When evaluating an expression with a missing referenced field, this w…

…ill be replaced with [fieldname]. When calculating data defined values, this will be counterproductive. A field with a data defined value might work well in the expression builder (because fiels are all present) but fail in the legend, because it's rendered with no feature and hence fields. With this change, we can at least use `try("fieldname", 0)` to fallback to a default value.
qgis · Dec 7, 2019 · e1b016b · e1b016b
1 parent 3bf8d96
commit e1b016b
Showing 1 changed file with 2 additions and 1 deletion.
diff --git a/src/core/expression/qgsexpressionnodeimpl.cpp b/src/core/expression/qgsexpressionnodeimpl.cpp
@@ -1270,7 +1270,8 @@ QVariant QgsExpressionNodeColumnRef::evalNode( QgsExpression *parent, const QgsE
         return feature.attribute( mName );
     }
   }
-  return QVariant( '[' + mName + ']' );
+  parent->setEvalErrorString( QStringLiteral( "Column '%1' not found" ).arg( mName ) );
+  return QVariant();
 }
 
 QgsExpressionNode::NodeType QgsExpressionNodeColumnRef::nodeType() const