Dear Developers,
to implement a QgsExpressionFunction I like to access the "layer" from a
QgsExpressionContext "context".
Using the C++ API this can be done with the QgsExpressionUtils:
QgsVectorLayer *vl = QgsExpressionUtils::getVectorLayer(
context->variable( QStringLiteral( "layer" ) ), parent );
(several example exist qgsexpressionfunction.cpp)
In Python the QgsExpressionUtils are not available and accessing the
"layer" attribute raises a
TypeError "unable to convert a C++ 'QPointer<QgsMapLayer>' instance to a
Python object"
Example:
from qgis.core import QgsExpressionContext, QgsVectorLayer,
QgsExpressionContextUtils
lyr = QgsVectorLayer('point?crs=epsg:4326&field=id:integer', 'dummy',
'memory')
context = QgsExpressionContext()
context.appendScope(QgsExpressionContextUtils.layerScope(lyr))
print(f'name={context.variable("layer_name")}')
# this raises: TypeError: unable to convert a C++
'QPointer<QgsMapLayer>' instance to a Python object
lyr2 = context.variable('layer')
Is there another way to get the QgsMapLayer instance in Python?
If not, I could open an issue to bring the QgsExpressionUtils into the
QGIS Python API as well.
--
--
Benjamin Jakimow, Doctoral Researcher
Earth Observation Lab | Geography Department | Humboldt-Universität zu
Berlin
e-mail: benjamin.jaki...@geo.hu-berlin.de
phone: +49 (0) 30 2093 6894
mobile: +49 (0) 157 5656 8477
fax: +49 (0) 30 2093 6848
mail: Unter den Linden 6 | 10099 Berlin | Germany
room: 2'222
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