developer mailing list [email protected] List info: https://lists.osgeo.org/mailman/listinfo/qgis-developer Unsubscribe: https://lists.osgeo.org/mailman/listinfo/qgis-developer
