#35395: Python 3.12 filter bug with Django 4.2.11
---------------------------------+--------------------------------------
Reporter: Tim Richardson | Owner: nobody
Type: Bug | Status: closed
Component: Template system | Version: 4.2
Severity: Normal | Resolution: needsinfo
Keywords: | Triage Stage: Unreviewed
Has patch: 0 | Needs documentation: 0
Needs tests: 0 | Patch needs improvement: 0
Easy pickings: 0 | UI/UX: 0
---------------------------------+--------------------------------------
Comment (by Tim Richardson):
Fair enough.
The code below is the slice_filter from the 4.2.11 source.
If you run this code in python 3.11,
the filter returns
{}
(an empty dict)
If you run it in python 3.12, it raises an exception.
{{{
Traceback (most recent call last):
File "/app/./test_filter.py", line 21, in <module>
r = slice_filter({},":5")
^^^^^^^^^^^^^^^^^^^^^
File "/app/./test_filter.py", line 13, in slice_filter
return value[slice(*bits)]
~~~~~^^^^^^^^^^^^^^
}}}
KeyError: slice(None, 5, None)
{{{
def slice_filter(value, arg):
"""
Return a slice of the list using the same syntax as Python's list
slicing.
"""
try:
bits = []
for x in str(arg).split(":"):
if not x:
bits.append(None)
else:
bits.append(int(x))
return value[slice(*bits)]
except (ValueError, TypeError) as e:
return value # Fail silently.
if __name__ == "__main__":
r = slice_filter({},":5")
print(f"{r=}")
}}}
I will try to work out why, but this at least describes the problem I
think.
--
Ticket URL: <https://code.djangoproject.com/ticket/35395#comment:3>
Django <https://code.djangoproject.com/>
The Web framework for perfectionists with deadlines.
--
You received this message because you are subscribed to the Google Groups
"Django updates" group.
To unsubscribe from this group and stop receiving emails from it, send an email
to [email protected].
To view this discussion on the web visit
https://groups.google.com/d/msgid/django-updates/0107018f05b269b7-e2c74100-d71d-45db-b937-25ca43f034b9-000000%40eu-central-1.amazonses.com.
