properly get layer from a QgsProcessingFeatureSourceDefinition (#41457)

Fixes #40055
qgis · Feb 11, 2021 · e9de56a83bc440d3207489bc4ddeb53a751eaa45 · e9de56a
1 parent 8eb93c5
commit e9de56a83bc440d3207489bc4ddeb53a751eaa45
Unified Split

Showing with 9 additions and 1 deletion.

+9 −1 src/gui/processing/qgsprocessingwidgetwrapperimpl.cpp
diff --git a/src/gui/processing/qgsprocessingwidgetwrapperimpl.cpp b/src/gui/processing/qgsprocessingwidgetwrapperimpl.cpp
@@ -2008,7 +2008,15 @@ 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 )