Change Output name in Script - QGIS 3

Question

I exported a model as script using the properly option in QGIS 3.8. The Script works perfectly, but the output name is changed.

Heres a print of the model:

enter image description here

I want that the script ́s output name, from the Processing Toolbox, be the one that I designed in the def initAlgorithm section, and not the given default name.

In the model, the output name comes as "output_name" and in the script, the output name comes as "Reprojected".

Here's the converted Script:

 from qgis.core import QgsProcessing
 from qgis.core import QgsProcessingAlgorithm
 from qgis.core import QgsProcessingMultiStepFeedback
 from qgis.core import QgsProcessingParameterVectorLayer
 from qgis.core import QgsProcessingParameterFeatureSink
 from qgis.core import QgsCoordinateReferenceSystem
 import processing
 class Rename(QgsProcessingAlgorithm):
 def initAlgorithm(self, config=None):
 self.addParameter(QgsProcessingParameterVectorLayer('shapeinput', 'ShapeInput', types=[QgsProcessing.TypeVector], defaultValue=None))
 self.addParameter(QgsProcessingParameterFeatureSink('Output_name', 'output_name', type=QgsProcessing.TypeVectorAnyGeometry, createByDefault=True, defaultValue=None))
 def processAlgorithm(self, parameters, context, model_feedback):
 # Use a multi-step feedback, so that individual child algorithm progress reports are adjusted for the
 # overall progress through the model
 feedback = QgsProcessingMultiStepFeedback(1, model_feedback)
 results = {}
 outputs = {}
 # Reproject Layer
 alg_params = {
 'INPUT': parameters['shapeinput'],
 'TARGET_CRS': QgsCoordinateReferenceSystem('EPSG:4326'),
 'OUTPUT': parameters['Output_name']
 }
 outputs['ReprojectLayer'] = processing.run('native:reprojectlayer', alg_params, context=context, feedback=feedback, is_child_algorithm=True)
 results['Output_name'] = outputs['ReprojectLayer']['OUTPUT']
 return results
 def name(self):
 return 'rename'
 def displayName(self):
 return 'rename'
 def group(self):
 return 'rename'
 def groupId(self):
 return ''
 def createInstance(self):
 return Rename()

• Joseph: This was also asked before: How to rename the result of a QGIS Processing Algorithm
• Rob: I have got exactly the same problem, any answers yet?
• Herbert Santos: Unfortunately not, I recently realized that the shape created has the properties @layer_name. This is the element that needs to be changed, but I haven't found how to change it by python.

Answer 1

An alternative solution that worked for me, as parameters['Output_name'] is type QgsProcessingOutputLayerDefinition with the attribute destinationName, add this line before calling the processing algorithm:

parameters['Output_name'].destinationName = 'your_preferred_name'

• Herbert Santos: Hello @Andrson, I came across this solution a few years ago, but I forgot to update it here. It's definitely the best way

Answer 2

If you need change the name of output, you can do it with this:

alg_params = {
 'INPUT': parameters['shapeinput'],
 'TARGET_CRS': QgsCoordinateReferenceSystem('EPSG:4326'),
 'OUTPUT': 'memory:Name_layer'
 }

or if you want to save, only do this:

path = '/home/shade/Desktop/name_layer.shp' #or something like that
alg_params = {
 'INPUT': parameters['shapeinput'],
 'TARGET_CRS': QgsCoordinateReferenceSystem('EPSG:4326'),
 'OUTPUT': path
 }

For example:

I run the next code:

layer = iface.activeLayer()
alg_params = {
 'INPUT': layer,
 'TARGET_CRS': QgsCoordinateReferenceSystem('EPSG:4326'),
 'OUTPUT': 'memory:Name_layer'
}
result = processing.run('native:reprojectlayer', alg_params)
print ('This is the result: {}'.format(result['OUTPUT']))
print ('This is the name of the layer result: {}'.format(result['OUTPUT'].name()))

And obtained this:

enter image description here

• Herbert Santos: The first option generated no result. But processed OK
• Jhon Galindo: In this case results['Output_name'] saved the qgsvectorlayer, if you need the name, use results['Output_name'].name()
• Joseph: @JhonGalindo - I think the OP means that if you run the script from the Processing Toolbox, the output name is given the default name. Your code works from the Python Console, now test if it works by converting it into a script which you run from the toolbox.
• Herbert Santos: Exactly! That is what I mean. Results from the toolbox, not from the console.
• Jhon Galindo: thanks @Joseph , I will do it and I will try to show the answer here.

Answer 3

I got it!

the last step of the script must be executed inside a variable, here called step1
so it respects the name entered in the output parameter through the last block

from qgis.core import QgsProcessing
from qgis.core import QgsProcessingAlgorithm
from qgis.core import QgsProcessingMultiStepFeedback
from qgis.core import QgsProcessingParameterVectorLayer
from qgis.core import QgsProcessingParameterFeatureSink
from qgis.core import QgsFeatureSink
import processing
class Model(QgsProcessingAlgorithm):
 def initAlgorithm(self, config=None):
 self.addParameter(QgsProcessingParameterVectorLayer('vetor', 'vetor', defaultValue=None))
 self.addParameter(QgsProcessingParameterFeatureSink('exit', 'exit name', type=QgsProcessing.TypeVectorPolygon, createByDefault=True, defaultValue=None))
 def processAlgorithm(self, parameters, context, model_feedback):
 feedback = QgsProcessingMultiStepFeedback(1, model_feedback)
 results = {}
 outputs = {}
 step1 = processing.run("native:buffer", {
 'DISSOLVE': False,
 'DISTANCE': 10,
 'END_CAP_STYLE': 0,
 'INPUT': parameters['vetor'],
 'JOIN_STYLE': 0,
 'MITER_LIMIT': 2,
 'SEGMENTS': 5,
 'OUTPUT': 'memory:'
 }, context=context, feedback=feedback)['OUTPUT']
 """here the output name is changed"""
 source = step1
 (sink, dest_id) = self.parameterAsSink(parameters,'exit',context,source.fields(),source.wkbType(),source.sourceCrs())
 features = source.getFeatures()
 for current, feature in enumerate(features):
 sink.addFeature(feature, QgsFeatureSink.FastInsert)
 return results
 def name(self):
 return 'rename'
 def displayName(self):
 return 'rename'
 def group(self):
 return ''
 def groupId(self):
 return ''
 def createInstance(self):
 return Model()

• Herbert Santos: @Rob here´s the solution