3

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()
3
  • This was also asked before: How to rename the result of a QGIS Processing Algorithm
    – Joseph
    Sep 5, 2019 at 15:17
  • I have got exactly the same problem, any answers yet?
    – Rob
    Apr 14, 2020 at 13:19
  • 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. Apr 30, 2020 at 13:38

2 Answers 2

1

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

9
  • The first option generated no result. But processed OK Sep 5, 2019 at 14:39
  • In this case results['Output_name'] saved the qgsvectorlayer, if you need the name, use results['Output_name'].name() Sep 5, 2019 at 14:46
  • 1
    @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.
    – Joseph
    Sep 6, 2019 at 9:54
  • 1
    Exactly! That is what I mean. Results from the toolbox, not from the console. Sep 6, 2019 at 11:58
  • 1
    thanks @Joseph , I will do it and I will try to show the answer here. Sep 6, 2019 at 15:52
1

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()
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