Getting file path from vector layer object in QGIS processing framework

Question

Using the QGIS 3.x processing framework, given a vector layer as an input parameter:
(from the default script template)
def initAlgorithm(self, config=None):
    self.addParameter(
        QgsProcessingParameterFeatureSource(
            self.INPUT,
            self.tr('Input layer'),
            [QgsProcessing.TypeVectorAnyGeometry]
        )
    )

How can I get this layer's source file path as a string (assuming that only file-based data sources will be used, not PostGIS tables, etc)?

dericke · Accepted Answer

I was able to get the path using the following:

self.parameterDefinition('INPUT').valueAsPythonString(parameters['INPUT'], context)

user3226314 · Answer

This answer might be slightly off-topic, but your question keeps coming up whilst searching for the graphical modeller answers.
If you are using the graphical modeller, you can use the parameter function to retrieve a string of any inputs. Unfortunately, you can not use the shorthand @parameter way of accessing data if you want a string.
The below code takes a folder input (outputFolder) and a vector layer input (dataFile) and produces a new string. i.e.:
IF outputFolder == c:temp  AND dataFile == c:someFolderanotherFolderSomeData.csv THEN
the output will be c:tempsomeData.gpkg
concat(
    @outputFolder,
    '',
    replace(file_name(to_string(parameter('dataFile'))),'.csv','.gpkg')
)

Getting file path from vector layer object in QGIS processing framework

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