properly get layer from a QgsProcessingFeatureSourceDefinition (qgis#…

…41457) Fixes qgis#40055
nyalldawson · Feb 19, 2021 · 6b6cfa4 · 6b6cfa4
1 parent 20024d8
commit 6b6cfa4
Showing 1 changed file with 9 additions and 1 deletion.
diff --git a/src/gui/processing/qgsprocessingwidgetwrapperimpl.cpp b/src/gui/processing/qgsprocessingwidgetwrapperimpl.cpp
@@ -2005,7 +2005,15 @@ void QgsProcessingExpressionWidgetWrapper::setParentLayerWrapperValue( const Qgs
     context = tmpContext.get();
   }
 
-  QgsVectorLayer *layer = QgsProcessingParameters::parameterAsVectorLayer( parentWrapper->parameterDefinition(), parentWrapper->parameterValue(), *context );
+  QVariant val = parentWrapper->parameterValue();
+  if ( val.canConvert<QgsProcessingFeatureSourceDefinition>() )
+  {
+    // input is a QgsProcessingFeatureSourceDefinition - get extra properties from it
+    QgsProcessingFeatureSourceDefinition fromVar = qvariant_cast<QgsProcessingFeatureSourceDefinition>( val );
+    val = fromVar.source;
+  }
+
+  QgsVectorLayer *layer = QgsProcessingParameters::parameterAsVectorLayer( parentWrapper->parameterDefinition(), val, *context );
   if ( !layer )
   {
     if ( mFieldExpWidget )