From 6b6cfa4dce4d494fd43ffdf5178f464f221bed0e Mon Sep 17 00:00:00 2001
From: Julien Cabieces <julien.cabieces@oslandia.com>
Date: Thu, 11 Feb 2021 02:26:43 +0100
Subject: [PATCH] properly get layer from a
 QgsProcessingFeatureSourceDefinition (#41457)

Fixes #40055
---
 src/gui/processing/qgsprocessingwidgetwrapperimpl.cpp | 10 +++++++++-
 1 file changed, 9 insertions(+), 1 deletion(-)

diff --git a/src/gui/processing/qgsprocessingwidgetwrapperimpl.cpp b/src/gui/processing/qgsprocessingwidgetwrapperimpl.cpp
index 9336575030c2..ceab9a133414 100644
--- 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 )