maybe after get it source (QgsProcessingAlgorithm.asSource => https://qgis.org/api/classQgsProcessingFeatureSource.html and from there with layer name you can find registered layer that not necessarily is a rendered layer.
Luigi Pirelli ************************************************************************************************** * LinkedIn: https://www.linkedin.com/in/luigipirelli * Stackexchange: http://gis.stackexchange.com/users/19667/luigi-pirelli * GitHub: https://github.com/luipir * Book: Mastering QGIS3 - 3rd Edition <https://www.packtpub.com/eu/application-development/mastering-geospatial-development-qgis-3x-third-edition> * Hire a team: http://www.qcooperative.net ************************************************************************************************** On Thu, 6 Aug 2020 at 18:56, 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. > > Thanks, > > Calvin > _______________________________________________ > 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
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