properly get layer from a QgsProcessingFeatureSourceDefinition (#41457)

Fixes #40055
qgis · Feb 19, 2021 · fb032754dd2fa6cb75dd3af72c4be219850c7b43 · fb03275
1 parent 8bab4c5
commit fb032754dd2fa6cb75dd3af72c4be219850c7b43
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
@@ -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 )