Fix data defined expressions cannot use fields in vector tile renderer

qgis · Sep 14, 2020 · c192b04 · c192b04
1 parent 229c240
commit c192b04
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Showing 2 changed files with 14 additions and 4 deletions.
diff --git a/src/core/symbology/qgssymbollayer.cpp b/src/core/symbology/qgssymbollayer.cpp
@@ -246,7 +246,10 @@ int QgsSymbolLayer::renderingPass() const
 
 QSet<QString> QgsSymbolLayer::usedAttributes( const QgsRenderContext &context ) const
 {
-  QSet<QString> columns = mDataDefinedProperties.referencedFields( context.expressionContext() );
+  // calling referencedFields() with ignoreContext=true because in our expression context
+  // we do not have valid QgsFields yet - because of that the field names from expressions
+  // wouldn't get reported
+  QSet<QString> columns = mDataDefinedProperties.referencedFields( context.expressionContext(), true );
   return columns;
 }
 

diff --git a/src/core/vectortile/qgsvectortilebasicrenderer.cpp b/src/core/vectortile/qgsvectortilebasicrenderer.cpp
@@ -122,10 +122,17 @@ void QgsVectorTileBasicRenderer::startRender( QgsRenderContext &context, int til
   // figure out required fields for different layers
   for ( const QgsVectorTileBasicRendererStyle &layerStyle : qgis::as_const( mStyles ) )
   {
-    if ( layerStyle.isActive( tileZoom ) && !layerStyle.filterExpression().isEmpty() )
+    if ( layerStyle.isActive( tileZoom ) )
     {
-      QgsExpression expr( layerStyle.filterExpression() );
-      mRequiredFields[layerStyle.layerName()].unite( expr.referencedColumns() );
+      if ( !layerStyle.filterExpression().isEmpty() )
+      {
+        QgsExpression expr( layerStyle.filterExpression() );
+        mRequiredFields[layerStyle.layerName()].unite( expr.referencedColumns() );
+      }
+      if ( layerStyle.symbol() )
+      {
+        mRequiredFields[layerStyle.layerName()].unite( layerStyle.symbol()->usedAttributes( context ) );
+      }
     }
   }
 }