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Feature source parameters cannot be used for vector layer algorithm
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parameters in models

There's not a 1:1 correlation here - vector layers can be used
as inputs for sources, but sources cannot be used when a
full vector layer is required.
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nyalldawson committed Jun 27, 2017
1 parent e171fe3 commit 4f096a6
Showing 1 changed file with 1 addition and 5 deletions.
6 changes: 1 addition & 5 deletions python/plugins/processing/gui/
Expand Up @@ -1077,16 +1077,12 @@ def createWidget(self):
return BatchInputSelectionPanel(self.param, self.row, self.col, self.dialog)
self.combo = QComboBox()
layers = self.dialog.getAvailableValuesOfType(QgsProcessingParameterRasterLayer, QgsProcessingOutputRasterLayer)
tables = self.dialog.getAvailableValuesOfType(QgsProcessingParameterVectorLayer, OutputTable)
layers = self.dialog.getAvailableValuesOfType(QgsProcessingParameterFeatureSource, QgsProcessingOutputVectorLayer)
tables = self.dialog.getAvailableValuesOfType(QgsProcessingParameterVectorLayer, QgsProcessingOutputVectorLayer)
if self.param.flags() & QgsProcessingParameterDefinition.FlagOptional:
self.combo.addItem(self.NOT_SELECTED, None)
for table in tables:
self.combo.addItem(self.dialog.resolveValueDescription(table), table)
for layer in layers:
self.combo.addItem(self.dialog.resolveValueDescription(layer), layer)

widget = QWidget()
layout = QHBoxLayout()
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