#35735: For python 3.9+ class property may not be accessible by Django's
template
system
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Reporter: Fabian Braun | Owner: Fabian Braun
Type: Bug | Status: new
Component: Template system | Version: dev
Severity: Normal | Resolution:
Keywords: | Triage Stage: Unreviewed
Has patch: 1 | Needs documentation: 0
Needs tests: 0 | Patch needs improvement: 0
Easy pickings: 0 | UI/UX: 0
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Comment (by Fabian Braun):
Python (since 3.9) resolves
[https://docs.python.org/3/reference/datamodel.html#object.__class_getitem__
MyClass["something"]] this way (taken from the docs linked - adapted for a
class):
{{{
def subscribe(cls, x):
"""Return the result of the expression 'cls[x] if cls is a class'"""
metaclass = type(cls)
# If the metaclass of cls defines __getitem__, call
metaclass.__getitem__(cls, x)
if hasattr(metaclass, '__getitem__'): # This is also true for python
pre-3.9
return metaclass.__getitem__(cls, x)
# New in Python 3.9:
# Else, if obj is a class and defines __class_getitem__, call
obj.__class_getitem__(x)
elif hasattr(cls, '__class_getitem__'):
# Instead of TypeError the __class_getitem__ class method returns
a GenericAlias object
return cls.__class_getitem__(x)
# Else, raise an exception - this will let Django's template system
try a property next
else:
raise TypeError(
f"'{cls.__name__}' object is not subscriptable"
)
}}}
--
Ticket URL: <https://code.djangoproject.com/ticket/35735#comment:10>
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