Accessing fields of created features in your PyQGIS

Question

I'm new using PyQGIS 3 and I'm wondering how to access the field information from a layer generated by script.

Inside my processAlgorithm function I have the following:

alg_params = {
    'EXPRESSION': parameters[self.IN_EXPRESSION],
    'INPUT': parameters[self.IN_SHP],
    'OUTPUT': parameters[self.OUT_SHP]
}

final_output = processing.run('native:extractbyexpression', alg_params, context=context, feedback=feedback, is_child_algorithm=True)['OUTPUT']

output_lyr = QgsVectorLayer(final_output, "output_testing", "ogr")    

fields = output_lyr.fields()
fieldnames = [field.name() for field in fields]
print(fieldnames)

return {"result": final_output }

The algorithm itself is returning the right output, however I'm not able to access the fields during inside the code, the print is returning me a empty list, however my final_output DO have fields.

Someone, could give me a glimpse of what I've missed?

Answer 1

Normally, when using a vector tool, processing.run with TEMPORARY_OUTPUT returns ["OUTPUT": QgsVectorLayer]. If is_child_algorithm=True is added, it returns a temporary layer id.

The reference of a temporary layer can be obtained by means of context.temporaryLayerStore(). You can add the following lines to your processing script to get the temporary layer's reference.

import os 
# ... other imports

class ExampleProcessingAlgorithm(QgsProcessingAlgorithm):
    # ... other stuffs

    def processAlgorithm(self, parameters, context, feedback):    
        # ... previous lines

        final_output = processing.run('native:extractbyexpression',
                                      alg_params, context=context, feedback=feedback,
                                      is_child_algorithm=True)['OUTPUT']

        # check output type
        if isinstance(final_output, str):            
            # check if a file path
            if os.path.isfile(final_output):
                output_lyr = QgsVectorLayer(final_output, "output_testing", "ogr")                            
            else: # if not a path, then it is a temporary layer's ID
                output_lyr = context.temporaryLayerStore().mapLayers()[final_output]        
        # possible QgsVectorLayer if "is_child_algorithm=True" is not used
        elif isinstance(final_output, QgsVectorLayer):
            output_lyr = final_output        
        else:
            raise ValueError('A very specific bad thing happened.')
        
        print([f.name() for f in output_lyr.fields()])
    
        return {"OUTPUT": final_output}

Answer 2

Use final_output['OUTPUT'] since the output is saved in a dictionary:

output_lyr = QgsVectorLayer(final_output['OUTPUT'], "output_testing", "ogr")