#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 django-updates+unsubscr...@googlegroups.com. 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.